ICSE Class 10 Mathematics Solved Paper 2026

(a) −1

Explanation:

Given, (x + 3), 1, (3x − 7) and −5 are in proportion.

\frac{(\mathrm{x}+3)}{1}=\frac{(3\mathrm{x}-7)}{-5}

⇒ −5(x + 3) = 3x − 7

⇒ −5x − 15 = 3x − 7

⇒ −8x = 8

⇒ x = −1

(c) 18%

Explanation:

Let the rate of GST be x%.

GST amount = \frac{\mathrm{Rate}}{100} × Marked price

⇒ 2160 = \frac{\mathrm{x}}{100} × 12000

⇒ 2160 = 120x

⇒ x = \frac{2160}{120} = 18

Therefore, the rate of GST is 18%.

(d) 9

Explanation:

The mode is the value that occurs most frequently in the data.

Here, 9 occurs 3 times, which is more frequent than any other number.

Therefore, the mode is 9.

(d) Both A and B are same (₹ 18,000)

Explanation:

For A:

Amount invested = ₹ 1,500 × 12 = ₹ 18,000

For B:

Amount invested = ₹ 1,200 × 15 = ₹ 18,000

Therefore, both A and B invested the same amount, ₹ 18,000.

(a) y = 6

Explanation:

A line parallel to the x-axis has slope m = 0 and the given y-intercept is c = 6.

Using y = mx + c:

y = 0x + 6

Therefore, y = 6.

(b) ₹ 15

Explanation:

Face value = ₹ 20

Dividend = 9%

Return = 12%

Dividend per share = \frac{9}{100} × 20 = ₹ 1.80

Market value = \frac{1.80\times100}{12} = ₹ 15

Therefore, she bought each share for ₹ 15.

(c) √3 : 2

Explanation:

Total surface area of sphere S₁ = 4πR²

Total surface area of solid hemisphere S₂ = 3πr²

Since the total surface areas are equal:

4πR² = 3πr²

⇒ 4R² = 3r²

⇒ \frac{R^2}{r^2}=\frac{3}{4}

⇒ \frac{\mathrm{R}}{\mathrm{r}}=\frac{\sqrt{3}}{2}

Therefore, R : r = √3 : 2.

(d) 130°

Explanation:

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Therefore, ∠ABC = ∠CDE = 65°.

The angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the remaining part of the circle.

∠AOC = 2 × ∠ABC

x = 2 × ∠ABC

⇒ x = 2 × 65° = 130°.

(c) Real, distinct and irrational

Explanation:

For 3x² − 6x − 3 = 0:

a = 3, b = −6 and c = −3

Discriminant, D = b² − 4ac

⇒ D = (−6)² − 4 × 3 × (−3)

⇒ D = 36 + 36 = 72

Since D > 0, the roots are real and distinct. Since 72 is not a perfect square, the roots are irrational.

(b) A is false and R is true.

Explanation:

The possible outcomes on throwing a die are {1, 2, 3, 4, 5, 6}. Therefore, the reason is true.

There is no outcome greater than 6.

Probability of getting a number greater than 6 = 0\over6 = 0.

Therefore, the assertion is false.

(a) 3 : 8

Explanation:

For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding altitudes.

\frac{\mathrm{Area}_1}{\mathrm{Area}_2}=\left(\frac{\mathrm{Altitude}_1}{\mathrm{Altitude}_2}\right)^2

⇒ \frac{9}{64}=\left(\frac{\mathrm{Altitude}_1}{\mathrm{Altitude}_2}\right)^2

Taking the positive square root:

⇒ \frac{\mathrm{Altitude}_1}{\mathrm{Altitude}_2}=\frac{3}{8}

Therefore, the ratio is 3 : 8.

(b) −16

Explanation:

Let P(x) = x³ + 7x² + 3x + 2.

By the Remainder Theorem, the remainder when P(x) is divided by x + 2 is P(−2).

P(−2) = (−2)³ + 7(−2)² + 3(−2) + 2

⇒ P(−2) = −8 + 28 − 6 + 2

⇒ P(−2) = 16

Therefore, −16 must be added to make the remainder zero.

(a) (x, y)

Explanation:

In a right-angled triangle, the midpoint of the hypotenuse is equidistant from all three vertices.

