Mathematics Part I Solutions Solutions for Class 10 Math Chapter 3 Second Degree Equations are provided here with simple step-by-step explanations. These solutions for Second Degree Equations are extremely popular among Class 10 students for Math Second Degree Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 10 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 58:

Question 1:

2000 rupees is invested is a scheme giving interest compounded annually. After two years the amount became 2205 rupees. What is the rate of interest?

Answer:

Amount, A = Rs.2205

Principal, P = Rs.2000

Time, t = 2 years

Let the rate of interest be R% per annum.

Thus, the rate of interest is 5% per annum.

Page No 58:

Question 2:

A pavement 2 metres wide runs around a square ground. The total area of the ground and the pavement is 1225 square metres. What is the area of the ground?

Answer:

Let the side of the square ground be x m.

Width of the pavement = 2 m

As the pavement is made around all the sides of the square ground, 4 m (= 2 m + 2 m) will be added to each side of the square ground.

Side of the square ground with pavement = (x + 4) m

Total area of the ground and pavement = 1225 m2

As the square garden with pavement is also square in shape, (x + 4)2 = 1225

x + 4 =

x + 4 = 35 m

x = (35 − 4) m

x = 31 m

Side of the square ground = 31 m

Area of the square ground = (31 m)2 = 961 m2

Page No 58:

Question 3:

A box is to be made by cutting off small squares from the corners of a square of thick paper and turning up the tabs.

The height of the box should be 5 centimetres and it should have a capacity of litre. What should be the length of a side of the square cut off? And the length of a side of the original square?

Answer:

The box is made from a square paper by cutting off small squares from each of the corners

Length of the box = Breadth of the box = a cm

Height of the box = 5 cm

Height of the box = Length of sides of the small squares which are cut off from the paper

Length of sides of the small squares = 5 cm

Capacity of the box =

We know that 1 L = 1000 cm3

Volume of the box =

Length × Breadth × Height = 500 cm3

a × a × 5 = 500 cm3

(a)2 = 100 

a =

As length cannot be negative, a = 10.

Length and breadth of the box = 10 cm

Length of the side of the original square = (10 + 5 + 5) cm = 20 cm



Page No 59:

Question 1:

The common difference of an arithmetic sequence is 1 and the product of the first and the third terms is 143. What are the first three terms?

Answer:

Common difference of an arithmetic sequence = 1

Let the second term be x.

First term = x 1

Third term = x + 1 

Product of the first term and the third term = 143

(x 1)(x + 1) = 143

If x = 12, then the second term = 12

First term = x 1 = 12 1 = 11

Third term = x + 1 = 12 + 1 = 13

If x = 12, then the second term = 12

First term = x 1 = 12 1 = 13

Third term = x + 1 = 12 + 1 = 11

Therefore, the first three terms of the sequence are 11, 12, and 13 or 13, 12, and 11.



Page No 65:

Question 1:

The length of a rectangle is 10 centimetres more than its breadth; and its area is 144 square centimetres. What are the length and breadth?

Answer:

Let the breadth of the rectangle be x cm.

Length of the rectangle = (x + 10) cm

Area of the rectangle = 144 cm2

Length × Breadth = 144 cm2

x(x + 10) = 144

x2 + 10x = 144

Adding 25 to both the sides to make L.H.S a perfect square:

x2 + 10x + 25 = 144 + 25

x2 + 2 × 5 × x + (5)2 = 169

(x + 5)2 = 169 {Using a2 + b2 + 2ab = (a + b)2}

As the side of a rectangle cannot be negative, the breadth of the rectangle is 8 cm.

Length of the rectangle = (8 + 10) cm = 18 cm

Page No 65:

Question 2:

How many terms of the arithmetic sequence 5, 7, 9, …. should be added to get 140?

Answer:

Let the terms of the sequence be x1, x2, x3, …

We know that the terms of an A.P. are of the form a + b, 2a + b, 3a + b, …

The given sequence is 5, 7, 9, …

First term, x1 = a + b = 5

Second term, x2 = 2a + b = 7

a + a + b = 7

a + 5 = 7

a = 7 − 5 = 2

b = 5 − a = 5 − 2 = 3

Let the sum of n terms of the given sequence be 140.

Let the nth term of the sequence be xn.

xn = an + b = 2n + 3

Sn = 140

Adding 4 to both the sides to make L.H.S a perfect square:

+ 4 = 140 + 4

{Using a2 + b2 + 2ab = (a + b)2}

Since n is a natural number, it cannot be negative.

n = 10

Thus, 12 terms of the given sequence should be added to get 140.



Page No 66:

Question 1:

An isosceles triangle as shown below is to be made:

The height should be 2 metres less than the base and the area should be 12 square metres. What should be the lengths of the sides.

