CBSE Class 10 Mathematics Solved Question Paper 2014

$$\begin{gathered} \mathrm{K}, 2\mathrm{K}-1, 2\mathrm{K}+1 \mathrm{are} \mathrm{in} \mathrm{Arithmetic} \mathrm{Progression} \\ 2\mathrm{x}(2\mathrm{k}-1) = \mathrm{k}+2\mathrm{k}+1 \\ 4\mathrm{k}-2=3\mathrm{k}+1 \\ \mathrm{k}=3 \end{gathered}$$

It is given that AB = 5 and BC = 12

Using pythagoras theorem

$$\begin{gathered} {\mathrm{AC}}^2 = {\mathrm{AB}}^2 + {\mathrm{BC}}^2 \\ = 5^2 + {12}^2 \\ = 169 \\ \mathrm{Thus} \mathrm{AC} = 13 \end{gathered}$$

We know that  two tangents drwan to a circlefrom the same point that is exterior to

the  circle are of equal iengths.

Thus AM = AQ = a

Similarly  MB = BP = b  and  PC = CQ = c

We know AB = a+b = 5

BC = b+c = 12 and  AC = a+c = 13

Solving simultaneously we get  a = 3,  b =2,  c = 10

We also know that the tangent is perpendicular to the radius.

Thus OMBP is a square with side b

Hence the length of the radius of the circle inscribed in the right angled triangle is 2 cm.

$$\begin{gathered} \mathrm{TA}=\mathrm{TP}\Rightarrow \angle \mathrm{TPA} = \angle \mathrm{TPA} \\ \mathrm{TB}= \mathrm{TP}\Rightarrow \angle \mathrm{TBP}=\angle \mathrm{TPB} \\ \therefore \angle \mathrm{TAP}+\angle \mathrm{TBP}=\angle \mathrm{TPA}+\angle \mathrm{TPB} = \angle \mathrm{APB} \\ \angle \mathrm{TAP}+\angle \mathrm{TBP}+\angle \mathrm{APB} = 180° [\because \mathrm{Sum} \mathrm{of} ...180°] \\ \therefore \angle \mathrm{APB}+\angle \mathrm{APB}=180° \\ \therefore 2\angle \mathrm{APB}=180° \\ \therefore \angle \mathrm{APB}=90° \end{gathered}$$

There are in all 2³ = 8 combinations or outcomes for the gender of the 3 children

The 8 combinations are as follows

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Thus the probability of having at least one boy in a family is $\frac{7}{8}$

$$\begin{gathered} \mathrm{Given}: \angle \mathrm{AOB} \mathrm{is} \mathrm{given} \mathrm{as} 90° \\ ∆\mathrm{AOB} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{trianglesince} \mathrm{OA}=\mathrm{OB} \\ \mathrm{Therefore} \angle \mathrm{OAB}= \angle \mathrm{OBA}= 45° \\ \mathrm{Thus} \angle \mathrm{AOP}=45° \mathrm{and} \angle \mathrm{BOP}= 45° \\ \mathrm{Hence} ∆\mathrm{AOP} \mathrm{and} ∆\mathrm{BOP} \mathrm{also} \mathrm{are} \mathrm{isosceles} \mathrm{triangles} \\ \mathrm{Thus} \mathrm{let} \mathrm{AP}=\mathrm{PB}=\mathrm{OP}=\mathrm{x} \\ \mathrm{Using} \mathrm{pythagoras} \mathrm{theorem} \\ {\mathrm{x}}^2 + {\mathrm{x}}^2 = {10}^2 \\ \mathrm{Thus} 2{\mathrm{x}}^2 = 100 \\ \mathrm{x}=5\sqrt{2} \\ \mathrm{Hence} \mathrm{length} \mathrm{of} \mathrm{chord} \mathrm{AB} = 2\mathrm{x} = 10\sqrt{2} \end{gathered}$$

Let AB be the tower and BC be the distance between tower and car. Let $\mathrm{\theta }$ be the angle of depression of the car.

According to the given information,

$$\begin{gathered} \mathrm{In} ∆\mathrm{ABC}, \\ \mathrm{tan\theta } = \frac{\mathrm{BC}}{\mathrm{AB}} \left( \mathrm{Using} \left(1\right) \mathrm{and} \tan 30° = \frac{1}{\sqrt{3}}\right) \\ \therefore \mathrm{BC} = \frac{150}{\sqrt{3}} = \frac{150\sqrt{3}}{3} = 50\sqrt{3} \\ \mathrm{Hence}, \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{tower} \mathrm{and} \mathrm{car} \mathrm{is} 50\sqrt{3} . \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{multiple} \mathrm{of} 4 \mathrm{between} 1 \mathrm{and} 15 \mathrm{are} 4, 8, 12. \mathrm{the} \mathrm{probability} \mathrm{of} \mathrm{getting} \mathrm{a} \\ \mathrm{multiple} \mathrm{of} 4 = \frac{3}{15} = \frac{1}{5} \end{gathered}$$

We se that AB= 4 units and BC=3 units 

Using pythagoras theorem 

AC² = AB² + BC² 

      = 4² + 3²

AC² = 25

Thus AC= 5 Units

Hence length of the diagonal of the rectangle is 5 units