CBSE Class 10 - Maths

1) Find the ratio of corresponding coefficients of the equations.

2) Relate the exact condition for finding the relation between the pair of lines.

1) Write the coefficients of the equations by comparing the standard form of the linear equations.

2) Find the ratios of the corresponding coefficients.

3) Find the relation between the pair of given lines by comparing the ratios of coefficients of the linear equations.

If $a_1$x +$b_1$y + $c_1$= 0 and $a_2$x+ $b_2$y + $c_2$= 0 are two linear equations.

If $\frac{a_1}{a_2}$ ≠ $\frac{b_1}{b_2}$, then the above equations represent intersecting lines.

If$\frac{a_1}{a_2}$ = $\frac{b_1}{b_2}$, then the equations represent coincident lines.

If $\frac{a_1}{a_2}$ = $\frac{b_1}{b_2}$≠$\frac{c_1}{c_2}$, then the equations represent parallel lines.

The given equations are :-

9x - 10y = 21 or 9x -10y - 21= 0

$\frac{3}{2}$x -$\frac{5}{3}$ y = $\frac{7}{2}$ or $\frac{3}{2}$x -$\frac{5}{3}$ y - $\frac{7}{2}$ = 0

When these equations are compared, $a_1$= 9, $b_1$= - 10 and $c_1$= - 21

Similarly, $a_2$= $\frac{3}{2}$, $b_2$= $\frac{-5}{3}$ and $c_2$= $\frac{-7}{2}$

$\frac{a_1}{a_2}$ = $\frac{9}{\frac{3}{2}}$ = 6

$\frac{b_1}{b_2}$= $\frac{-10}{\frac{-5}{3}}$= 6

and $\frac{c_1}{c_2}$ = $\frac{-21}{\frac{-7}{2}}$= 6

Hence, $\frac{a_1}{a_2}$ = $\frac{b_1}{b_2}$= $\frac{c_1}{c_2}$

Therefore, the given pair of equations represents coincident lines.

1) Write the correct formula to find the $n^{\mathrm{th}}$ term of the A.P.

2) Find the difference between the ${17}^{\mathrm{th}}$ term and the ${10}^{\mathrm{th}}$ term.

1) Find the ${17}^{\mathrm{th}}$ term and the ${10}^{\mathrm{th}}$ term of an A.P.

2) Find the difference between the ${17}^{\mathrm{th}}$ term and the ${10}^{\mathrm{th}}$ term.

3) Equate the difference to 7.

4) Simplify to find the common difference (d) between the terms of the A.P.

For an A.P. its nth term or $a_n$= a + (n − 1)d where a is the first term and d is the common difference between the terms.

${17}^{\mathrm{th}}$ term of the A.P. or $a_{17}$= a + (17 - 1)d = a + 16d

${10}^{\mathrm{th}}$ term of the A.P. or $a_{10}$= a + (10 - 1)d = a + 9d

Given that the difference between the ${17}^{\mathrm{th}}$ term and the ${10}^{\mathrm{th}}$ term = 7

⇒(a + 16d) − (a + 9d) = 7

⇒7d = 7

⇒d = 1

Therefore, the common difference (d) between the terms of the A.P. is 1.

1) Use the correct trigonometric ratios of complimentary angles.

1) Use the trigonometric ratios of complementary angles such that all the trigonometric ratios are cancelled.

2) Apply the reciprocals of trigonometric ratios such that all the trigonometric ratios are cancelled.

3) Simplify to get the value of the expression.

According to trigonometric ratios of complementary angles, cos θ = sin (90° - θ) and cot θ = tan (90° - θ)

$\frac{7}{2}\times \frac{\cos 70°}{\sin 20°}$ + $\frac{3}{2}\times \frac{\cos 55°\cos \mathrm{ec}35°}{\tan 5°\tan 25°\tan 45°\tan 85°\tan 65°}$

= $\frac{7}{2}\times \frac{\cos 70°}{\sin (90°-70°)}+\frac{3}{2}\times \frac{\cos 55°\cos \mathrm{ec}(90°-55°)}{\tan 5°\tan 25°\tan 45°\tan (90°-5°)\tan (90-25°)}$

= $\frac{7}{2}$× $\frac{\cos 70°}{\cos 70°}$ + $\frac{3}{2}$× $\frac{\cos 55°\cos \mathrm{ec}(90°-55°)}{\tan 5°\tan 25°\times 1\times \cot 5°\times \cot 25°}$

= $\frac{7}{2}$ + $\frac{3\times \cos 55°\tan 5°\tan 25°}{2\times \tan 5°\tan 25°\cos 55°}$ [cot θ = $\frac{1}{\tan \theta }$ and sec θ = $\frac{1}{\cos \theta }$]

= $\frac{7}{2}$ + $\frac{3}{2}$

= $\frac{10}{2}$

= 5.

Hence, the value of the given expression is 5.

Use Pythagoras theorem, to prove that the given points form a right angled triangle.

Find the diameter of the circle by multiplying the radius by 2.

1) Find the lengths of sides joining the given points using the distance formula.

2) Verify if the lengths of two sides are equal and the sides satisfy Pythagoras theorem.

3) Find the radius of the circle using the distance formula and find the diameter of the circle.

4) Equate the diameter of the circle to 20 units to find the value of α.

The coordinates of the given points are A(- 2, 5), B(3, -4) and C(7, 10).