Answer:

Let the base of the triangle be x m.

Height of the triangle = (x − 2) m

Area of the triangle = 12 m2

× Base × Height = 12 m2

Adding 1 to both the sides to make L.H.S a perfect square:

As length cannot be negative, x = 6 m.

Base of the triangle = 6 m

Height of the triangle = (6 − 2) m = 4 m

Page No 66:

Question 2:

From a rectangular sheet of paper, two squares are cut off as shown below:

The remaining rectangle should have area 24 squares centimetres. What should be the lengths of the sides of the squares?

Answer:

Let the length of the side of one square be x cm.

Length of the side of two squares = (x + x) cm = 2x cm

Length of the remaining rectangle = (16 − 2x) cm

Breadth of the remaining rectangle = Length of the side of square = x cm

Area of the remaining rectangle = 24 cm2 

Length × Breadth = 24 cm2 

(16 − 2x)x = 24

16x 2x2 = 24

8x x2 = 12

x2 8x = 12

Adding 16 to both the sides to make L.H.S a perfect square:

x2 8x + 16 = 12 + 16

x2 2 × 4 × x + (4)2 = 4

(x − 4)2 = 4 {Using a2 + b2 2ab = (a b)2}

Thus, the length of the sides of the squares can be 6 cm or 2 cm.

It can be observed from the figure that the length of the remaining rectangle should be less than the length of the side of the square. So, x cannot be equal to 2.

Hence, the length of the sides of the square should be 6 cm.

Page No 66:

Question 3:

A pole 2.6 metres high leans against a wall, with its fool 1 metre away from the wall.

When this distance way slightly increased, the top of the pole slide down the wall by the same distance. How far was the foot of the pole shifted?

Answer:

Length of the pole, x = 2.6 m

Distance of the foot of pole from the wall, y = 1 m

Let the length of the wall be z m.

By Pythagoras theorem:

z2 + y2 = x2 

z2 + (1)2 = (2.6)

z2 + 1 = 6.76

z2 = 5.76

As length cannot be negative, z = 2.4.

Length of the wall = 2.4 m

It is given that when the distance of the pole from the floor is slightly increased, then the top of the pole slide down the wall by the same distance.

Let the distance of the pole from the floor be increased by a m.

Distance of the foot of pole from the wall, y = (a + 1) m

Length of the wall = z a = (2.4 a) m

Again by Pythagoras theorem:

(2.4 − a)2 + (1 + a)2 = (2.6)2 

5.76 + a2 − 4.8a + 1 + a2 + 2a = 6.76 

2a2 − 2.8a = 6.76 5.76 1

2a2 − 2.8a = 0

a2 − 1.4a = 0

Adding 0.49 to both the sides to make L.H.S a perfect square:

a2 − 1.4a + 0.49 = 0.49

a2 − 2 × 0.7 × a + (0.7)2 = 0.49

(a − 0.7)2 = 0.49

As a cannot be 0, a = 1.4.

Thus, the foot of the pole is shifted by 1.4 m.



Page No 67:

Question 1:

The perimeter of a rectangle is 28 metres and its diagonal is 10 metres. What are the lengths of its sides?

Answer:

Let the length and breadth of the rectangle be l and b respectively.

Diagonal of the rectangle = 10 m

As the interior angles of a rectangle are 90°, the length, breadth and diagonal of the rectangle form a right-angled triangle.

Using Pythagoras theorem: 

l2 + b2 = 102 

l2 + b2 = 100 … (1)

Perimeter of the rectangle = 28 m

2(l + b) = 28

l + b = 14 … (2)

Squaring both sides:

(l + b)2 = (14)2

l2 + b2 + 2lb = 196

100 + 2lb = 196 (From (1))

2lb = 196 − 100 

2lb = 96

lb = 48

l =

Putting the value of l in equation (2):

+ b = 14

Adding 49 to both the sides to make L.H.S a perfect square:

b2 − 14b + 49 = 48 + 49

b2 − 2 × 7 × b + (7)2 = 1

(b − 7)2 = 1 {Using a2 + b2 2ab = (a b)2}

b − 7 =

b − 7 = 1

b = 7 1

b = 7 + 1 or b = 7 − 1

b = 8 or 6

Put the value of b in equation (2):

If b = 8 then l + 8 = 14 

l = 14 − 8 = 6 

If b = 6 then l + 6 = 14 

l = 14 − 6 = 8 

Thus, the length and the breadth of the rectangle are 8 m and 6 m respectively.

Page No 67:

Question 2:

Find a pair of numbers with sum 4 and product 2.

Answer:

Let the first number be x.