Since C is the midpoint of hypotenuse AB:

CA = CO = CB

The coordinates of C are given as (x, y).

Therefore, the point equidistant from A, O and B is (x, y).

(d) AB is not possible

Explanation:

The order of A is 2 × 2 and the order of B is 1 × 2.

For the product AB to exist, the number of columns of A must equal the number of rows of B.

Columns of A = 2, while rows of B = 1.

Since 2 ≠ 1, the product AB is not possible.

(c) Both A and R are true and R is the correct explanation of A.

Explanation:

Here, first term a = 6 and common ratio r = −2.

The nth term of a G.P. is Tₙ = arⁿ⁻¹.

T₉ = 6(−2)⁸

Since (−2)⁸ is positive, T₉ is positive.

Therefore, both A and R are true, and R correctly explains A.

Let the first term be a and the common difference be d.

The nth term of an A.P. is Tₙ = a + (n − 1)d.

For the 4th term:

a + 3d = 60 …(1)

For the 7th term:

a + 6d = 114 …(2)

Subtracting (1) from (2):

3d = 54

⇒ d = 18

Substituting d = 18 in (1):

a + 3(18) = 60

⇒ a + 54 = 60

⇒ a = 6

Therefore, the first term is 6 and the common difference is 18.

Now, Sₙ = \frac{\mathrm{n}}{2}[2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}]

S₁₀ = \frac{10}{2}[2(6)+(10-1)(18)]

⇒ S₁₀ = 5[12 + 162]

⇒ S₁₀ = 5 × 174

⇒ S₁₀ = 870

Therefore, the sum of the first 10 terms is 870.

Given:

A = \begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix} and B = \begin{bmatrix}-1 & \mathrm{a} \\ 3 & -5\end{bmatrix}

AB = \begin{bmatrix}3(-1)+1(3) & 3\mathrm{a}+1(-5) \\ 5(-1)+3(3) & 5\mathrm{a}+3(-5)\end{bmatrix}

⇒ AB = \begin{bmatrix}0 & 3\mathrm{a}-5 \\ 4 & 5\mathrm{a}-15\end{bmatrix}

But AB = \begin{bmatrix}\mathrm{b} & 7 \\ 4 & 5\end{bmatrix}

⇒ \begin{bmatrix}0 & 3\mathrm{a}-5 \\ 4 & 5\mathrm{a}-15\end{bmatrix} = \begin{bmatrix}\mathrm{b} & 7 \\ 4 & 5\end{bmatrix}

Comparing corresponding elements,

b = 0

and 3a − 5 = 7

⇒ 3a = 12

⇒ a = 4

Verification using the bottom-right element:

5a − 15 = 5

⇒ 5(4) − 15 = 5

⇒ 20 − 15 = 5

Therefore, a = 4 and b = 0.

Since AB is a diameter, the angle in a semicircle is a right angle.

Therefore, ∠ACB = 90°.

By the tangent-chord theorem:

∠DBC = ∠BAC = x

and ∠ABE = ∠BCA = y

Angles on the straight line DBE give:

x + 90° + y = 180°

⇒ x + y = 90° …(1)

In triangle ADB:

∠ADB + ∠DAB + ∠ABD = 180°

⇒ 32° + x + (x + 90°) = 180°

⇒ 2x + 122° = 180°

⇒ 2x = 58°

⇒ x = 29°

Substituting x = 29° in (1):

29° + y = 90°

⇒ y = 61°

Therefore, x = 29° and y = 61°.

Let P(x) = kx³ + 3x² − 11x − 6.

Since P(x) leaves a remainder of 6 when divided by x + 1, by the Remainder Theorem:

P(−1) = 6

⇒ k(−1)³ + 3(−1)² − 11(−1) − 6 = 6

⇒ −k + 3 + 11 − 6 = 6

⇒ −k + 8 = 6

⇒ −k = −2

⇒ k = 2

Therefore, P(x) = 2x³ + 3x² − 11x − 6.

Now, P(2) = 2(2³) + 3(2²) − 11(2) − 6

⇒ P(2) = 16 + 12 − 22 − 6 = 0

Therefore, x − 2 is a factor.

