Class 9 SELINA Solutions Maths Chapter 29: Assertion-Reasoning Type Questions

Correct option: (c) Both A and R are true.

If the denominator of a rational number can be expressed as 2^m × 5ⁿ, where m and n both are whole numbers, the rational number has a terminating decimal expansion.

Since the denominator 400 can be expressed as 2⁴ × 5², the given rational number has a decimal expansion.

Hence, both assertion and reason are true.

Correct option: (b) A is false, R is true.

If x is an irrational number,  is also irrational.

Hence, the assertion is false.

The reciprocal of every irrational number is always irrational.

Hence, the reason is true.

Correct option: (a) A is true, R is false.

Hence, the assertion is true.

The sum of two irrational numbers need not be an irrational number.

Hence, the reason is false.

Correct option: (c) Both A and R are true

S.I. for one year = Rs. 600

Then, C.I. at the end of second year = C.I. for first year + Interest on it for one year

= Rs. 200 + 5% of Rs. 200

= Rs. 210

Hence, both the assertion and the reason are true.

Correct option: (d) Both A and R are false.

Let the sum = Rs. 12000

C.I. for 1^st year

And, amount of 1^st year = Rs. (12000 + 1200) = Rs. 13200

The sum for 2^nd year = Rs. 13200

C.I. for 2^nd year

And, amount of 2^nd year = Rs. (13200 + 1320) = Rs. 14520

The sum for 3^rd year = Rs. 14520

C.I. for 3^rd year

And, amount of 3^rd year = Rs. 14520 + 10% of Rs. 14520

Hence, the assertion is false.

Given, amount at the end of five years, which is not possible.

Hence, the reason is false.

Correct option: (d) Both A and R are false

The sum or the Principal amount is greater than the compound interest.

Hence, the assertion is false.

The sum = Final Amount – Compound Interest

Hence, the reason is false.

Correct option: (a) A is true, R is false.

Let the sum = P

C.I. for 1^st year

And, amount of 1^st year = P + 0.1P = Rs. 1.1P

The sum for 2^nd year = Rs. 1.1P

C.I. for 2^nd year

And, amount of 2^nd year = 1.1P + 0.11P = Rs. 1.21P = Rs. 1.21 × the sum lent

Hence, the assertion is true.

C.I. for second year = Rs. 0.11P = Rs. 0.11 × the sum lent

Hence, the reason is false.

Correct option: (d) Both A and R are false

The amount on Rs. 4000 at 10% p.a compounded annually for years 

Hence, the assertion is false.

When the total time is not complete number of years, for example time = m years and n months, and rate is r% p.a. (compound annually), then

Hence, the reason is false.

Correct option: (d) Both A and R are false.

When the interest is compounded yearly, the amount is given by

, where P = Principal or sum, r = rate of interest compounded yearly, n = number of years

Then, for A = Rs. 5000, r = 10%, n = 2 years

Hence, the assertion is false.

Hence, the reason is false.

Correct option: (b) A is false, R is true.

When the interest is compounded yearly, the amount is given by

, where P = Principal or sum, r = rate of interest compounded yearly, n = number of years

Then, for A = Rs. x, r = 8%, n = 5 years

Then, sum lent

Hence, the assertion is false.

The equation given in reason (R) is correct.

Hence, the reason is true.

Correct option: (c)

(2x)³ – (3y)³ = (2x – 3y)³ + 3(2x)(3y)(2x – 3y)

⇒ 8x³ – 27y³ = 10³ + 18(xy)(10)

⇒ –27y³ + 8x³ = 1000 + 180(16) = 1000 + 2880 = 3880

Hence, the assertion is true.

(x – y)(x² + xy + y²) = x³ + x²y + xy² – x²y – xy² – y³

= x³ – y³

Hence, the reason is true.

Correct option: (d) Both A and R are false.

Hence, both the assertion and the reason are false.

Correct option: (b) A is false R is true.

a² – 8a – 1 = 0

Hence, the assertion is false.

