Mix Test 12

Given that a is a positive integer such that the least prime factor of a is 3.

Then a must be an odd integer.

(Because if a is an even integer than least prime factor of a is 2.)

Similarly, b is a positive integer such that the least prime factor is 5.

Then b must be also an odd integer.

Since, odd + odd = even.

\(\therefore\) a + b is an even integer which has the least prime factor 2.

Hence, the least prime factor of (a + b) is 2.

Given that \(\alpha , \beta\) are the zeros of the polynomial f(x) \(=\mathrm{x^2 - 5x + k}\) and \(\alpha - \beta =1\)

\(\therefore \) sum of zeros and product of zeros are \(\frac{-(-5)}{1}\) and \(\frac{k}{1}\)

\(\therefore \) \(\alpha + \beta = 5\) and \(\alpha \beta = k\)

We know that \((\alpha-\beta)^2\) \(=(\alpha + \beta)^2-4\alpha \beta\) = 25 - 4k  (\(\because\) \(\alpha + \beta = 5\) & \(\alpha + \beta = k\))

\(\Rightarrow\) 25 – 4k = \((\alpha-\beta)^2\) \(=1^2=1\)  \((\because \alpha - \beta =1)\)

\(\Rightarrow\) 4k = 25 – 1 = 24 

\(\Rightarrow\) \(k = \frac{24}{4}=6\)

Hence, the value of k is 6.

Given that length and breadth of rectangle is \(l\) m and b m respectively.

Therefore, the area of rectangle is \(l\times b \,m^2\)

Length and breadth of original rectangle is \(l\) m and b m respectively.

Given that length of rectangle is increased by 20 %.

\(\therefore\) Length of modified rectangle is \(l + \frac{20}{100}l\) \(=l + \frac{1}{5}l = \frac{6}{5}l \,m\)

Hence, the new length is \(\frac{6l}{5}m\).

Length and breadth of original rectangle is \(l \) m and b m respectively.

Given that breadth of rectangle is decreased by 20%.

\(\therefore\) Breadth of modified rectangle \(=b-\frac{20}{100}b = b - \frac{1}{5}b=\frac{4b}{5}m\)

Hence, the new breadth is \(\frac{4b}{5}m\).

Length and breadth of original rectangle is \(l\) m and b m respectively.

Given that length of rectangle is increased by 20 %.

\(\therefore\) Length of modified rectangle is \(l + \frac{20}{100}l\) \(=l + \frac{1}{5}l\) \(=\frac{6}{5}l\,m\)

Hence, the new length is \(\frac{6l}{5}m\)

Given that breadth of rectangle is decreased by 20%.

\(\therefore\) Breadth of modified rectangle = \(b-\frac{20}{100}b\) \(=b-\frac{1}{5}b = \frac{4b}{5}m\)

Hence, the new breadth is \(\frac{4b}{5}m\)

\(\therefore\) The area of modified rectangle is \(\frac{6l}{5}\times \frac{4b}{5}\) \(=\frac{24lb}{25}m^2\)

Hence, the new area of rectangle is \(\frac{24lb}{25}m^2\).

Length and breadth of original rectangle is \( l\) m and b m respectively.

Given that length of rectangle is increased by 20 %.

\(\therefore\) Length of modified rectangle is \(l+\frac{20}{100}l =l + \frac{1}{5}l = \frac{6}{5}l\,m\)

Hence, the new length is \(\frac{6l}{5}m\)

Given that breadth of rectangle is decreased by 20%.

\(\therefore\) Breadth of modified rectangle \(=b - \frac{20}{100}b\) \(=b-\frac{1}{5}b = \frac{4b}{5}m\)

Hence, the new breadth is \(\frac{4b}{5}m\)

\(\therefore\) The area of modified rectangle is \(\frac{6l}{5}\times \frac{4b}{5}=\frac{24lb}{25}m^2\)

Hence, the new area of rectangle is \(\frac{24lb}{25}m^2\)

\(\because\) The original area of rectangle is \(l\)b \(m^2\) and the new area of rectangle after modification is \(\frac{24lb}{25}m^2\)

\(\therefore\) The change in area \(= lb-\frac{24lb}{25}=\frac{lb}{25}\)

\(\therefore\) Percentage change in area \(=\frac{The\,new\,area\,of\,modified\,rectangle}{The\,area\,of\,original\,rectangle}\times 100\)

\(=\frac{lb}{25}\times \frac{100}{lb}=4\%\)

The ascending order of first 8 prime numbers is 2, 3, 5, 7, 11, 13, 17, 19.