Dividing 2x³ + 3x² − 11x − 6 by x − 2 gives:

Now, 2x² + 7x + 3

= 2x² + 6x + x + 3

= 2x(x + 3) + 1(x + 3)

= (2x + 1)(x + 3)

Hence, the complete factorisation is:

2x³ + 3x² − 11x − 6 = (x − 2)(2x + 1)(x + 3).

Diameter = 7 cm

Therefore, radius r = 7/2 = 3.5 cm.

Height of the cylindrical part = 7 cm.

Height of the conical cap = 7 cm.

(a) From the given arrangement, the minimum height of the cylindrical box is:

7 + 7 = 14 cm

Therefore, the minimum height of the cylindrical box is 14 cm.

(b) The liquid is contained in the hemispherical and cylindrical parts.

Volume of hemisphere = \frac{2}{3}\pi \mathrm{r}^3

= \frac{2}{3}\times\frac{22}{7}\times(3.5)^3

= 89.83 cm³

Volume of cylinder = πr²h

= \frac{22}{7}\times(3.5)^2\times7

= 269.5 cm³

Total volume = 89.83 + 269.5

= 359.33 cm³

Therefore, the bottle can contain approximately 359 cm³ of liquid medicine.

Steps of construction :

1. Draw a line segment AB = 5 cm.
2. At point A with radius = 5 cm draw an arc.
3. At point B with radius = 5 cm draw another arc, cutting previous arc at point C.
4. Join AC and BC.
5. Construct perpendicular bisectors of AB and BC, let the bisectors meet at point O.
6. With O as centre and OA as radius draw a circumcircle.
7. Construct angle bisector of ∠ABC, mark the point as D where angle bisector intersects circumcircle.
8. Join AD and CD.

Since, points A, B, C and D lie on the circumcircle, thus ABCD is a cyclic quadrilateral.

L.H.S. (sec θ − cos θ)(cosec θ − sin θ)

= (\frac{1}{\cos \theta} − cos θ)(\frac{1}{\sin \theta} − sin θ)

= \frac{1-\cos^2 \theta}{\cos \theta} × \frac{1-\sin^2 \theta}{\sin \theta}

= \frac{\sin^2 \theta}{\cos \theta} × \frac{\cos^2 \theta}{\sin \theta}

= sin θ cos θ = R.H.S.

Hence, proved (sec θ − cos θ)(cosec θ − sin θ) = sin θ cos θ.

(a) Marked price = ₹ 24,000

Discount = 10%

Discount = \frac{10}{100} × ₹ 24,000
= ₹ 2,400

Discounted price = ₹ 24,000 − ₹ 2,400
= ₹ 21,600

(b) GST = 12% of ₹ 21,600

GST = \frac{12}{100} × ₹ 21,600

= ₹ 2,592

Amount paid by the customer = ₹ 21,600 + ₹ 2,592

= ₹ 24,192.

(a) In ΔADE and ΔABC:

∠ADE = ∠ABC (corresponding angles)

∠AED = ∠ACB (corresponding angles)

Therefore, ΔADE ~ ΔABC by AA similarity.

Given AD : DB = 2 : 3.

Let AD = 2x and DB = 3x.

Then AB = AD + DB = 5x.

\frac{\mathrm{DE}}{\mathrm{BC}}=\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{2\mathrm{x}}{5\mathrm{x}}=\frac{2}{5}

Therefore, DE : BC = 2 : 5.

(b) In ΔDFE and ΔCFB:

∠DFE = ∠CFB (vertically opposite angles)

∠DEF = ∠CBF (alternate interior angles)

Therefore, ΔDFE ~ ΔCFB by AA similarity.

(c) Let area of ΔCFB = x square units.

⇒ \frac{16}{\mathrm{x}}=\left(\frac{2}{5}\right)^2=\frac{4}{25}

⇒ 4x = 16 × 25

⇒ x = \frac{16\times25}{4} = 100

Therefore, area of ΔCFB = 100 square units.

(a) From the graph:

150–160 cm = 9 students

160–170 cm = 4 students

Number of students whose height is 150 cm and above = 9 + 4 = 13.

(b) The modal height indicated by the graph is 137 cm.

(c) Total number of students = 6 + 2 + 9 + 14 + 12 + 9 + 4 = 56.

(a) Let P(x, 0) divide AB internally in the ratio m : n.