The statement given in reason (R) is correct.

Hence, the reason is true.

Correct option: (a) A is true, R is false.

If a + b + c = 0, then a³ + b³ + c³ = 3abc

Hence, the reason is false.

Now, a + b + c = 2 + (–6) + 4 = 0

Hence, the assertion is true.

Correct option: (b) A is false, R is true.

5x³(x² – 4) can be further factorised as 5x³(x – 2)(x + 2).

Hence, the assertion is false.

The factorisation shown in the reason (R) is correct.

Hence, the reason is true.

Correct option: (b) A is false, R is true.

+b(3y – 2x) = –b(2x – 3y)

But, +b(3y – 2x)² ≠ –b(2x – 3y)²

Hence, the assertion is false and the reason is true.

Correct option: (b) A is false, R is true.

a³ + b³ = (a + b)(a² – ab + b²) is correct.

Hence, the reason is true.

113³ + 15³ = (113 + 15)(113² – 113 × 15 + 15²) = (128)(11299)

Since 98 is not the factor of 113³ + 15³, it is not divisible by 98.

Hence, the assertion is false.

Correct option: (c) Both A and R are true

The statement given in reason (R) is correct.

Hence, the reason is true.

For, 3x² – 8x – 15,

a = 3, b = –8 and c = –15

Then, b² – 4ac = (–8)² – 4(3)(–15) = 64 + 180 = 244, which is not a perfect square.

So, 3x² – 8x – 15 is not factorisable.

Hence, the assertion is true.

Correct option: (d) Both A and R are false.

Speed of stream

Hence, the assertion is false.

If the speed of the boat in still water is x km/hr and the speed of the stream is y km/hr, then x – y = 30 and x + y = 50 ⇒ x = 40 and y = 10.

Hence, the reason is false.

Correct option: (b) A is false, R is true.

The subtraction shown in reason (R) is correct.

That is, 3x – 3y = 3(91 – 80)

Hence, the reason is true.

And, 3x + 3y ≠ 3(91 – 80)

Hence, the assertion is false.

Correct option: (c) Both A and R are true.

ax + by = c …..(i)

bx + ay = d ….(ii)

Adding (i) and (ii),

Subtracting (ii) from (i),

Hence, the reason is true.

For equations, 217x + 131y = 913 and 131x + 217y = 827,

Hence, the assertion is true.

Correct option: (b) A is false, R is true.

Hence, the assertion is false.

And,

Hence, the reason is true.

Correct option: (c) Both A and R are true.

25^{2025 – 2020 – 2 – 2 – 1} = 25^{2025 – 2025}

                                   = 25⁰

                                   = 1              (∵ any number raised to power zero is 1)       

Hence, both the assertion and reason are true.

Correct option: (c) Both A and R are true.

If a^x = a^y, then x = y.

⇒ 2x = 3y

⇒ 2x – 3y = 0

Hence, both the assertion and the reason are true.

Correct option: (d) Both A and R are false.

Hence, both the assertion and the reason are false.

Correct option: (a) A is true, R is false.

Hence, the assertion is true.

 where a, b and x all are positive.

Hence, the reason is false.

Correct option: (b) A is false, R is true.

log_x(m × n × p) = log_xm + log_xn + log p is incorrect as log p = log₁₀ p

Hence, the assertion is false.

The statement given in reason (R) is correct.

Hence, the reason is true.

Correct option: (c) Both A and R are true.

Hence, both assertion and reason are true.

Correct option: (d) Both A and R are false.

log₁₀ 5x = log₁₀ x + log₁₀5 ≠ log₁₀ x × log₁₀ 5

Hence, both assertion and reason are false.

Correct option: (c) Both A and R are true.

In ∆AOB and ∆COB,

OA = OC                   (Given)

∠AOB = ∠COB       (Right angle)

OB = OB                   (Common side)

∴ ∆AOB ≌ ∆COB     (SAS congruence)

Hence, the assertion is true.

