CBSE Class X

Question 2:

Write first four terms of the A.P. when the first term a and the common difference d are given as follows

(i) a = 10, d = 10                     (ii) a = − 2, d = 0                            (iii) a = 4, d = − 3

(iv) a = − 1 d = 1/2                (v) a = − 1.25, d = − 0.25

Answer:

(i) a = 10, d = 10

Let the series be a₁, a₂, a₃, a₄, a₅ …

a₁ = a = 10

a₂ = a₁ + d = 10 + 10 = 20

a₃ = a₂ + d = 20 + 10 = 30

a₄ = a₃ + d = 30 + 10 = 40

a₅ = a₄ + d = 40 + 10 = 50

Therefore, the series will be 10, 20, 30, 40, 50 …

First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) a = −2, d = 0

Let the series be a₁, a₂, a₃, a₄ …

a₁ = a = −2 

a₂ = a₁ + d = − 2 + 0 = −2

a₃ = a₂ + d = − 2 + 0 = −2

a₄ = a₃ + d = − 2 + 0 = −2

Therefore, the series will be −2, −2, −2, −2 …

First four terms of this A.P. will be −2, −2, −2 and −2.

(iii) a = 4, d = −3

Let the series be a₁, a₂, a₃, a₄ …

a₁ = a = 4

a₂ = a1 + d = 4 − 3 = 1

a₃ = a2 + d = 1 − 3 = −2

a₄ = a3 + d = − 2 − 3 = −5

Therefore, the series will be 4, 1, −2 −5 …

First four terms of this A.P. will be 4, 1, −2 and −5.

(iv) a = −1, d = 1/2

Let the series be a₁, a₂, a₃, a₄ …

(v) a = −1.25, d = −0.25

Let the series be a₁, a₂, a₃, a₄ …

a₁ = a = −1.25

a₂ = a₁ + d = − 1.25 − 0.25 = −1.50

a₃ = a₂ + d = − 1.50 − 0.25 = −1.75

a₄ = a₃ + d = − 1.75 − 0.25 = −2.00

Clearly, the series will be 1.25, −1.50, −1.75, −2.00 ……..

First four terms of this A.P. will be −1.25, −1.50, −1.75 and −2.00.

Question 1:

In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent
metre.

(iv)The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Answer:

(i) It can be observed that

Taxi fare for 1^st km = 15

Taxi fare for first 2 km = 15 + 8 = 23

Taxi fare for first 3 km = 23 + 8 = 31

Taxi fare for first 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.

(ii) Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time.

In other words, after every stroke, only 1 − (1/4) = 3/4 th  part of air will remain.

Therefore, volumes will be

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

(iii) Cost of digging for first metre = 150

Cost of digging for first 2 metres = 150 + 50 = 200

Cost of digging for first 3 metres = 200 + 50 = 250

Cost of digging for first 4 metres = 250 + 50 = 300

Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

(iv) We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be 

Therefore, after every year, our money will be

 Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

Question 5:

Is it possible to design a rectangular park of perimeter 80 and area 400 m²? If so find its length and breadth.

Answer

Let the length and breadth of the park be l and b.

Perimeter = 2 (l + b) = 80

l + b = 40

Or, b = 40 − l

Area = l × b = l (40 − l) = 40l − l²

40l − l² = 400

l² − 40l + 400 = 0

Comparing this equation with

al² + bl + c = 0, we obtain

a = 1, b = −40, c = 400

Discriminate = b² − 4ac = (− 40)² −4 (1) (400) = = 1600 − 1600 = 0

As b² − 4ac = 0,

Therefore, this equation has equal real roots. And hence, this situation is possible.

Root of this equation,

l =−(b/2a)

Therefore, length of park, l = 20 m

And breadth of park, b = 40 − l = 40 − 20 = 20 m

Question 4:

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the
product of their ages in years was 48.

Answer:

Let the age of one friend be x years.

Age of the other friend will be (20 − x) years.