Using the section formula for the y-coordinate,

y = \frac{\mathrm{m}\mathrm{y}_2+\mathrm{n}\mathrm{y}_1}{\mathrm{m}+\mathrm{n}}

Since P lies on the x-axis, y = 0.

0 = \frac{\mathrm{m}(10)+\mathrm{n}(-2)}{\mathrm{m}+\mathrm{n}}

⇒ 10m − 2n = 0

⇒ 10m = 2n

⇒ \frac{\mathrm{m}}{\mathrm{n}}=\frac{2}{10}=\frac{1}{5}

Therefore, P divides AB in the ratio 1 : 5.

(b) Using the section formula for the x-coordinate,

x = \frac{\mathrm{m}\mathrm{x}_2+\mathrm{n}\mathrm{x}_1}{\mathrm{m}+\mathrm{n}}

Here, m = 1, n = 5, x₁ = −10 and x₂ = 2.

x = \frac{1(2)+5(-10)}{1+5}

x = \frac{2-50}{6}

x = \frac{-48}{6} = −8

Since P lies on the x-axis, y = 0.

Therefore, P = (−8, 0).

⇒ (x − 2)² − 5x − 3 = 0

⇒ x² – 4x + 4 − 5x − 3 = 0

⇒ x² – 9x + 1 = 0

Comparing x² – 9x + 1 = 0 with ax² + bx + c = 0 , we get :

a = 1, b = -9 and c = 1

\mathrm{x}=\frac{-\mathrm{b}\pm\sqrt{\mathrm{b}^2-4\mathrm{ac}}}{2\mathrm{a}}

⇒ \mathrm{x}=\frac{9\pm\sqrt{77}}{2}

⇒ \mathrm{x}=\frac{9+8.775}{2} or \mathrm{x}=\frac{9-8.775}{2}

⇒ x = 8.8875 or x = 0.1125

Correct to 3 significant figures, x = 8.89 or x = 0.113.

(a) Market price per share = Nominal value + Premium

= ₹100 + ₹25 = ₹125

Money invested = 120 × ₹125 = ₹15,000.

(b) Let the rate of dividend be R%.

Dividend = Number of shares × \frac{\mathrm{R}}{100} × Nominal value

₹1,080 = 120 × \frac{\mathrm{R}}{100} × ₹100

1,080 = 120R

R = \frac{1080}{120} = 9

Rate of dividend = 9%.

(c) Rate of return = \frac{\text{Dividend income}}{\text{Investment}}\times100

= \frac{1080}{15000}\times100

= 7.2%.

Here, class size i = 8 and assumed mean A = 28.

Mean = A + \left(\frac{\sum \mathrm{ft}}{\sum \mathrm{f}}\right)\times \mathrm{i}

= 28 + \left(\frac{-22}{100}\right)\times8

= 28 + 8(-0.22)

= 28 – 1.76

= 26.24

Let the numbers be x and x + 5.

\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}+5}=\frac{3}{10}

⇒ \frac{2\mathrm{x}+5}{\mathrm{x}(\mathrm{x}+5)}=\frac{3}{10}

⇒ 10(2x + 5) = 3x(x + 5)

⇒ 20x + 50 = 3x² + 15x

⇒ 3x² − 5x − 50 = 0

⇒ 3x² + 10x − 15x − 50 = 0

⇒ x(3x + 10) − 5(3x + 10) = 0

⇒ (x − 5)(3x + 10) = 0

x = 5 or x = \frac{-10}{3}

Since x is natural, x = 5.

The other number = x + 5 = 10.

Therefore, the numbers are 5 and 10.

Let height of building (BC) = h m and height of flagpole (CD) = x m.

tan 31° = \frac{\mathrm{h}}{100}

h = 100 tan 31° = 60.09 m

tan 33° = \frac{\mathrm{h}+\mathrm{x}}{100}

h + x = 100 tan 33° = 64.94 m

x = 64.94 − 60.09

= 4.85 m

≈ 5 m

Therefore, the height of the flagpole is 5 m.

The cumulative-frequency table for the given continuous distribution is:

(a) Total number of students, N = 100

Median position = \frac{\mathrm{N}}{2} = \frac{100}{2} = 50

From the ogive, the median weight is approximately 56 kg.