Using the distance formula, we get

AB = $\sqrt{{[3-(-2\left)\right]}^2+{(-4-5)}^2}$= $\sqrt{5^2+{(-9)}^2}$ = $\sqrt{25+81}$ = $\sqrt{106}$

BC = $\sqrt{(7-3{)}^2+[10-(-4){]}^2}$= $\sqrt{4^2+{14}^2}$ = $\sqrt{16+196}$ = $\sqrt{212}$

CA = $\sqrt{(-2-7{)}^2+(5-10{)}^2}$= $\sqrt{(-9{)}^2+(-5{)}^2}$= $\sqrt{81+25}$= $\sqrt{106}$

The triangle ABC is isosceles since AB = CA.

${\mathrm{AB}}^2+{\mathrm{CA}}^2$= 106 + 106 = 212 = ${\mathrm{BC}}^2$

∴ According to Pythagoras theorem, the triangle is right-angled at A.

Hence, the points (−2, 5), (3, −4) and (7, 10) are the vertices of a right-angled triangle.

OR

The center of the circle is given as (2α−1, 7). The point through which the circle passes through is given as (−3, −1).

Using distance formula, the radius of the circle or OP

= $\sqrt{[-3-(2\alpha -1){]}^2+(-1-7{)}^2}$

= $\sqrt{(-2-2\alpha {)}^2+(-8{)}^2}$

= $\sqrt{(2+2\alpha {)}^2+64}$

Since the diameter of circle is given as 20 units and diameter = 2 × radius or 2 × OP

2 × OP = 20 units = 2 $\sqrt{(2+2\alpha {)}^2+64}$

⇒ $\sqrt{(2+2\alpha {)}^2+64}$= 10

⇒ (2 + 2α ${)}^2$+ 64 = 100

⇒ (2 + 2α ${)}^2$ = 36

⇒ 2 + 2α = ± 6

If 2 + 2α = 6 , then 2α = 4 ⇒

⇒ α = 2 units

If 2 + 2α = - 6 , then 2α = - 8 ⇒

⇒ α = - 4 units

∴ α = 2 or - 4 units.

1) Find the ratio in which the point C divides AB using the given condition.

2) Assume the A(1, 1) as the first point and B(2, -3) as the second point.

1) Find the ratio in which the point C divides AB using the given condition.

2) Consider A as the first point and B as the second point.

3) Use the section formula to find the coordinates of point C.

Assume that the coordinates of C are x and y. Point C can be represented as C(x, y).

Given that 3AC = CB⇒AC:CB = 1:3

Coordinates of the point on a line joining the points ($x_1,y_1)\mathrm{and}(x_2,y_2)$in the ratio of m:n are

$(\frac{m{x}_2+n{x}_1}{m+n}$, $\frac{m{y}_2+n{y}_1}{m+n})$

Using the section formula,

C(x, y) = $[\frac{1\times 2+3\times 1}{1+3}$, $\frac{1\times (-3)+3\times 1}{1+3}]$ = $[\frac{2+3}{4}$, $\frac{-3+3}{4}]$ = $[\frac{5}{4},0]$

∴ The coordinates of point C are ($\frac{5}{4}$, 0)

1) Find the squares of the general form of the odd positive integers using the algebraic identity.

2) Write 7 + 3$\sqrt{2}$using integers as $\frac{p}{q}$.

1) Write the general form of the positive odd integers as 8n + r where 0≤ r <8 and x is an integer.

2) Find the square of the each form and show that it is of the form 8m + 1.

OR

1) Assume that 7 + 3$\sqrt{2}$is a rational number and write it in the form of $\frac{p}{q}$.

2) Prove that it is not a rational number after simplification.

Assume a positive integer as ‘a’ and assume b = 8.

It can be represented as a = 8n + r where 0≤ r <8 and x is an integer.

Hence 'a' will be (8n + 1) or (8n + 2) or (8n + 3) or (8n + 4) or (8n + 5) or (8n + 6) or (8n + 7).

Positive odd integer is of the form (8n + 1) or (8n + 3) or (8n + 5) or (8n + 7).

Consider the above 4 positive odd integer forms separately.

Case 1: If a = 8n + 1

$a^2$=${(8n+1)}^2$= ${64n}^2$+ 16n + 1 = 8(8$n^2$+ 2n) + 1 = 8m + 1 where m = (8$n^2$ + 2n) is an integer.

Case 2: If a = 8n +3

$a^2$= ${(8n+3)}^2$= ${(64n}^2+48n+9)$= 8(8n + 6 + 1) + 1 = 8m + 1 where m = (8$n^2$+ 6n + 1) is an integer.

Case 3: If a = 8n + 5

$a^2$ = (8n + 5${)}^2$= 64$n^2$+ 80 n + 25 = 8(8$n^2$ + 10n + 3) + 1 = 8m + 1 where m = (8$n^2+10x+3$) is an integer.

Case 4: If a = 8n + 7

$a^2$= (8n + 7${)}^2$= (64$n^2$+ 112n + 49) = 8(8$n^2$+ 14n + 6) + 1 = 8m + 1 where m = (8$n^2$ + 14n + 6) is an integer.

From all these cases, it can be noticed that the square of a positive odd integer is of the form 8m + 1 where m is any integer.

OR

Assume that 7 + 3$\sqrt{2}$is a rational number.

