Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 5 Second Degree Equations Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.
SSLC Maths Chapter 5 Second Degree Equations Important Questions and Answers
Second Degree Equations Class 10 Extra Questions Kerala Syllabus
Second Degree Equations Class 10 Kerala Syllabus Extra Questions
Question 1.
Sum of a number and its reciprocal is \(\frac{26}{5}\). The number is
(a) 5
(b) 1
(c) 2
(d) 6
Answer:
(a) 5
Question 2.
Solutions of the equation (x – 1)2 = 100 are
(a) 9, -10
(b) 10, -9
(c) 10, -10
(d) 11, -9
Answer:
(d) 11, -9
Question 3.
A small square is drawn inside a big square as shown in the figure.
The space between the squares is 2 and area of the inner square is 81
a) If the side of the outer square is x then write an equation.
Answer:
(x – 4)2 = 81
b) Find the side of the outer square.
Answer:
x – 4 = 9, – 9
x – 4 = 9, x = 13
Side of the outer square is 13 cm
Question 4.
One side of a rectangle is 4 more than the other side. Area of the rectangle is 140 sq.cm
a) If the small side is x then write the equation.
Answer:
(x + 4)x = 140
b) Find the sides of the rectangle.
Answer:
x(x + 4) = 140
x2 + 4x = 140, x2 + 4x + 4 = 144,
(x + 2)2 = 144, x + 2 = 12, x = 10
Sides are 10 cm and 14 cm
Question 5.
Two chords of a circle intersect at a point inside the circle.The intersecting point divides one chords into the segments of length 9 cm and 4 cm.The length of other chord is 15 cm
a) If one segment of the second chord of length 15 cm is x then what is the length of the other segment?
b) Form a second degree equation connecting length of the segments
c) Find the length of the segments on either side of the intersecting point on the second chord.
Answer:
a) 15 – x
b) x(15 – x) = 9 × 4, -x2 + 15x = 36, x2 – 15x = -36
c) Solving x = 12. Length of the segments are 12, 3
Question 6.
Sum of the area of two squares is 80.The sum of its perimetres is 48.
a) If the sides are x and y then what is x + y?
b) Form an equation representing the sum of the areas
c) Find the side of the the squares.
Answer:
a) Sides are x and y. 4x + 4y = 48, x + y = 12
b) Sides are x and 12 – x
x2 +(12 – x)2 = 80
x2 + 144 – 24x + x2 = 80
2x2 – 24x = -64,
x2 -12x = -32
c) x2 – 12x + 36 = -32 + 36,
(x – 6)2 = 4,
x – 6 = 2, x = 8 cm,
y = 12 – 8 = 4 cm
Question 7.
The perpendicular sides of a right triangle differ by 2. Area of the triangle is 24
a) Form an equation on x connecting the sides and area ?
b) Find the sides of the triangle
Answer:
a) Perpendicular sides are x and x + 2
\(\frac{1}{2}\) × x(x + 2) = 24, x2 + 2x = 48, x2 + 2x +1 = 49
b) (x + 1 )2 = 49, x + 1 = 7, x = 6
Perpendicular sides are 6 and 8.
Hypotenuse is \(\sqrt{6^2+8^2}\) = 10
Question 8.
Manju’s age after 15 years will be the square of his age before 15 years.
a) If the present age is x then what is the age before and after 15 years.
b) Write the equation connecting the given conditions.
c) Find the present age
Answer:
a) x – 15 and x + 15
b) (x – 15)2 = (x + 15), x2 -30x + 225
= x + 15, x2 -31x + 210 = 0
c) Solving x = 21. Present age is 21 years
Question 9.
Sum of the squares of three consecutive integers is 14. If the middle number is x then
a) What are the other numbers?
b) Form an equation
c) Find the numbers.
Answer:
a) x – 1, x + 1
b) (x – 1)2 + x2 + (x + 1)2 = 14
3x2 + 2 = 14, 3x2 = 12, x2 = 4, x = 2
c) Numbers are 1, 2, 3
Question 10.