(b) Cumulative frequency at 60 kg = 70

Number of students weighing 60 kg or more = 100 − 70 = 30

Percentage = \frac{30}{100}\times100 = 30%

Therefore, 30% of the students weigh 60 kg or more.

(c) Number of students above the required weight = 20% of 100 = 20

Therefore, the cumulative frequency below the required weight = 100 − 20 = 80.

From the ogive, the weight corresponding to cumulative frequency 80 is approximately 63 kg.

Therefore, the weight above which 20% of the students lie is approximately 63 kg.

(a) Time = 2 years = 24 months

Rate of interest = 8% per annum

Monthly deposit made by Vinay = ₹ 300

Interest = \frac{\mathrm{P}\times\mathrm{n}(\mathrm{n}+1)}{2}\times\frac{\mathrm{r}}{12\times100}

⇒ Interest = \frac{300\times24\times25}{2}\times\frac{8}{1200}

⇒ Interest = 300 × 300 × \frac{8}{1200}

⇒ Interest = ₹ 600

Therefore, interest earned by Vinay is ₹ 600.

(b) Interest earned by Rohit = ₹ 600 + ₹ 800 = ₹ 1400

Let Rohit deposit ₹ P per month.

⇒ 1400 = \frac{\mathrm{P}\times24\times25}{2}\times\frac{8}{1200}

⇒ 1400 = 2P

⇒ P = \frac{1400}{2} = ₹ 700

Therefore, Rohit deposits ₹ 700 per month.

For a G.P., nth term = arⁿ⁻¹.

Given,

Fourth term: ar³ = 16 …(1)

Seventh term: ar⁶ = 128 …(2)

(a) Dividing equation (2) by equation (1):

\frac{\mathrm{a}\mathrm{r}^{6}}{\mathrm{a}\mathrm{r}^{3}}=\frac{128}{16}

⇒ r³ = 8

⇒ r = ∛8 = 2

Therefore, the common ratio is 2.

(b) Substituting r = 2 in equation (1):

a(2)³ = 16

⇒ 8a = 16

⇒ a = \frac{16}{8} = 2

Therefore, the first term is 2.

From graph, on reflecting O, A, B, C on x-axis we get,

O(0, 0) ⇒ O(0, 0)

A(2, 3) ⇒ D(2, -3)

B(5, 3) ⇒ E(5, -3)

C(3, 0) ⇒ C(3, 0)

(b) From graph, on reflecting O, A, B, C trough origin we get,

O(0, 0) ⇒ O(0, 0)

A(2, 3) ⇒ I(-2, -3)

B(5, 3) ⇒ J(-5, -3)

C(3, 0) ⇒ H(-3, 0)

(c) From graph, on reflecting O, A, B, C on y-axis we get,

O(0, 0) ⇒ O(0, 0)

A(2, 3) ⇒ F(-2, 3)

B(5, 3) ⇒ G(-5, 3)

C(3, 0) ⇒ H(-3, 0)

−1 < \frac{2\mathrm{x}-3}{3} − \frac{\mathrm{x}}{5} ≤ 1

Taking LCM 15:

−1 < \frac{10\mathrm{x}-15-3\mathrm{x}}{15} ≤ 1

⇒ −1 < \frac{7\mathrm{x}-15}{15} ≤ 1

Solving the left-hand inequality:

−15 < 7x − 15

⇒ 0 < 7x

⇒ x > 0

Solving the right-hand inequality:

7x − 15 ≤ 15

⇒ 7x ≤ 30

⇒ x ≤ \frac{30}{7}

Therefore, 0 < x ≤ \frac{30}{7}.

Solution set = \left(0,\frac{30}{7}\right]

(a) From the graph:

A(4, 8), B(−1, 2) and C(6, 2).

(b) Mid-point of AC = \left(\frac{4+6}{2},\frac{8+2}{2}\right)

⇒ Mid-point of AC = (5, 5)

Slope of AB = \frac{8-2}{4-(-1)}

⇒ Slope of AB = \frac{6}{5}

The required line is parallel to AB, so its slope is also 6/5.

Using the point-slope form through (5, 5):

y − 5 = \frac{6}{5}(x − 5)

⇒ 5(y − 5) = 6(x − 5)

⇒ 5y − 25 = 6x − 30

⇒ 6x − 5y − 5 = 0

Therefore, the required equation is 6x − 5y − 5 = 0.