Total prime numbers is n = 8.

\(\therefore \)  \(Median=\frac{\frac{n}{2}^{th}term + \left(\frac{n}{2}+1\right)^{th}term}{2}\)  

\(=\frac{4^{th}term + 5^{th}term}{2}\)

\(=\frac{7+11}{2}=\frac{18}{2}=9\) \(\left(\because \frac{n}{2}=\frac{8}{2}=4\right)\)

Hence, the median of first 8 prime numbers is 9.

Let the event E be the event that family having at least one boy.

Total number of children in the family = 3.

\(\therefore\) Sample space = {BBB, BBG, GBB, BGG, GBG, GGB, GGG}.

Total number of possible outcomes = \(2^3\) = 8.

Outcomes favorable to event E = {BBB, BBG, BGB, GBB, BGG, GBG, GGB}.

Total outcomes favorable to event E = 7.

Now, probability of having at least one by 

\(=\frac{No.\,of\,outcomes\,favourable\,to\,event\,E}{Total\,no.\,of\,possible\,outcomes}\) \(=\frac{7}{8}\)

Hence, the probability of having at least one boy in a family of 3 children is \(\frac{7}{8}\).

Given that point A(4, 3) lies on a circle whose centre is O(2, 3).

\(\therefore\) Distance between points O(2, 3) & A(4, 3) is equals to the radius of the circle.

\(\therefore\) Radius of the circle is r \(=\sqrt{(4-2)^2+(3-3)^2}\)  (By distance formula)

\(\Rightarrow\) \(r=\sqrt{2^2+0}=2\,unit\)

Hence, the radius of the circle is r = 2 unit

Now, the equation of the circle whose centre is O(2, 3) and radius 2 is

\((\mathrm x-2)^2+(y-3)^2=2^2\)

\(\Rightarrow \mathrm{x^2 - 4x + 4 + y^2-6y+ 9 = 4}\)  (\(\because\) \((a-b)^2 = a^2+b^2-2ab\))

\(\Rightarrow \mathrm{x^2 + y^2-4x - 6y + 9=0}\)  ...........(1)

Since, point B(x, 5) lie on the circle whose equation given by equation (1), therefore, point B(x, 5) satisfies equation (1).

\(\therefore\) \(\mathrm{x^2 + 5^2 - 4x - 30 + 9} = 0\)

\(\Rightarrow \mathrm{x^2 - 4x - 21 + 25 = 0}\)

\(\Rightarrow \mathrm{x^2 - 4x + 4 = 0}\)

\(\Rightarrow (\mathrm x-2)^2 = 0\)

\(\Rightarrow \mathrm x- 2 = 0\) \(\Rightarrow \mathrm x = 2\)

Hence, the value of x is x = 2.

Let first term of AP is a and common difference of AP is d.

We know that \(n^{th}\) term of AP is given by \(T_n = a+(n-1)d\)

Now, given that \(4^{th}\) term of AP is 11. i.e., \(T_4 = 11\)

\(\Rightarrow \) a + (4 –1)d = 11

\(\Rightarrow \) a + 3d = 11

Let first term of AP is a and common difference of AP is d.

Now, given that the sum of \(5^{th}\) and \(7^{th}\) terms of AP is 34. i.e., \(T_5 + T_7 = 34\)

\(\therefore\) a + (5 – 1)d + a + (7 – 1)d = 34

\(\Rightarrow \) 2a + 4d + 6d = 34

\(\Rightarrow \) a + 5d = 17

Let first term of AP is a and common difference of AP is d.