Sum of the two numbers = 4

Second number = 4 − x 

Product of the numbers = 2

x(4 − x) = 2

4x x2 = 2

x2 4x = 2

Adding 4 to both the sides to make L.H.S a perfect square:

x2 4x + 4 = 2 + 4

x2 2 × 2 × x + (2)2 = 2

(x − 2)2 = 2 {Using a2 + b2 2ab = (a b)2}

Page No 67:

Question 3:

The sum of a number and its reciprocals is. What are the numbers?

Answer:

Let the number be x.

Reciprocal of the number =

Sum of the number and its reciprocal =

Adding to both the sides to make L.H.S a perfect square:

Thus, the numbers are

Page No 67:

Question 4:

Thirty candies were distributed among some children. Relishing the sweet, the whiz-kid among them said, “If we were one short, we would all get one more.” How many kids were there?

Answer:

Let the number of kids present there be x.

Total number of candies = 30

Number of candies each kid will get =

According to the question: 

Adding to both sides to make L.H.S a perfect square:

As the number of kids cannot be negative, x = 6.

Thus, there was 6 kids.

Page No 67:

Question 5:

There are two taps opening into a tank. If both are opened, the tank would be full in 12 minutes. The time taken for it to fill with only the small tap open, is 10 minutes more than the time to fill with only the large tap open. What is the time taken to fill the tank with only the small tap open?

Answer:

Let the time taken by the large tap to fill the tank be x minutes.

Tank filled by the large tap in 1 minute =

Time taken by the small tap to fill the tank = (x + 10) minutes

Tank filled by the small tap in 1 minute =

Time taken by both the taps to fill the tank = 12 minutes

Tank filled by both the taps in 1 min =

Adding 49 to both the sides to make L.H.S a perfect square:

x2 + 14x + 49 = 120 + 49

x2 + 2 × 7 × x + (7)2 = 169

(x + 7)2 = 169 {Using a2 + b2 + 2ab = (a + b)2}

x + 7 =

x + 7 =

x = 7

x = 7 + 13 or 7 13

x = 20 or x =

As time cannot be in negative, x = 20.

Therefore, the time taken by the small tap to fill the tank is (20 + 10) minutes = 30 minutes



Page No 70:

Question 1:

A rectangle of perimeter 10.75 metres and area 5.8 square metres is to be made. What should be the lengths of its sides?

Answer:

Let the length and breadth of the rectangle be l and b respectively.

Perimeter of the rectangle = 10.75 m

2(l + b) = 10.75

l + b =

l + b = 5.375 … (1)

Area of the rectangle = 5.8 m2

lb = 5.8

Putting the value of b in equation (1):

Comparing the above equation with general quadratic equation:

a = 1, b = 5.375 and c = 5.8

From (1), we get:

b = 1.49 or 3.88

Thus, the length and the breadth of the rectangle are 3.88 m and 1.49 m respectively.

Page No 70:

Question 2:

The distance travelled by an object thrown upwards in t seconds is 30t − 4.9t2 metres. After how much time would it fall down? At what all times would it be 20 metres above the ground?

Answer:

Distance travelled by an object thrown upwards in t seconds = (30t − 4.9t2) m

When the thrown object comes back to the ground then the displacement becomes zero.

30t − 4.9t2 = 0

4.9t2 − 30t = 0

Comparing the above equation with general quadratic equation:

a = 4.9, b = 30 and c = 0

The time taken by the object to fall down = 6.12 seconds

If the distance is 20 m, then 30t − 4.9t2 = 20.

⇒ − 4.9t2 + 30t 20 = 0

4.9t2 − 30t + 20 = 0

Comparing the above equation with general quadratic equation:

a = 4.9, b = 30 and c = 20

Thus, the object would be 20 m above the ground after 0.76 seconds and 5.36 seconds.

Page No 70:

Question 3:

In each of the second degree polynomials given below, find the numbers to be taken as x to get zero:

(i) x2 − 5x + 6

(ii) x2 − 2x − 1

(iii) x2 + 5x + 6

(iv) 2x2 − 7x − 15

(v) x2 + x − 6

(vi) 9x2 + 12x + 4

(vii) x2x − 6

Answer:

(i)

The value of x obtained by equating the given polynomial to 0 is the required value of x.

Comparing the above equation with general quadratic equation: 

a = 1, b = 5 and c = 6


 

(ii)

The value of x obtained by equating the given polynomial to 0 is the required value of x.

Comparing the above equation with general quadratic equation: 

a = 1, b = 2 and c = 1


 

(iii)

The value of x obtained by equating the given polynomial to 0 is the required value of x.

Comparing the above equation with general quadratic equation: 

a = 1, b = 5 and c = 6


 

(iv)

The value of x obtained by equating the given polynomial to 0 is the required value of x.

Comparing the above equation with general quadratic equation: 

a = 2, b = 7 and c = 15


 

(v)

The value of x obtained by equating the given polynomial to 0 is the required value of x.