Given, radius r = 7 cm and height h = 24 cm.

Slant height of the cone:

l = √(r² + h²)

⇒ l = √(7² + 24²)

⇒ l = √(49 + 576)

⇒ l = √625 = 25 cm

(a) Total surface area of the sphere = 4πr²

⇒ 4 × \frac{22}{7} × 7² = 616 cm²

Total surface area of the cylinder = 2πr(h + r)

⇒ 2 × \frac{22}{7} × 7 × (24 + 7) = 1364 cm²

Total surface area of the cone = πr(r + l)

⇒ \frac{22}{7} × 7 × (7 + 25) = 704 cm²

Total surface area of the toy = 616 + 1364 + 704 = 2684 cm²

Therefore, the total surface area is 2684 cm².

(b) Cost of painting = 2684 × ₹ 0.50 = ₹ 1342

Therefore, the cost of painting is ₹ 1342.

Given, x = \frac{5\mathrm{a}\mathrm{b}}{\mathrm{a}-\mathrm{b}}

(a) Dividing both sides by a:

\frac{\mathrm{x}}{\mathrm{a}}=\frac{1}{\mathrm{a}}\left(\frac{5\mathrm{a}\mathrm{b}}{\mathrm{a}-\mathrm{b}}\right)

⇒ \frac{\mathrm{x}}{\mathrm{a}}=\frac{5\mathrm{b}}{\mathrm{a}-\mathrm{b}}

(b) Applying componendo and dividendo:

\frac{\mathrm{x}+\mathrm{a}}{\mathrm{x}-\mathrm{a}}=\frac{5\mathrm{b}+(\mathrm{a}-\mathrm{b})}{5\mathrm{b}-(\mathrm{a}-\mathrm{b})}

⇒ \frac{\mathrm{x}+\mathrm{a}}{\mathrm{x}-\mathrm{a}}=\frac{\mathrm{a}+4\mathrm{b}}{6\mathrm{b}-\mathrm{a}}

(a) Total number of families = 300

Let E be the event of selecting a family with one girl child.

Number of favourable outcomes = 165

P(E) = \frac{\mathrm{Number of favourable outcomes}}{\mathrm{Total number of outcomes}}

P(E) = \frac{165}{300}

P(E) = \frac{11}{20}

Hence, the probability of selecting a family with one girl child is \frac{11}{20}.

(b) Let A be the event of selecting a family with one or more girl child.

Number of favourable outcomes = 165 + 95 = 260

P(A) = \frac{260}{300}

P(A) = \frac{13}{15}

Hence, the probability of selecting a family with one or more girl child is \frac{13}{15}.

(c) Let B be the event of selecting a family with no boy child.

A family has no boy child when it has two girl children.

Number of favourable outcomes = 95

P(B) = \frac{95}{300}

P(B) = \frac{19}{60}

Hence, the probability of selecting a family with no boy child is \frac{19}{60}.

Given, ∠CBQ = 40°

In a circle, the angle between a tangent and a chord through the point of contact is equal to the angle in the opposite (alternate) segment of the circle.

∠BAC = ∠CBQ = 40°

x = 40°

Since, AB = AC.

∠ABC = ∠BCA [Angles opposite to equal sides of a triangle are equal]

In triangle ABC,

∠ABC + ∠BAC + ∠BCA = 180°

2∠ABC + 40° = 180°

2∠ABC = 180° – 40°

2∠ABC = 140°

∠ABC = 70°.

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∠BOC = 2∠BAC

y = 2x

y = 80°.

In triangle OBC,

OB = OC (Radii of same circle)

∠OBC = ∠OCB (Angles opposite to equal sides in a triangle are equal)

By angle sum property of triangle,

∠OBC + ∠OCB + ∠BOC = 180°

2∠OBC + 80° = 180°

2∠OBC = 100°

∠OBC = 50°

From figure,

w = ∠ABC – ∠OBC = 70° – 50° = 20°.

We know that,

Sum of opposite angles of a cyclic quadrilateral is 180°.

In cyclic quadrilateral ABCD,

∠ABC + ∠ADC = 180°

70° + z = 180°

z = 180° – 70° = 110°.

Hence, x = 40°, y = 80°, z = 110°, w = 20°.