4 years ago, age of 1^st friend = (x − 4) years

And, age of 2^nd friend = (20 − x − 4) = (16 − x) years

Given that,

(x − 4) (16 − x) = 48

16x − 64 − x² + 4x = 48

− x² + 20x − 112 = 0

x² − 20x + 112 = 0

Comparing this equation with ax² + bx + c = 0, we obtain a = 1, b = −20, c = 112

Discriminant = b² − 4ac = (− 20)² − 4 (1) (112) = 400 − 448 = −48

As b² − 4ac < 0,

Therefore, no real root is possible for this equation and hence, this situation is not possible.

Question 3:

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²?

If so, find its length and breadth.

Answer:

Let the breadth of mango grove be l.

Length of mango grove will be 2l.

Area of mango grove = (2l) (l) = = 2l²

2l² = 800

l² = 800/2 = 400

l² − 400 = 0

Comparing this equation with al² + bl + c = 0, we obtain a = 1 b = 0, c = 400

Discriminant = b² − 4ac = (0)² − 4 × (1) × (− 400) = 1600

Here, b² − 4ac > 0

Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

l = ±20

However, length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

Question 2:

Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(I) 2x² + kx + 3 = 0    (II) kx (x − 2) + 6 = 0

Answer:

We know that if an equation ax² + bx + c = 0 has two equal roots, its discriminant

(b² − 4ac) will be 0.

(I) 2x² + kx + 3 = 0

Comparing equation with ax² + bx + c = 0, we obtain a = 2, b = k, c = 3

Discriminant = b² − 4ac = (k)²− 4(2) (3) = k² − 24

For equal roots,

Discriminant = 0

k² − 24 = 0

k² = 24

k = ±√24 = ±2√6

(II) kx (x − 2) + 6 = 0

or kx² − 2kx + 6 = 0

Comparing this equation with ax² + bx + c = 0, we obtain a = k, b = −2k, c = 6

Discriminant = b² − 4ac = (− 2k)² − 4 (k) (6) = = 4k² − 24k

For equal roots,

b² − 4ac = 0

4k² − 24k = 0

4k (k − 6) = 0

Either 4k = 0 or k = 6 = 0

k = 0 or k = 6

However, if k = 0, then the equation will not have the terms ‘x²’ and ‘x’.

Therefore, if this equation has two equal roots, k should be 6 only.

Question 1:

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;

(I) 2x² −3x + 5 = 0                              (ii) 3x² – 4√3 x + 4 = 0

(III) 2x² − 6x + 3 = 0

Answer:

We know that for a quadratic equation ax² + bx + c = 0, discriminant is b² − 4ac.

(A) If b² − 4ac > 0 → two distinct real roots

(B) If b² − 4ac = 0 → two equal real roots

(C) If b² − 4ac < 0 → no real roots

(I) 2x² −3x + 5 = 0

Comparing this equation with ax² + bx + c = 0, we obtain a = 2, b = −3, c = 5

Discriminant = b² − 4ac = (− 3)2 − 4 (2) (5) = 9 − 40 = −31

As b² − 4ac < 0,

Therefore, no real root is possible for the given equation.

 (ii) 3x² – 4√3 x + 4 = 0

Comparing this equation with ax² + bx + c = 0, we obtain a = 3, b = −4√3, c = 4

Discriminant = b² − 4ac = (−4√3)² − 4(3) (4)

= 48 − 48 = 0

As b² − 4ac = 0,

Therefore, real roots exist for the given equation and they are equal to each other.

And the roots will be

(III) 2x² − 6x + 3 = 0

Comparing this equation with ax² + bx + c = 0, we obtain a = 2, b = −6, c = 3

Discriminant = b² − 4ac = (− 6)² − 4 (2) (3) = 36 − 24 = 12

As b² − 4ac > 0,

Therefore, distinct real roots exist for this equation as follows.

Question 11:

Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer:

Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively and their areas will be x² and y² respectively.

It is given that

4x − 4y = 24

x − y = 6

x = y + 6

However, side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m

Question 10:

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration
the time they stop at intermediate stations). If the average speeds of  the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Answer:

Let the average speed of passenger train be x km/h.

Average speed of express train = (x + 11) km/h

It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

Speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

Question 9:

Two water taps together can fill a tank in  The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer:

Let the time taken by the smaller pipe to fill the tank be x hr.

Time taken by the larger pipe = (x − 10) hr

Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 

It is given that the tank can be filled in   hours by both the pipes together. Therefore,

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours.  As in this case, the time taken by the larger pipe will be negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 − 10 =15 hours respectively.