We know that \(n^{th}\) term of AP is given by \(T_n=a+(n-1)d\)

Now, given that \(4^{th}\) term of AP is 11. i.e., \(T_4 =11\)

\(\Rightarrow\) a + (4 –1)d = 11

\(\Rightarrow\) a + 3d = 11  ..........(1)

Again, given that the sum of \(5^{th}\) and \(7^{th}\) terms of AP is 34. i.e., \(T_5+T_7 = 34\)

\(\therefore\) a + (5 – 1)d + a + (7 – 1)d = 34

\(\Rightarrow\) 2a + 4d + 6d = 34

\(\Rightarrow\) a + 5d = 17  .......(2)

Now, subtracting equation (1) from equation (2), we get (a + 5d) – (a + 3d) = 17 – 11

\(\Rightarrow\) 2d = 6

\(\Rightarrow\) \(d=\frac{6}{2}=3\)

Hence, the common difference of AP is d = 3.

Now, putting d = 3 in equation (1), we get a + 3\(\times\)3 = 11

\(\Rightarrow\) a = 11– 9 =2.

Hence, the first term of AP is a = 2.

Let first term of AP is a and common difference of AP is d.

We know that \(n^{th}\) term of AP is given by \(T_n = a + (n-1)d\)

Now, given that \(4^{th}\) term of AP is 11. i.e, \(T_4 =11\)

\(\Rightarrow a + (4-1)d = 11\)

\(\Rightarrow a + 3d = 11\) ...........(1)

Again, given that the sum of \(5^{th}\) and \(7^{th}\) terms of AP is 34. i.e., \(T_5 + T_7 = 34\)

\(\therefore\) a + (5 – 1)d + a + (7 – 1)d = 34

\(\Rightarrow 2a + 4d + 6d = 34\)

\(\Rightarrow a + 5d = 17\)  ............(2)

Now, subtracting equation (1) from equation (2), we get (a + 5d) – (a + 3d) = 17 – 11

\(\Rightarrow 2d = 6\)

\(\Rightarrow d = \frac{6}{2}=3\)

Hence, the common difference of AP is d = 3.

Let first term of AP is a and common difference of AP is d.

We know that \(n^{th}\) term of AP is given by \(T_n = a + (n-1)d\)

Now, given that \(4^{th}\) term of AP is 11. i.e, \(T_4 =11\)

\(\Rightarrow\) a + (4 –1)d = 11

\(\Rightarrow\) a + 3d = 11  ............(1)

Again, given that the sum of \(5^{th}\) and \(7^{th}\) term of AP is 34. i.e, \(T_5 + T_7 = 34\)

\(\therefore\) a + (5 – 1)d + a + (7 – 1)d = 34

\(\Rightarrow\) 2a + 4d + 6d = 34

\(\Rightarrow\) a + 5d = 17 ...........(2)

Now, subtracting equation (1) from equation (2), we get (a + 5d) – (a + 3d) = 17 – 11

\(\Rightarrow\) 2d = 6 \(\Rightarrow\) d = \(\frac{6}{2}=3\)

Hence, the common difference of AP is d = 3.

Now, putting d = 3 in equation (1), we get a + 3\(\times\) 3 = 11

\(\Rightarrow\) a = 11– 9 =2.

Hence, the first term of AP is a = 2.

\(T_{50}=a +(50-1)d\)

= a + 49 d

= 2 + 49 × 3

= 2 + 147 = 149. (\(\because\) a = 2, d = 3)

Let the numbers are a and b.

Given that difference between both numbers is 5.

\(\therefore\) a - b = 5  ...........(1)

Given that difference between their squares is 65.

\(\therefore\) \(a^2 - b^2 = 65\) 

\(\Rightarrow\) (a – b)( a + b) = 65  (\(\because\) (a - b)(a + b) = \(a^2 - b^2\))

\(\Rightarrow\) 5(a + b) = 65

\(\Rightarrow\) a + b \(=\frac{65}{5}=13\)  ...........(2)

Now, adding equations (1) & (2), we get (a – b) + ( a + b) = 5 + 13

\(\Rightarrow\) 2a = 18

\(\Rightarrow\) a \(=\frac{18}{2}=9\)

Now, putting a = 9 in equation (1), we get 9 – b = 5 ⇒ b = 9 – 5 = 4.

Hence, the numbers are 4 and 9.