A, B, C, D are the numbers on a calendar page as given below
a) If A = x write B,C and D
b) If A × C = 48 then form an equation
c) Find x
d) Find B, C and D
e) Can A + B+ C + D = 25 in any squares in the calendar? Justify your answer
Answer:
a) B = x + 1, C = x + 8, D = x + 1
b) x(x + 8) = 48, x2 + 8x = 48
c) (x + 4)2 = 64, x + 4 = 8, x = 4
d) Numbers are A = 4, B = 5, C = 12, D = 11
e) No. A + B + C + D always a multiple of 4.
Question 11.
Sum of a number and its square is 30. Which of the following is the number?
(a) -5
(b) 1
(c) 2
(d) -6
Answer:
(d) -6
Question 12.
x + √x = 6 in the form ax2 + bx + c = 0 is
(a) x2 + 12x + 6 = 0
(b) x2 – 13x + 36 = 0
(c) x2 – 10x – 12 = 0
(d) x2 + 13x – 36 = 0
Answer:
(b) x2 – 13x + 36 = 0
Question 13.
Sum of the areas of two squares is 130. Side of one square is 2 more than the side of the other square .
a) If the side of the small square is x then what is the side of the big square ?
b) Form a second degree equation using the condition.
Answer:
a) Side of the big square is x + 2
b) x2 + (x + 2)2 = 130
x2 + x2 + 4x + 4 = 130,
2x2 + 4x + 4 – 130 = 0,
2x2 + 4x – 126 = 0,
x2 + 2x – 63 =0
Question 14.
The product of two consecutive even numbers is 360
a) If the odd number in between these numbers is x then write the numbers .
b) Form an equation using the given condition.
c) Find the numbers.
Answer:
a) Numbers are x – 1, x + 1
b) (x – 1)(x + 1) = 360, x2 – 1 = 360 .
c) x2 = 361, x = \(\sqrt{361}\) = 19.
Numbers 19 – 1 = 18, 19 + 1 = 20
Question 15.
Consider the arithmetic sequence 5, 9, 13, 17, 21…
a) Write the algebraic form of this sequence.
b) What is the position of the term in the sequence whose square is 625?
c) Is 36 a term of this sequence. How can you realize it ?
d) What is the position of 49 in this sequence ?
Answer:
a) xn = dn + (f – d)
= 4n + (5 – 4)
= 4n + 1
b) (4n + 1)2 = 625,
4n + 1 = \(\sqrt{625}\) = 25,
4n = 24,
n = 6
c) All terms are odd numbers. The even number 36cannot be a term of this sequence
d) 4n + 1 = 49,
4n = 48,
n = 12.
12th term is 49
Question 16.
Hypotenuse of a right angled triangle is 1 less than twice its small side.Third side is 1 more than its small side
a) If the small side is x what is the length of other two sides.
b) Form an equation connecting the length of the sides.
c) Calculate the length of the sides of the triangle.
Answer:
a) Hypotenuse = 2x – 1, Third side = x + 1
b) (2x – 1)2 = x2 + (x + 1)2,
4x2 – 4x + 1 = x2 + x2 + 2x + 1
2x2 – 6x = 0
c) x = 3. Sides are :
Hypotenuse, 2x – 1 = 6 – 1 = 5 cm.
Other two sides are 3cm, 4cm.
Question 17.
Length of a rectangle is 4 more than its breadth. Area of the rectangle is 357 sq.cm
a) If the breadth is x then what is its length?
b) Write an equation connecting length, breadth and area.
c) Find the lenghth and breadth of the rectangle.
Answer:
a) Length x + 4
b) x(x + 4) = 357, x2 + 4x = 357
c) x2 + 4x + 4 = 357 + 4 = 361,
(x + 2)2 = 361,
x + 2 = \(\sqrt{361}\) = 19,
x = 19 – 2 = 17
Breadth 17cm, length 17 + 4 = 21 cm
Question 18.
Sum of the areas of two squares is 468sq.cm.The difference between the perimetres is 24cm.
a) If the small side is x then what is the length of the big side ?
b) What is the perimetre of the big square?
c) Write the length of the sides the squares in x
d) Form a second degree equation and find the length of the small square.
e) Find the length of the big square.