It can be represented using integers p and q (q ≠0) as, 7 + 3$\sqrt{2}$= $\frac{p}{q}$

⇒$\sqrt{2}$√= $\frac{1}{3}$ ($\frac{p}{q}$ - 7)

p and q are rational numbers. So, $\frac{1}{3}$($\frac{p}{q}$ - 7) should also be rational and $\sqrt{2}$√should also be a rational number. But we know that $\sqrt{2}$ is an irrational number and so our assumption is wrong.

So, our assumption is false and (7 + 3$\sqrt{2}$) is not a rational number.

1) Find the remainder by the long division method of polynomials.

2) Equate both the terms of the remainder to zero.

1) Divide the given polynomial by 2$x^2$-5.

2) Equate the remainder to zero to find the values of 'a' and 'b'.

p(x) = 6 $x^4$+ 8 $x^3$−5 $x^2$+ ax + b

q(x) = 2 $x^2$ - 5

Dividing p(x) by q(x)

We get the remainder as (a + 20)x + (b + 25).

As p(x) is exactly divisible by q(x), remainder should be 0.

∴(a + 20)x + (b + 25) = 0

⇒a + 20 = 0 and b + 25 = 0

⇒ a + 20 = 0 , a = - 20 and

⇒ b + 25 = 0, b = - 25

Therefore, the value of a = -20 and b = - 25.

Find the relation between the${19}^{\mathrm{th}}$ term and the${29}^{\mathrm{th}}$ term of the progression.

1) Equate the $9^{\mathrm{th}}$ term of the progression to zero and find the value of 'a' in terms of d.

2) Find ${19}^{\mathrm{th}}$ term and the ${29}^{\mathrm{th}}$ term in terms of d.

3) Find the relation between ${19}^{\mathrm{th}}$ term and the ${29}^{\mathrm{th}}$ term.

Let a be the $1^{\mathrm{st}}$ term and d be the common difference of the A.P.

$n^{\mathrm{th}}$ term of A.P. = $a_n$= a + (n - 1)d

∴ $9^{\mathrm{th}}$ term of A.P. = $a_9$= a + (9 - 1)d = a + 8d

Given that, $a_9$= 0

∴ $a_9$= a + 8d = 0

⇒a = - 8d

${19}^{\mathrm{th}}$ term of A.P. = $a_{19}$= a + (19 - 1)d = a + 18d = - 8d + 18d = 10d

${29}^{\mathrm{th}}$ term of A.P. = $a_{29}$= a + (29 - 1)d = a + 28d = - 8d + 28d = 20d

∴ $a_{29}$= 20d = 2 × 10d = 2 ×$a_{19}$

Hence, the ${29}^{\mathrm{th}}$ term of the A.P. is double the ${19}^{\mathrm{th}}$ term of the A.P.

Mark the point at 6 cm from the center of the circle either to the left or right of the circle.

Join the point at 6 cm to points of intersection of circles exactly.

Draw a circle of given radius and mark a point O at 6 cm from the center.

Draw the perpendicular bisector to the line and mark the point at which it intersects the circle.

Draw another circle with the point of intersection as center.

Mark the points of intersections of circles and join with the point marked in step 1.

Draw a circle of radius 3 cm and mark the center point as O.

Mark a point P at a distance of 6 cm from O and join O and P.

Draw the perpendicular bisector to the line OP.

Mark the point at which it intersects the circle as Q.

With Q as center and QP as radius, draw another circle.

This will intersect the first circle at 2 points. Mark the points as M and N.

Join M with P to obtain line MP and N with P to obtain the line NP. MP and NP are the tangents to the circle with center O.

On measuring, the length of the tangents, MP and NP is 5.20 cm.

1) Consider the exact corresponding sides of the triangles.

2) Apply converse of basic proportionality theorem to prove that the lines are parallel.

1) Consider △ABC and △QPC, and apply basic proportionality theorem.

2) Consider △BDC and △PQR, and apply basic proportionality theorem.

3) Use transitive property and converse of basic proportionality theorem to prove that line QR is parallel to AD.

Consider △ABC and △QPC

PQ || BA and hence according to the basic proportionality theorem, we get

$\frac{\mathrm{CP}}{\mathrm{PB}}$ = $\frac{\mathrm{CQ}}{\mathrm{QA}}$ --------------------- (1)

Consider △BDC and △PQR

PR || BD and hence according to the basic proportionality theorem, we get

$\frac{\mathrm{CP}}{\mathrm{PB}}$ = $\frac{\mathrm{CR}}{\mathrm{RD}}$ -----------------------(2)

Taking (1) and (2) together we get

$\frac{\mathrm{CQ}}{\mathrm{QA}}$ = $\frac{\mathrm{CR}}{\mathrm{RD}}$

∴Considering △ADC and △QRC, from above, using converse of proportionality theorem, we get QR|| AD.

1) Each point of trisection divides the side of the triangle in the ratio of 1:2 or 2:1.

2) Substitute the equations of the Pythagoras theorem of one triangle in the other.

1) Apply Pythagoras theorem in △ABD, △ABE and △ABC.

2) Substitute the value of BD according the trisection ratio as $\frac{\mathrm{BC}}{3}$.

3) Substitute the values of ${\mathrm{AC}}^2$and ${\mathrm{AD}}^2$ in ${3\mathrm{AC}}^2+5{\mathrm{AD}}^2$and simplify to get $8{\mathrm{AE}}^2$.

Given: BC is trisected at D and E.