The congruency rule for two right-angled triangles given in reason (R) is correct.

Hence, the reason is true.

Correct option: (c) Both A and R are true.

The statements given in both assertion (A) and reason (R) are correct.

Hence, both assertion and reason are true.

Correct option: (a) A is true, R is false.

Since the opposite sides of a quadrilateral ABCD are parallel, it is a parallelogram.

In ∆ADN and ∆CBM,

∠NAD = ∠MCB       (alternate angles)

AD = BC                    (Opposite sides of a parallelogram ABCD)

∠ADN = ∠CBM  (since ∠ADC = ∠ABC, opposite angles of a parallelogram)

∴ ∆ADN ≌ ∆CBM     (ASA congruence)

⇒ AN = CM              (CPCT))

⇒ AN + MN = CM + MN

⇒ AM = CN

Hence, the assertion is true.

Based on ∠NDC = ∠MBC, it cannot be proved that AM = CN.

Hence, the reason is false.

Correct option: (c) Both A and R are true.

In ∆BEC and ∆CDB,

∠BEC = ∠CDB       (right angle)

∠EBC = ∠DCB        (Given)

BC = BC                    (Common side)

∴ ∆BEC ≌ ∆CDB    (AAS congruence)

⇒ CE = BD             (CPCT)

Hence, both assertion and reason are true.

Correct option: (b) A is false, R is true.

In ∆PQR and ∆PSR,

PQ = PS     (Given)

RQ = RS     (Given)

PR = PR     (Common)

∴ ∆PQR ≌ ∆PSR      (SSS congruence)

⇒ ∠PQR = ∠PSR     (CPCT)

Now, the opposite angles of a kite are not supplementary.

⇒ ∠PQR + ∠PSR ≠ 180° ⇒ ∠PQR = PSR ≠ 90°

Hence, the assertion is false and the reason is true.

Correct option: (b) A is false, R is true.

Given, AD = CD

⇒ ∠CAD = ∠ACD = 40°    (Angles opposite to equal sides are equal)

Also, AD = BD

⇒ ∠BAD = ∠ABD = x   (Angles opposite to equal sides are equal)

Now, in ∆ABC, by angle sum property,

∠A + ∠B + ∠C = 180°

⇒ (x + 40°) + x + 40° = 180°

⇒ x = 50°

Hence, the assertion is false.

The value of x obtained in reason (R) is correct.

Hence, the reason is true.

Correct option: (c) Both A and R are true.

By linear pair axiom, ∠A = 75°, ∠B = 45° and ∠C = 60°.

In a triangle, the side opposite to the greater angle is longer than the side opposite to the smaller angle.

Since ∠A > ∠C > ∠B,

⇒ BC > AB > AC

Hence, both assertion and reason are true.

Correct option: (c) Both A and R are true.

The sum of the lengths of any two sides of a triangle is always greater than the third side.

Here, AB + BC = 8 cm + 12 cm = 20 cm and AC = 25 cm

⇒ AB + BC ≯ AC

Thus, a triangle cannot be formed with these points.

Hence, both assertion and reason are true.

Correct option: (c) Both A and R are true.

Construction: Join BD.

In right-angled ∆BAD,

BD² = AB² + AD² = 16² + 12² = 400

⇒ BD = 20 cm

In ∆BCD, by Mid-point Theorem,

Hence, both the assertion and the reason are true.

Correct option: (b) A is false, R is true.

Since, ∠ADE = ∠ABC = 90°

⇒ DE ‖ BC and AB and AC are transversals.

⇒ CE : EA = 3 : 5

Hence, the assertion is false and the reason is true.

Correct option: (b) A is false, R is true.

In right-angled ∆ACP,

BC² = BP² – CP² = 13² – 12² = 25

Hence, the assertion is false.

The solution shown in reason (R) is correct.

Hence, the reason is true.

Correct option: (b) A is false, R is true.