Given that 3x = cosecθ and \(\frac{3}{\mathrm x}\) = cotθ

Now, \(\mathrm{cosec}^2\theta=\mathrm{(3x)^2=9x^2}\)

And \(\cot^2\theta = \left(\frac{3}{\mathrm x}\right)^2=\frac{9}{\mathrm x^2}\)

We know that \(\mathrm{cosec}^2\theta -\cot^2\theta =1\)

\(\therefore\) \(9\mathrm x^2-\frac{9}{\mathrm x^2}=1\) 

\(\Rightarrow \) \(9\left(\mathrm x^2 - \frac{1}{\mathrm x^2}\right)=1\)

\(\Rightarrow 3\mathrm{\left(x^2-\frac{1}{x^2}\right)}=\frac{1}{3}\)

Hence, the value of \(\mathrm{3\left(x^2-\frac{1}{x^2}\right)}\) is \(\frac{1}{3}\).

Given that speed of the stream is x = 2 km/hour.

Speed of boat in still water is y = 10 km/hour.

Now, speed downstream = speed of boat in still water + speed of the stream = x + y = 2 + 10

= 12 km/hour. (\(\because\) work is in favour of stream and x = 2 & y = 10)

Given that speed of the stream is x = 2 km/hour.

Speed of boat in still water is y = 10 km/hour.

Now, speed upstream = speed of boat in still water – speed of the stream = y – x = 10 – 2

= 8 km/hour. (\(\because\) work done is against stream and x = 2 & y = 10)

Given that speed of the stream is x = 2km/hour.

Speed of boat in still water is y = 10 km/hour.

Total time = 1 hour 40 minutes \(=1\frac{40}{60}\)hours \(=1\frac{2}{3}\) \(=\frac{5}{3}hours\)

Distance covered by boat is S = 8 km in both ways downstream and upstream.

Now, speed downstream = speed of boat in still water + speed of the stream = x + y = 2 + 10

= 12 km/hour.  (\(\because\) work is in favour of stream and x = 2 & y = 10)

Time taken to cover 8km downstream \(=\frac{8}{speed\,down\,stream}\) \(=\frac{8}{12}=\frac{2}{3}hour.\)

Given that speed of the stream is x = 2km/hour.

Speed of boat in still water is y = 10 km/hour.

Total time = 1 hour 40 minutes \(=1\frac{40}{60}\) hours \(=1\frac{2}{3}\) \(=\frac{5}{3}\) hours

Distance covered by boat is S = 8 km in both ways downstream and upstream

Now, speed upstream = speed of boat in still water – speed of the stream = y – x = 10 – 2

= 8 km/hour. (\(\because\) work done is against stream and x = 2 & y = 10)

Time taken to cover 8 km up–stream \(=\frac{8}{speed\,up\,stream}\) \(=\frac{8}{8}\) = 1 hour.

Given that \(\frac{8}{\mathrm x+2}+\frac{8}{\mathrm x-2}=\frac{5}{3}\)

\(\Rightarrow \mathrm{\frac{8(x-2)+8(x+2)}{(x+2)(x-2)}} =\frac{5}{3}\)

\(\Rightarrow \mathrm{\frac{8x-16+8x+16}{x^2-4}} = \frac{5}{3}\)  (\(\because\) (a + b)(a - b) = \(a^2\) - \(b^2\))

\(\Rightarrow \mathrm{3\times 16x = 5(x^2-4)}\) (By cross multiplication)

\(\Rightarrow \mathrm{48x=5x^2-20}\)

\(\Rightarrow \mathrm{5x^2-48x-20=0}\)

Let one zero of the given polynomial is \(\alpha\).

Given that one zero of given polynomial is reciprocal of the other.

\(\therefore\) The other zero of given polynomial is \(\frac{1}{\alpha}\)

Hence, the zeros of the given polynomial \((a^2+9)\mathrm x^2+13\mathrm x + 6a\) are \(\alpha\) and \(\frac{1}{\alpha}\).

Now, the product of the zeros is \(\alpha.\frac{1}{\alpha}=\frac{6a}{a^2+9}\)

\(\Rightarrow \frac{6a}{a^2+9}=1\)

\(\Rightarrow a^2+9 = 6a\)

\(\Rightarrow a^2-6a + 9 = 0\)

\(\Rightarrow (a-3)^2 = 0\)  (\(\because (a-b)^2\) \(=a^2-2ab+b^2)\)

\(\Rightarrow a = 3\)

Hence, the value of a is 3.