∴BD = DE = EC = $\frac{\mathrm{BC}}{3}$

⇒ BE = $\frac{2\mathrm{BC}}{3}$

Consider ∆ABD. Using Pythagoras Theorem, we get

${\mathrm{AB}}^2$+ ${\mathrm{BD}}^2$= ${\mathrm{AD}}^2$

${\mathrm{AB}}^2+{\left(\frac{\mathrm{BC}}{3}\right)}^2={\mathrm{AD}}^2$

${\mathrm{AB}}^2+\frac{\left({\mathrm{BC}}^2\right)}{9}={\mathrm{AD}}^2$------------------ (1)

Consider ∆ABE. Using Pythagoras theorem, we get

${\mathrm{AB}}^2+{\mathrm{BE}}^2={\mathrm{AE}}^2$

${\mathrm{AB}}^2+{\left(\frac{2\mathrm{BC}}{3}\right)}^2={\mathrm{AE}}^2$

${\mathrm{AB}}^2+\frac{{4\mathrm{BC}}^2}{9}={\mathrm{AE}}^2$ -----------------------(2)

Consider ∆ABC. Using Pythagoras theorem, we get

${\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2$ -----------------------(3)

We have to prove that 3 ${\mathrm{AC}}^2$+ 5 ${\mathrm{AD}}^2$= 8 ${\mathrm{AE}}^2$

${3\mathrm{AC}}^2+5{\mathrm{AD}}^2=3({\mathrm{AB}}^2+{\mathrm{BC}}^2)+5\{{\mathrm{AB}}^2+\left(\frac{{\mathrm{BC}}^2}{9}\right)\}$

= $3{\mathrm{AB}}^2+3{\mathrm{BC}}^2+5{\mathrm{AB}}^2+\frac{\left({5\mathrm{BC}}^2\right)}{9}$

= ${8\mathrm{AB}}^2+\frac{\left({32\mathrm{BC}}^2\right)}{9}$

= ${8[\mathrm{AB}}^2+\frac{\left({4\mathrm{BC}}^2\right)}{9}$]

From (2), we know that ${\mathrm{AB}}^2+\frac{{4\mathrm{BC}}^2}{9}={\mathrm{AE}}^2$.

Hence substituting in above, we get

$3{\mathrm{AC}}^2+5{\mathrm{AD}}^2=8{\mathrm{AE}}^2$

Hence proved.

OR

AD and AF are tangents drawn from point A to the circle. Similarly BD and BE are tangents from point B to the circle and CE and CF are tangents from point C

The tangents drawn to the circle from any point external to the circle are equal.

∴ AD = AF

BD = BE and

CE = CF

Let us take AD = a, BE = b and CF = c

AB = AD + BD = a + b = 12 cm ------------- (1)

BC = BE + CE = b + c = 8 cm --------------- (2)

AC = AF + CF = a + c = 10 cm -------------- (3)

AB + BC + AC = 2(a + b + c) = 12 + 8 + 10 = 30 cm

⇒a + b + c = 15 cm --------------- (4)

Substituting (1) in (4), we get c = (15) − (a + b) = 15 − 12 = 3 cm

Substituting (2) in (4), we get a =15 − (b + c) = 15 − 8 = 7 cm

Substituting (3) in (4), we get b = 15 − (a + c) = 15 − 10 = 5 cm

Thus, we get, AD = a = 7 cm, BE = b = 5 cm and CF = c = 3 cm.

1) Apply appropriate trigonometric identities.

2) Tangent is the ratio of sine and cosine.

1) Take ${\sin }^2\theta$ as a common factor in the numerator and ${\cos }^2\theta$ as a common factor in the denominator of the fraction of the second term of the equation.

2) Apply appropriate trigonometric identities and simplify.

${\sec }^2\theta -\frac{{\sin }^2\theta -{2\sin }^4\theta }{2{\cos }^4\theta -{\cos }^2\theta }$

⇒${\sec }^2\theta -\frac{{\sin }^2\theta -2{\sin }^4\theta }{{\cos }^2\theta (2{\cos }^2\theta -1)}$ [(${\cos }^2\theta =1-{\sin }^2\theta \left)\right]$

${\Rightarrow \sec }^2\theta -\frac{{\tan }^2\theta (1-2{\sin }^2\theta )}{{\cos }^2\theta (1-{\sin }^2\theta )-1}$

(tan θ = $\frac{\sin \theta }{\cos \theta }$)

⇒${\sec }^2\theta -\frac{{\tan }^2\theta (1-2{\sin }^2\theta )}{(1-2{\sin }^2\theta )}$

⇒${\sec }^2\theta -{\tan }^2\theta$

= 1 [${\sec }^2\theta =1+{\tan }^2\theta$ ]

Therefore, ${\sec }^2\theta$ - $\frac{{\sin }^2\theta -2{\sin }^4\theta }{2{\cos }^4\theta -{\cos }^2\theta }=1.$

1) Write the correct formula for the area of a triangle.

2) Substitute the corresponding coordinates in the formula.

1) Assume the three points as three vertices of a triangle.

2) Area of the triangle formed by three points is zero as the points are collinear.

3) Equate the area formed by three points to zero.

Since the points (x, y), (1, 2) and (7, 0) are collinear, the area of the triangle formed with these three points as vertices will be zero.