Question 8:

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find
the speed of the train. of the train.

Answer:

Let the speed of the train be x km/hr.

Time taken to cover 360 km = 360/x hr

According to the given question,

However, speed cannot be negative.

Therefore, the speed of train is 40 km/h

Question 7:

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Answer:

Let the larger and smaller number be x and y respectively.

According to the given question,

x² – y² = 180 and y² = 8x

⇒ x² – 8x = 180

⇒ x² – 8x – 180 = 0

⇒ x² – 18x + 10x – 180 = 0

⇒x(x – 18) + 10(x – 18) = 0

⇒(x – 18) (x + 10) = 0

⇒x = 18, −10

However, the larger number cannot be negative as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be 18 only.

x = 18

Therefore, the numbers are 18 and 12 or 18 and −12.

Question 6:

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find
the sides of the field.

Answer:

Let the shorter side of the rectangle be x m.

Then, larger side of the rectangle = (x + 30) m

However, side cannot be negative. Therefore, the length of the shorter side will be 90 m.

Hence, length of the larger side will be (90 + 30) m = 120 m

Question 5:

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in
English, the product of their marks would have been 210. Find her marks in the two subjects.

Answer:

Let the marks in Maths be x.

Then, the marks in English will be 30 − x.

According to the given question,

(x + 2) (30 – x – 3) = 210

(x + 2) (27 – x) = 210

⇒–x² + 25x  + 54 = 210

⇒ x² – 25x + 156 = 0

⇒x² – 12x – 13x + 156 = 0

⇒x(x – 12) – 13 (x – 12) = 0

⇒(x – 12) (x – 13) = 0

⇒x = 12, 13

If the marks in Maths are 12, then marks in English will be 30 − 12 = 18

If the marks in Maths are 13, then marks in English will be 30 − 13 = 17

Question 4:

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Answer:

Let the present age of Rehman be x years.

Three years ago, his age was (x − 3) years.

Five years hence, his age will be (x + 5) years.

It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3.

However, age cannot be negative.

Therefore, Rehman’s present age is 7 years.

Question 3:

Find the roots of the following equations:

Answer:

 

Question 2:

Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Answer:

(iv) 2x² +x +4 = 0

On comparing this equation with ax² + bx + c = 0, we obtain

a = 2, b = 1, c = 4

By using quadratic formula, we obtain

However, the square of a number cannot be negative.

Therefore, there is no real root for the given equation.

Question 1:

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x² – 7x + 3 = 0                             (ii) 2x² + x – 4 = 0
(iii) 4x² + 4√3x + 3 = 0                     (iv) 2x² + x + 4 = 0

Answer:

However, the square of a number cannot be negative.

Therefore, there is no real root for the given equation.

Question 6:

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.

Answer:

Let the number of articles produced be x.

Therefore, cost of production of each article = Rs (2x + 3)

It is given that the total production is Rs 90.

∴ x(2x + 3) = 90

⇒ 2x² + 3x  – 90 = 0

⇒ 2x² + 15x – 12x  – 90 = 0

⇒x(2x + 15) –6(2x + 15) = 0

⇒(2x + 15) (x – 6) = 0

Either 2x + 15 = 0 or x − 6 = 0, i.e., x = –15/2 or x = 6

As the number of articles produced can only be a positive integer, therefore, x can only be 6.

Hence, number of articles produced = 6

Cost of each article = 2 × 6 + 3 = Rs 15

Question 5:

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer:

Let the base of the right triangle be x cm.

Its altitude = (x − 7) cm

From pythagoras theorem,

Base² + Altitude² = Hypotenuse²

∴ x² + (x – 7)² = 13²

⇒ x² + x² +49 – 14x = 169

⇒ 2x² – 14x – 120 = 0

⇒ x² – 7x – 60 = 0

⇒ x² – 12x + 5x – 60 = 0

⇒x(x – 12) +5(x – 12) = 0

⇒(x – 12) (x + 5) = 0

Either x − 12 = 0 or x + 5 = 0, i.e., x = 12 or x = −5

Since sides are positive, x can only be 12.

Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 − 7) cm = 5 cm.