Answer:
a) Length of the big side be y, difference between the perimeter,
4y – 4x = 24,
4 (x – y) = 24,
x – y = 6,
y = x +6
b) 4x + 24
c) Side of the small square is x, Side of the big square is \(\frac{4 x+24}{4}\) = x + 6
d) x2 +(x + 6)2 = 468,
x2 + x2 + 12x + 36 = 468,
2x2 +12x = 432
x2 + 6x = 216,
x2 + 6x + 9 = 225,
(x + 3)2 = 225
(x + 3) = 15, x = 12
Side of the small square is 12 cm
e) Length of the big square is 12 + 6 = 18 cm
Question 19.
A two digit number is 4 times the sum of the digits. Also the number is 3 times the product of the digits.
a) Form an equation by taking x, y as the digits.
b) Make a second degree equation using the given condition.
c) Find the numbers.
Answer:
a) Digit in tens place x, digit in one’s place y Number is 10x + y
10x + y = 4(x + y)
10x + y = 3xy
b) 10x + y = 4x + 4y,
6x = 3y, y = 2x
10x + y = 3xy
⇒ 10x + 2x = 3x × 2x
12x = 6x2
6x2 – 12x = 0
c) x = 0, x =2.
Tens place cannot be 0.
Tens place = 2,
one’s place 2x = 4
Number = 24
Question 20.
A number whose square is decreased by 119 equal to 10 times 8 less than that number.
a) If x is the number then write the equation
b) Find the number.
Answer:
a) x2 – 119 = (x – 8) x 10
b) x2 – 10x = 119 – 80, x2 – 10x = 39 x2 – 10x + 25 = 64
⇒ (x – 5)2 = 64
x – 5 = 8 or x – 5 = -8
x = 13, x = – 3
Question 21.
A man is 5 times as old as his son. Sum of the squares of their ages is 2106
a) If x is son’s age then write the equation
b) Find the age of man and son?
Answer:
a) x2 + (5x)2 = 2106
x2 + 25x2 = 2106,
26x2 = 2106, x2 = 81,
x = 9
b) Son’s age 9 years , mans age 45years
Question 22.
Sum of a number and its square 9 times the next highest number.
a) If x is the number then write the equation
b) Find the number
Answer:
a) Let x be the number. x2 + x = 9 (x + 1)
x2 + x – 9x = 9,
x2 – 8x = 9
x2 – 8x + 16 = 9+ 16
⇒ (x – 4)2 = 25
b) x – 4 = 5 or -5, x = 9,-1
Question 23.
Length of a rectangle is 8 more then its breadth.
a) If the breadth is x then what is its length?
b) If the area is 240 sq.cm form a second degree equation.
c) Calculate the length and breadth
Answer:
a) x + 8
b) x(x + 8) = 240
c) x2 + 8x = 240,
x2 + 8x + 16 = 240 + 16
(x + 4)2 = 256, x + 4 = 16, x = 12
Sides are 12cm, 20cm
Question 24.
Consider the sequence of even numbers 2, 4, 6, 8
a) What is the algebraic form for the sum of first n even numbers?
b) How many terms from the beginning in the or-der makes the sum 210?
Answer:
a) This is the sequence of even numbers. Sum of the first n even numbers is n(n + 1)
b) n(n+ 1) = 210 × n2 + n = 210,
n2 + n + \(\frac{1}{4}\) = 210 + \(\frac{1}{4}\)
(n + \(\frac{1}{2}\))2 = 210 + \(\frac{1}{4}\)
(n + \(\frac{1}{2}\))2 = \(\frac{841}{4}\),
n + \(\frac{1}{2}\) = \(\frac{29}{2}\),
n = 14
Sum of first 14 terms is 210
Question 25.
The sum of the squares of two consecutive natural numbers is 313.
a) If one number is x then what is the other?
b) Form an equation using this condition.
c) Find the numbers.
Answer:
a) x + 1
b) x2 +(x + 1 )2 – 313
x2 + x2 + 2x + 1 = 313
2x2 + 2x – 312 = 0
x2 + x – 156 = 0
c) x = 12, -13, Numbers are 12, 13