Area of triangle with points ($x_1$, $y_1$), ($x_2$, $y_2$) , ($x_3$, $y_3$) =

$\frac{1}{2}${$x_1$($y_2$−$y_3$) + $x_2$($y_3$− $y_1$) + $x_3$($y_1$−$y_2$ )}

Area of the triangle formed by the given points

= $\frac{1}{2}${x(2 − 0) + 1(0 − y) + 7(y − 2)}

= $\frac{1}{2}${2x − y + 7y − 14}

= $\frac{1}{2}${2x + 6y − 14}

= x + 3y − 7

Thus, the area of the triangle is x + 3y − 7. Since this area is zero, we get x + 3y−7 = 0.

∴ x + 3y−7 = 0 is the equation which shows the relation between x and y, if the points have to be collinear.

1) Use the reflex angle of 90° to find the area of the top of the table.

2) Perimeter of the table top includes both the radii and the length of the arc of the sector.

OR

1) Half of the length of the side of the square is equal to the radius of the semicircle.

2) Area of the shaded part of the square = Area of the square - Area of the unshaded part of the square.

1) Find the reflex angle of ∠BOD which is the angle of the sector.

2) Use the formula to find the area of the sector to find the area of the table top.

OR

1) Find the area of the square.

2) Half of the side of the square is equal to the radius of the semicircle.

3) Find the area of the semicircle.

4) Subtract twice the area of the semicircle from the area of the square to find the area of the unshaded part of the circle.

(i) Area of table top

Area of a sector can be obtained by $\frac{\theta }{360°}\times \pi r^2$where ∅ is the angle of the sector and r is the radius.

Given that ∠BOD = 90°.

∴Reflex ∠BOD = 360° - 90° = 270°

Area of the table top = $\frac{\theta }{360°}\times \pi r^2$

where ∅ = 270° and r = radius of the sector = 60 cm (BO = BD = 60 cm)

= $\frac{270°}{360°}$ × 3.14 × 60 × 60

= $\frac{3}{4}$ × 3.14 × 3600

= 8478 ${\mathrm{cm}}^2$

Hence, the area of the table top = 8478 ${\mathrm{cm}}^2$

(ii) Perimeter of the table top = length of the arc BD + OB + OD

Length of the arc BD = $\frac{270°}{360°}\times 2\mathrm{\pi r}$

= $\frac{3}{4}$× 2 × 3.14 × 60

= 282.6 cm

∴ Perimeter of the table top = 282.6 + 60 + 60 = 402.6 cm

OR

Area of the shaded region = Area of the square ABCD - (area of the semicircle APD + area of the semicircle BPC)

Area of the square ABCD

=14 ×14 =196 ${\mathrm{cm}}^2$(since the length of the sides of the square = 14 cm)

Area of the semicircle APD = $\frac{\left(\pi r^2\right)}{2}$, where r = 7 cm. (AD = 14 cm, r = $\frac{\mathrm{AD}}{2}$)

Area = $\frac{(22\times 7\times 7)}{(7\times 2)}$ = 77 ${\mathrm{cm}}^2$

Area of semicircle BPC = 77 ${\mathrm{cm}}^2$since, r = 7 cm

∴ Area of the shaded region = 196 -(77 + 77)

= 196 -154

= 42 ${\mathrm{cm}}^2$

Hence, area of the shaded region of the figure = 42 ${\mathrm{cm}}^2$.

1) Find the exact number of number of cards in the box.

2) Find the number of odd numbers, perfect squares and the numbers divisible by 7 between 14 and 99.

1) Find the number of cards in the box which are numbered from 14 to 99.

2) Find the number of odd numbers between 14 and 99 and find the probability of the card being an odd number using the probability formula.

3) Find the number of odd numbers between 14 and 99 and find the probability of the card being a perfect square using the probability formula.

4) Find the numbers which are divisible by 7 between 14 and 99 and find the probability of the card being a number divisible by 7 using the probability formula.

The probability of an event = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$

Number of cards in box = 99 −14 + 1 = 86

Number of cards having odd number = $\frac{86}{2}$ = 43

Hence, probability of getting an odd number = $\frac{43}{86}$ = $\frac{1}{2}$

Number of cards having perfect squares = 6 (Perfect squares between 14 and 99 are 16, 25, 36, 49, 64 and 81)

Hence, probability of getting perfect square = $\frac{6}{86}$ = $\frac{3}{43}$

Number of cards with numbers divisible by 7 = 13

(Numbers divisible by 7 between 14 and 99 are 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91 and 98)

Hence, probability of getting number divisible by 7 = $\frac{13}{86}$.

1) If the total number of articles is x, then find the selling price of (x - 5) articles.

2) Equate the difference between the selling price of (x - 5) articles and Rs 900 to Rs 80.

OR

1) Subtract the number of years to find the age of son 2 years ago.

2) Add the number of years to find the age of the son 3 years later.

1) Assume the number of articles as 'x' and find the cost price of each article which is equal to Rs $\frac{900}{x}$.

2) Find the selling price of 1 articles which is equal to Rs ($\frac{900}{x}$ + 2).

3) Find the selling price of (x − 5) articles.

4) Equate the difference between the selling price of (x − 5) articles and Rs 900 to Rs 80.

5) Solve the quadratic equation obtained to find original number of articles.

OR

1) Assume the present age of the son as 'x'.

2) Find the age of the son and the father two years ago with the given condition.

3) Find the age present age of the father.

4) Find the age of the son and the the age of the father three years later.

5) Find the present age of the father.

6) Equate the present ages of the father to find the present age of the father by solving the quadratic equation obtained.

7) Obtain the present age of the father from the present age of the son.