ABCD is a rectangle.

⇒ AD = BC = 10 cm

PD = PC – CD = 15 – 5 = 10 cm

In right-angled ∆ADP,

AP² = AD² + PD² = 10² + 10² = 200

Hence, the assertion is false and the reason is true.

Correct option: (c) Both A and R are true.

Hence, both the assertion and the reason are true.

Correct option: (d) Both A and R are false.

Since ABCD is a rectangle ⇒ AB = DC.

And, DCEF is a parallelogram ⇒ DC = EF

Then, AB = EF.

But, for ABEF to be a rectangle, the other pair of opposite sides also has to be equal and each angle should be a right angle, which is not the case.

Therefore, ABEF is not a rectangle.

Hence, both the assertion and the reason are false.

Correct option: (c) Both A and R are true.

Construction: Join AC

∠BAX ∠A       (AX is the bisector of ∠A)

∠DCY ∠C       (CY is the bisector of ∠C)

⇒ ∠BAX = ∠DCY    (i)(since ∠A = ∠C, Opposite angles of a parallelogram)

Now, ∠BAC = ∠DCA    (ii)(Alternate angles)

Subtracting (ii) from (i),

∠CAX = ∠ACY

⇒ AX ‖ CY   (Alternate angles are equal)

Similarly, AY ‖ CX.

Therefore, AXCY is a parallelogram.

Hence, the assertion is true.

The statement given in the reason is correct.

Hence, the reason is true.

Correct option: (c) Both A and R are true.

Parallelogram ABEF and rectangle ABCD are on the same base AB and between the same parallels AB and CF.

Therefore, Area of parallelogram ABEF = Area of rectangle ABCD

Hence, the assertion is true.

The statement given in the reason is correct.

Hence, the reason is true.

Correct option: (c) Both A and R are true.

If the diagonals of a square are equal, the rhombus is a square.

Hence, the assertion is true.

Diagonals of a rhombus bisect each other at right angles.

The solution shown in the reason is correct.

As ∠ABC = 90° is proved, similarly, ∠BAD = ∠ADC = ∠BCD = 90° can be proved.

Hence, the reason is true.

Correct option: (c) Both A and R are true.

The statement given in the assertion for regular hexagon is correct.

Hence, the assertion is true.

The angle subtended by each side of a regular hexagon at the centre of its circumcircle is 60°.

⇒ OAB is an equilateral triangle

So, AB = OA = radius (r) of the circumcircle

Hence, the reason is true.

Correct option: (b) A is false, R is true.

The area of a triangle is half that of a parallelogram on the same base and between the same parallels.

Since square and ∆APB are on the same base AB and between the same parallels AB and DP, their areas cannot be equal.

Hence, the assertion is false.

The solution shown in the reason is correct.

Hence, the reason is true.

Correct option: (d) Both A and R are false.

The areas of triangles with the same vertex and bases along the same line are in the ratio of their bases.

∆ABD and ∆ABC have the same vertex A and their bases are along the same line.

∆AOB and ∆ABD have the same vertex B and their bases are along the same line.

Hence, both the assertion and the reason are false.

Correct option: (c) Both A and R are true.

arc AB : arc BC = 2 : 1

⇒ ∠AOB : ∠BOC = 2 : 1

Hence, the assertion is true.

The statement given in the reason is correct.

Hence, the reason is true.

Correct option: (a) A is true, R is false.

The statement given in assertion is correct.

Hence, the assertion is true.

The reason statement is incorrect as if the chords are parallel only, then the line through the mid-points of the chords will pass through the centre.

Hence, the reason is false.

Correct option: (c) Both A and R are true.

Arc APB : arc APBQC = 5 : 7

⇒ ∠AOB : reflex ∠AOC = 5 : 7

Hence, both the assertion and the reason are true.

Correct option: (d) Both A and R are false.

Two tangents drawn to a circle from an exterior point are equal in length.