Comparing the above equation with general quadratic equation: 

a = 1, b = 1 and c = 6


 

(vi)

The value of x obtained by equating the given polynomial to 0 is the required value of x.

Comparing the above equation with general quadratic equation: 

a = 9, b = 12 and c = 4


 

(vii)

The value of x obtained by equating the given polynomial to 0 is the required value of x.

Comparing the above equation with general quadratic equation: 

a = 1, b = 1 and c = 6



Page No 72:

Question 1:

Can the sum of the first few consecutive terms of the arithmetic sequence 5, 7, 9, ….be 140? What about 240?

Answer:

Let the first term of the A.P. be a + b and the common difference be a.

The given sequence is 5, 7, 9, …

a = 7 − 5 = 2

a + b = 5

2 + b = 5

b = 5 − 2 = 3 

The general term of an A.P. is given by

Let the sum of n terms of the given sequence be 140.

Sum of n terms =

Comparing the above equation with general quadratic equation: 

a = 2, b = 8 and c = 280

Discriminant of the above equation is: 

b2 − 4ac = (8)2 − 4(2)(280

= 64 + 2240 

= 2304

As the discriminant is positive, the above equation has two solutions.

As n is the number of terms, n cannot be negative.

n = 10

Therefore, the sum of 10 consecutive terms of the given sequence is 140.

Now, let the sum of m terms of the given sequence be 240.

Sum of m terms =

Comparing the above equation with general quadratic equation:

a = 2, b = 8 and c = 480

Discriminant of the above equation is: 

b2 − 4ac = (8)2 − 4(2)(480

= 64 + 3840 

= 3904

As the discriminant is positive, the above equation has two solutions.

As n is the number of terms, n cannot be negative or in decimals.

Therefore, 240 cannot be the sum of the first few consecutive terms of the given sequence.

Page No 72:

Question 2:

From the polynomial p(x) = x2 + x + 1, do we get p(x) = 0 for any x? What about p(x) = 1? And p(x) = −1?

Answer:

p(x) = x2 + x + 1

p(x) = 0

x2 + x + 1 = 0

Comparing the above equation with general quadratic equation: 

a = 1, b = 1 and c = 1

Discriminant of the above equation is: 

b2 − 4ac = (1)2 − 4(1)(1) = 1 − 4 = 3

Since, the discriminant is negative, p(x) 0 for any x.

p(x) = 1

x2 + x + 1 = 1

x2 + x + 1 1 = 0

x2 + x + 0 = 0

Comparing the above equation with general quadratic equation: 

a = 1, b = 1 and c = 0

Discriminant of the above equation is: 

b2 − 4ac = (1)2 − 4(1)(0) 

= 1 − 0 

= 1

p(x) = 1 for x = 1 or x = 0.

Now, p(x) = 1

x2 + x + 1 = 1

x2 + x + 1 + 1 = 0

x2 + x + 2 = 0

Comparing the above equation with general quadratic equation: 

a = 1, b = 1 and c = 2

Discriminant of the above equation is: 

b2 − 4ac = (1)2 − 4(1)(2) 

= 1 − 8 

= 7

Since the discriminant is negative, p(x) 1 for any x.

Page No 72:

Question 3:

From the expression, do we get 0, 1 or 2 for some number x?

Answer:

Let p(x) =

Putting p(x) = 0:

Comparing the above equation with general quadratic equation: 

a = 1, b = 0 and c = 1

Discriminant of the above equation is: 

b2 − 4ac = (0)2 − 4(1)(1) 

= 0 − 4 

= 4

Since the discriminant is negative, p(x) 0 for any x.

Putting p(x) = 1:

Comparing the above equation with general quadratic equation: 

a = 1, b = 1 and c = 1

Discriminant of the above equation is: 

b2 − 4ac = (1)2 − 4(1)(1) 

= 1 − 4 

= 4

Since the discriminant is negative, p(x) 1 for any x.

Putting p(x) = 2:

Comparing the above equation with general quadratic equation: 

a = 1, b = 2 and c = 1

Discriminant of the above equation is: 

b2 − 4ac = (2)2 − 4(1)(1) 

= 4 − 4 

= 0

Since the discriminant is zero, the above equation has only one solution.

p(x) = 2 for x = 1

Page No 72:

Question 4:

Prove that if a, b, c are positive numbers and if the equation ax2 + bx + c = 0 has solutions then they are negative numbers.

Answer:

Given quadratic equation is , where a, b and c are positive numbers.

If then x will surely be a negative value.

Now, for

Let us suppose that

Either a is negative or c is negative.

However, it is given that both a and c are positive values.

Thus, our supposition was wrong.

Hence, the solutions of the given quadratic equation are negative numbers.



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