Assume the total number of articles bought as ‘x’

Cost of x articles = Rs 900

∴ Cost price of 1 article =Rs $\frac{900}{x}$

Number of articles sold = (x - 5)

Selling price of 1 article = Rs $\frac{900}{x}$ + 2

Profit after selling (x − 5) articles = Rs 80

∴ (x − 5) × ($\frac{900}{x}$ + 2) − 900 = 80

(x − 5) × ($\frac{900}{x}$ + 2) = 980

(x − 5) × ($\frac{900}{x}$ + 2) = 980

900 − $\frac{4500}{x}$ + 2(x − 5) = 980

450 − $\frac{2250}{x}$ + x − 5 = 490

450x − 2250 + $x^2$ − 5x = 490x

$x^2$−45x − 2250 = 0

$x^2$−75x + 30x − 2250 = 0

x(x − 75) + 30(x − 75) = 0

(x − 75)(x + 30) = 0

∴x = 75, −30

Since number of articles can not be negative, x = 75

Hence, the number of articles bought by the trader = 75

OR

Let the son’s present age be ‘x’

Son’s age 2 years ago = (x − 2) years

Father’s age 2 years ago = 3(x − 2${)}^2$= 3($x^2$−4x + 4)

Father’s present age = 3($x^2$−4x + 4) + 2

Father’s age 3 years later from now = (x + 3)

Father’s age 3 years later from now = 4(x + 3) = 4x +12

Father’s present age = 4x + 12 − 3 = 4x + 9

⇒4x + 9 = 3($x^2$−4x + 4) + 2

⇒4x = 3$x^2$−12x + 12 + 2 − 9

⇒3$x^2$−16x + 5 = 0

⇒3$x^2$−15x − x + 5 = 0

⇒3x(x − 5) − 1(x − 5) = 0

⇒(x − 5)(3x − 1) = 0

x = 5, $\frac{1}{3}$

Since age cannot be a fraction, x = 5

∴ Son’s age = 5 years

Father’s age = 4x + 9 = 4×5 + 9 = 29 years.

1) Consider the ratio of the exact corresponding sides of the similar triangles.

2) Diagonal of a square is $\sqrt{2}$ times the length of the side.

3) Each angle in every equilateral triangle is equal to 60°.

1) Draw perpendiculars AP and XQ in each of the similar triangles ABC and XYZ.

2) Find the ratio of the areas of similar triangles.

3) Prove that ∆APB ~ ∆XQY.

4) Find the ratio of corresponding sides AP and XQ as $\frac{\mathrm{AB}}{\mathrm{XQ}}$.

5) Using the ratio of the corresponding sides of △ABC and △XYZ to prove the theorem.

6) Assume the side of the square as 'a' units then the side of the equilateral triangle drawn on the side is 'a'.

7) Diagonal of the square is√$\sqrt{2}$a, also the side of the equilateral triangle drawn on the diagonal is $\sqrt{2}$a.

8) Equate the ratio of the equilateral triangles to the ratio of the square of the corresponding sides to prove the theorem.

Consider two similar triangles ∆ABC and ∆XYZ.

To prove: $\frac{(\mathrm{Area}\mathrm{of}△\mathrm{ABC})}{(\mathrm{Area}\mathrm{of}△\mathrm{XYZ})}$= ${\left(\frac{\mathrm{AB}}{\mathrm{XY}}\right)}^2$ $$= ${\left(\frac{\mathrm{BC}}{\mathrm{YZ}}\right)}^2$= ${\left(\frac{\mathrm{AC}}{\mathrm{XZ}}\right)}^2$

Draw AP⊥BC and XQ⊥YZ.

Area of ∆ABC = $\frac{1}{2}$× base × height = $\frac{1}{2}$ × BC × AP ------(1)

Area of ∆XYZ = $\frac{1}{2}$× base × height = $\frac{1}{2}$ × YZ × XQ ------(2)

∴ $\frac{(\mathrm{Area}\mathrm{of}△\mathrm{ABC})}{(\mathrm{Area}\mathrm{of}△\mathrm{XYZ})}$= $\frac{(\mathrm{BC}\times \mathrm{AP})}{(\mathrm{YZ}\times \mathrm{XQ})}$ -------------(3)

∠APB = ∠XQY = 90°

∴ ∠ABP = ∠XYQ since ∆ABC ~ ∆XYZ

∴ ∆APB ~ ∆XQY based on AA similarity criterion

From above, we get

$\frac{\mathrm{AP}}{\mathrm{XQ}}$ = $\frac{\mathrm{AB}}{\mathrm{XY}}$ ------------ (4)

Since ∆ABC ~ ∆XYZ, $\frac{\mathrm{AB}}{\mathrm{XY}}$ = $\frac{\mathrm{BC}}{\mathrm{YZ}}$ = $\frac{\mathrm{AC}}{\mathrm{XZ}}$ -----------(5)

Using (4) and (5), we get

$\frac{\mathrm{AB}}{\mathrm{XY}}$ = $\frac{\mathrm{BC}}{\mathrm{YZ}}$ = $\frac{\mathrm{AC}}{\mathrm{XZ}}$ = $\frac{\mathrm{AP}}{\mathrm{XQ}}$ -----------------------------(6)

Substituting (6) in (3), we get

$\frac{(\mathrm{Area}\mathrm{of}△\mathrm{ABC})}{(\mathrm{Area}\mathrm{of}△\mathrm{XYZ})}$ = $\frac{(\mathrm{BC}\times \mathrm{AP})}{(\mathrm{YZ}\times \mathrm{XQ})}$ = $\frac{\mathrm{BC}}{\mathrm{YZ}}$ × $\frac{\mathrm{BC}}{\mathrm{YZ}}$= ${\left(\frac{\mathrm{BC}}{\mathrm{YZ}}\right)}^2$