Tangents AM and BN are drawn from different exterior points M and N respectively.

⇒ AM ≠ BN

Hence, both the assertion and the reason are false.

Correct option: (b) A is false, R is true.

For a class-interval 5-15,

Class-mark

Hence, the assertion is false.

The class-intervals obtained in reason, 0 – 10, 10 – 20 and 20 – 30 are correct.

Hence, the reason is true.

Correct option: (c) Both A and R are true.

Class size = 104 – 94 = 10

For class-mark 94,

For class-mark 104,

Hence, the assertion is true.

The equation shown in the reason is correct.

Hence, the reason is true.

Correct option: (b) A is false, R is true

Arranging data in ascending order: 16, 19, 25, 26, 28, 31, 32, 35

Number of observations = 8 (even)

Hence, the assertion is false.

The statement given in reason is correct.

Hence, the reason is true.

Correct option: (b) A is false, R is true.

The cumulative frequency of a class interval is the sum of frequencies of all classes up to that class.

Then,

Hence, the assertion is false and the reason is true.

Correct option: (a) A is true, R is false.

The median is the value of the middle observation when data is arranged in ascending or descending order.

Then, if each observation is increased by 7 and the result is multiplied by 3, the resulting median will be (42 + 7) × 3 = 147.

Hence, the assertion is true, and the reason is false.

Correct option: (a) A is true, R is false.

Mean of 5 observations = 30

Sum of 5 observations = 5 × 30

Mean of 4 observations = 31

Sum of 4 observations = 4 × 31

Then, the excluded observation = 5 × 30 – 4 × 31 = 150 – 124 = 26

Hence, the assertion is true and the reason is false.

Correct option: (c) Both A and R are true.

Let the radii of the circles be r₁ and r₂ respectively.

Then, π(r₁)² + π(r₂)² = 74π

⇒ (r₁)² + (r₂)² = 74

⇒ (r₁)² + (12 – r₁)² = 74

⇒ (r₁)² + 144 – 24r₁ + (r₁)² = 74

⇒ 2(r₁)² – 24r₁ + 70 = 0

⇒ (r₁)² – 12r₁ + 35 = 0

⇒ (r₁ – 7)(r₁ – 5) = 0

⇒ r₁ = 7 or r₁ = 5

⇒ r₂ = 5 or r₂ = 7

⇒ Difference between r₁ and r₂ = 2

Hence, the assertion is true.

The statement given in reason is correct.

Hence, the reason is true.

Correct option: (c) Both A and R are true.

Let the side of a square = x cm

⇒ Area of a square = x² sq. cm

New side of a square = (x + 5) cm

⇒ New area of a square = (x + 5)² sq. cm

Now, (x + 5)² – x² = 75

x² + 10x + 25 – x² = 75

10x = 50 ⇒ x = 5 cm

Hence, the assertion is true.

The area of a new square equals the sum of the area of the original square and 75 sq. cm. This gives the length of the side of a square as 5 cm.

Hence, the reason is true.

Correct option: (b) A is false, R is true.

Let the radius of the circle = r

Then, area of the circle = πr²

Now,

New radius = 2r

New area = π(2r)²

Hence, the assertion is false and the reason is true.

Correct option: (b) A is false, R is true.

The area of a parallelogram is the product of its base and the corresponding height. The height refers to the distance between the base and the side opposite to the base.

Hence, the assertion is false.

The formula for the area of a parallelogram given in reason is correct.

Hence, the reason is true.

Correct option: (b) A is false, R is true.

The volume of the cube whose edge measures ‘a’ cm = a³ cm³

If the ratio of the edges of three cubes = 2 : 3 : 4

Then, ratio of their volumes = 2³ : 3³ : 4³ = 8 : 27 : 64

Hence, the assertion is false and the reason is true.

Correct option: (b) A is false, R is true.

The volume of water that runs through the pipe in 2 minutes

= 21 cm² × 5km/h × 2minutes

Here, area, speed, and time are all in different units. To calculate volume, we must convert km/h to cm/s and minutes to seconds.