From (6), we get

$\frac{(\mathrm{Area}\mathrm{of}△\mathrm{ABC})}{(\mathrm{Area}\mathrm{of}△\mathrm{XYZ})}$= ${\left(\frac{\mathrm{AB}}{\mathrm{XY}}\right)}^2$ = ${\left(\frac{\mathrm{BC}}{\mathrm{YZ}}\right)}^2$= ${\left(\frac{\mathrm{AC}}{\mathrm{XZ}}\right)}^2$

To prove: Area of ∆AEC = $\frac{1}{2}$ (Area of ∆BFC)

Let the length of each side of the square be ‘a’

∴AC = EC = EA = a (since all the sides of an equilateral triangle are equal)

BC = $\sqrt{2}$ a (since BC is the diagonal)

∴BC = CF = BF = $\sqrt{2}$ a

All the angles of an equilateral triangle are 60° and all the sides of an equilateral triangle are equal.

Hence, all equilateral triangles are similar.

∴ ∆AEC ~ ∆BFC

∴ $\frac{(\mathrm{Area}\mathrm{of}△\mathrm{AEC})}{(\mathrm{Area}\mathrm{of}△\mathrm{BFC})}$= ${\left(\frac{\mathrm{AC}}{\mathrm{BC}}\right)}^2$= ${\left(\frac{\mathrm{EA}}{\mathrm{CF}}\right)}^2$= ${\left(\frac{\mathrm{EC}}{\mathrm{BF}}\right)}^2$= ${\left(\frac{a}{\sqrt{2}a}\right)}^2$= $\frac{a^2}{2a^2}$= $\frac{1}{2}$

Hence, area of ∆AEC= $\frac{1}{2}$ × Area of ∆BFC

Therefore, the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.

1) Angle of elevation is measured at the foot of the tower and the building.

2) Apply the appropriate trigonometric ratio of the angle of elevations.

1) Draw a figure showing the angles of elevation to the top of the tower CD and the building AB from the foot of the building and the tower.

2) Apply tangent to the angle of elevation of the tower and find the length of BD.

3) Apply tangent to the angle of elevation of the building and find the length of AB.

4) Substitute the value of BD and find the value of AB which is equal to the height of the building.

Let the building be AB and the tower be CD.

Length of CD = 50 m

Consider ∆CBD

$\frac{\mathrm{CD}}{\mathrm{BD}}$= tan 60° = $\sqrt{3}$

$\frac{50m}{\mathrm{BD}}$= $\sqrt{3}$

∴BD = $\frac{(50m)}{\sqrt{3}}$

Consider ∆ABD

$\frac{\mathrm{AB}}{\mathrm{BD}}$ = tan 30° = $\frac{1}{\sqrt{3}}$

AB = $\frac{\mathrm{BD}}{\sqrt{3}}$ = $\frac{(50m)}{\sqrt{3}}$ × $\frac{1}{\sqrt{3}}$ = $\frac{50}{3}$ m = 16 $\frac{2}{3}$ m

Thus, the height of the building is 16 $\frac{2}{3}$ m.

1) Find the volume of the shell in terms of π and 'r'.

2) Find the volume of the cone in terms of π.

3) Avoid calculation mistakes.

OR

1) Find the slant height of the bucket using the radii and the vertical height.

2) The area of the metal sheet used in making the bucket is the sum of the curved surface area of the bucket and the area of the bottom circular end of the bucket.

1) Assume the inner radius of the shell as 'r' and find the volume of the shell in terms of 'r'.

2) Find the volume of the cone using the formula for the volume of the cone.

3) Equate the volume of the shell and the cone to find the value of 'r'.

OR

1) Find the height of a bucket in the form of a frustum of a cone using the volume and the radii of the top and bottom circular ends.

2) Find the slant height of the bucket using the vertical height and the radii.

3) Find the area of the metal sheet used in making the bucket by adding the CSA of the bucket and the area of the bottom circular end of the bucket.

Let ‘r’ be the inner radius and ‘R’ be the external radius of the copper shell.

Given that the external diameter of the copper shell = 2R = 18 cm

∴R = 9 cm

Volume of the spherical copper shell = $\frac{4}{3}$ × π × ($R^3-r^3)$cu units

= $\frac{4}{3}$ π($9^3-r^3)$

= $\frac{4}{3}$ π(729 − $r^3$) ${\mathrm{cm}}^3$

The spherical copper shell is made into a solid cone.

Hence, volume of solid cone = volume of spherical shell

Volume of cone = $\frac{1}{3}$ π${(\mathrm{radius}\mathrm{of}\mathrm{cone})}^2$(height of cone) = $\frac{1}{3}$ π${\left(14\right)}^2$(4$\frac{3}{7}$) ${\mathrm{cm}}^3$

= $\frac{1}{3}$ π × ${\left(14\right)}^2$× $\frac{31}{7}$

∴ $\frac{4}{3}$ π(729 − $r^3$) = $\frac{1}{3}$ π × ${\left(14\right)}^2$× $\frac{31}{7}$

4(729 − $r^3$) = ${\left(14\right)}^2$× $\frac{31}{7}$

729 − $r^3$= 7 × 31

$r^3$−512 = 0

$r^3$= 512

Hence, r = cube root of 512 = 8 cm

Therefore, inner radius of the spherical copper shell = 8 cm and inner diameter = 16 cm.