Hence, the assertion is false.

The conversion shown in reason (R) for calculating volume is correct.

Hence, the reason is true.

Correct option: (b) A is false, R is true.

In triangle ABC, A + B + C = 180°

⇒ A + B = 180° – C

Hence, the assertion is false.

sec (90° – θ) = cosec θ is correct.

Hence, the reason is true.

Correct option: (c) Both A and R are true.

4 cos θ = 11 sin θ

Hence, both the assertion and the reason are true.

Correct option: (c) Both A and R are true.

Then,  is correct.

Hence, both the assertion and the reason are true.

Correct option: (c) Both A and R are true.

Then, 5sin A = 4cos A

Hence, both the assertion and the reason are true.

Correct option: (c) Both A and R are true.

2 cos 3A = 1

Hence, both the assertion and the reason are true.

Correct option: (b) A is false, R is true.

∠A = 30°

⇒ tan²2A – sec²2A = tan²(2×30°) – sec²(2×30°)

= tan² 60° – sec² 60°

Hence, the assertion is false and the reason is true.

Correct option: (c) Both assertion and reason are true.

Hence, both the assertion and the reason are true.

Correct option: (d) Both A and R are false.

Area of a rhombus

Hence, both the assertion and the reason are false.

Correct option: (c) Both A and R are true.

5 cos 40°. cosec 50°

= 5 cos 40° × cosec (90° – 40°)

= 5 cos 40° × sec 40°

Hence, both the assertion and the reason are true.

Correct option: (b) A is false, R is true.

cosec (90° – 3A) = 1

⇒ cosec (90° – 3A) = cosec 90°

⇒ (90° – 3A) = 90°

⇒ A = 0°

Hence, the assertion is false and the reason is true.

Correct option: (c) Both A and R are true.

x = a is the equation of a line parallel to the y-axis and at a distance of ‘a’ units from it.

Therefore, the statements given in both the assertion and the reason are correct.

Hence, both the assertion and the reason are true.

Correct option: (a) A is true, R is false.

The statement given in the assertion is correct.

Hence, the assertion is true.

All points on the line y = a, the abscissa (x-coordinate) value = 0.

Hence, the reason is false.

Correct option: (b) A is false, R is true.

If ax + by + c = 0 is the general form of the equation of a line, then the coefficient of x is the slope of the given line.

Then, x – y = 8

⇒ y = x – 8

∴ Slope = coefficient of x = 1 is correct.

Hence, the assertion is false and the reason is true.

Correct option: (b) A is false, R is true.

(2x – 3y, 8) = (2, x + 2y)

⇒ 2x – 3y = 2  ….(i)   and x + 2y = 8   …..(ii)

Multiplying equation (i) by 2 and equation (ii) by 3,

4x – 6y = 4      ….(iii)

3x + 6y = 24  ….(iv)

Adding equations (iii) and (iv),

7x = 28 ⇒ x = 4

From (ii), 4 + 2y = 8 ⇒ y = 2

Hence, the assertion is false and the reason is true.

Correct option: (c) Both A and R are true.

The number of these articles manufactured and sold to the breakeven point indicates a no-profit, no-loss situation.

⇒ Cost price = Selling price

⇒ 5(30 + x) = 8x

⇒ 150 + 5x = 8x

⇒ x = 50

Hence, both the assertion and the reason are true.

Correct option: (b) A is false, R is true.

The value of ‘a’ is the y-coordinate.

The solution shown in the reason (R) is correct.

Hence, the assertion is false and the reason is true.

Correct option: (b) A is false, R is true.

AB = 4

⇒ 4x² = 16

⇒ x² = 4

⇒ x = ±2

⇒ B = (2, 0), (–2, 0)

Hence, the assertion is false and the reason is true.

Correct option: (d) Both A and R are false.

Hence, both the assertion and the reason are false.