OR

Let the top radius of the bucket be ‘r’ and the bottom radius be ‘R’

r = 20 cm and R = 12 cm

Let the height of the bucket be ‘h’

Volume of the bucket = $\frac{1}{3}$ × π × ($r^2$ + $R^2$ + rR) × h = 12308.8 ${\mathrm{cm}}^3$

$\frac{1}{3}$ × π × ($r^2$ + $R^2$ + rR) × h = 12308.8

$\frac{1}{3}$ × 3.14 × (${20}^2$+ ${12}^2$+ 20 × 12) × h = 12308.8

784 h = $\frac{(12308.8\times 3)}{\left(3.14\right)}$

h = $\frac{11760}{784}$ = 15 cm

Let ‘s’ be the slant height of the frustum of the cone.

s = $\sqrt{\left[\right(h^2+{\left(r-R\right)}^2]}$=$\sqrt{[{15}^2+{\left(20-12\right)}^2]}$ = $\sqrt{\left(225+64\right)}$= $\sqrt{289}$= 17 cm

Area of the metal sheet used in making the bucket

= CSA of the bucket + Area of the bottom circular end of the bucket

= π(r + R)s + π$R^2$$$sq units

= 3.14 × (20 + 12) × 17 + (3.14 ×${12}^2$) $$

= 3.14 × 32 × 17 + (3.14 ×144) $$

= 3.14(544 + 144) $$

= 3.14 × 688$$

= 2160.32 ${\mathrm{cm}}^2$

Therefore, area of the metal sheet used in making the bucket is 2160.32 ${\mathrm{cm}}^2$.

1) Model class is the class with highest frequency.

2) Median class is the class in which the middle value of the sum of the frequencies ($\frac{N}{2}$) falls.

3) The class marks for each interval can be calculated by using the formula: $\frac{\mathrm{upper}\mathrm{class}lim\mathrm{it}+\mathrm{lower}\mathrm{class}lim\mathrm{it}}{2}$.

1) Find the model class of the data which has the highest number of students.

2) Find the mode using the formula: l + $\frac{(f_1-f_0)}{(2f_1-f_0-f_2)}$ × h, where l = lower class limit of modal class,

$f_1$= frequency of modal class, $f_0$= frequency of the class preceding the modal class, $f_2$= frequency of the class succeeding the modal class, h = class size.

3) Median class is the class in which the middle value of the sum of the frequencies ($\frac{N}{2}$) falls.

4) Find the median using the formula: l + $\frac{(\frac{n}{2}-\mathrm{cf})}{f}$× h, where l = lower limit of median class, h = class size, f = frequency of median class, cf = cumulative frequency of the class preceding median class.

3) Find the mean of the data using the formula: Mean = a + $\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)$× h, where a = assumed mean of the data, $f_i$ = frequencies, $u_i$=$\frac{d_i}{h}$ and $d_i$= ($x_i$- a) and $x_i$= $\frac{\left(\mathrm{upper}\mathrm{class}lim\mathrm{it}+\mathrm{lower}\mathrm{class}lim\mathrm{it}\right)}{2}$.

(a) Mode

The maximum frequency is 33 and the class having the maximum frequency is 45−55.

∴ modal class = 45−55.

Lower class limit (l) of modal class = 45

Frequency ( $f_1$) of the modal class = 33

Frequency ( $f_0$) of the class preceding the modal class = 31

Frequency ( $f_2$) of the class succeeding the modal class = 17

Class size (h) = 10

Mode = l + $\frac{(f_1-f_0)}{(2f_1-f_0-f_2)}\times h$

= 45 + $\frac{(33-31)}{\left[2\right(33)-31-17]}\times 10$

= 45 + $\frac{2}{(66-48)}\times 10$

= 45 + $\frac{2}{18}$× 10

= 45 + $\frac{10}{9}$

= 45 + 1.11

= 46.11(approximately)

Therefore, mode of the data is 46.11

(b) Median

Calculate the cumulate frequency of the data in another column:-

We obtain 'n' as 100

$\frac{n}{2}$ = 50

Cumulative frequency (cf) just greater than 50 = 71

71 falls under the class 45-55. Hence median class = 45-55.

Lower limit (l) of median class = 45

Class size (h) = 10

Frequency (f) of median class = 33

Cumulative frequency (cf) of class preceding median class = 38

Median = l + $\frac{(\frac{n}{2}-cf)}{f}\times h$ = 45 + $\frac{(50-38)}{33}\times 10$ = 45 + $\frac{40}{11}$

= 45 + 3.63

= 48.63 (approximately)

Therefore, median of the data is 48.63 (approximately)

(c) Mean

The class marks for each interval can be calculated by using the formula.

$x_i$= $\frac{(\mathrm{Upper}\mathrm{class}lim\mathrm{it}+\mathrm{Lower}\mathrm{class}lim\mathrm{it})}{2}$

Considering 60 as assumed mean (a), $d_i$, $u_i$ and $f_i{u}_i$ can be calculated as follows:

Mean = a + $\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)$ × h

= 60 + $\left(\frac{-103}{100}\right)$ × 10 = 60 - $\frac{103}{10}$ = 60 - 10.3 = 49.7

Therefore, mean of the data is 49.7

∴ Mode = 46.11, median = 48.63 and mean = 49.7