Telangana Board SSC Class 10 Maths 2016 Question with Solutions PART A
PART A
SECTION - I
1. Find the value of log5 125.
Solution:
Let log5 125 = x
⇒ 5x = 125
⇒ 5x = 53
⇒ x = 3
Therefore, log5 125 = 3
2. If A = {1, 1/4, 1/9, 1/16, 1/25}, then write A in set-builder form.
Solution:
A = {1, 1/4, 1/9, 1/16, 1/25}
= {1/1², 1/2², 1/3², 1/4², 1/5²}
The set-builder form: A = {1/x²: x ≤ 5, x ∈ N}
3. Write an example for a quadratic polynomial that has no zeroes.
Solution:
x² + x + 11 is one of the polynomials which do not have zeroes.
4. If b²- 4ac > 0 in ax²+ bx + c = 0, then what can you say about roots of the equation? (a ≠ 0)
Solution:
Given,
ax² + bx + c = 0
And
b²- 4ac > 0
Hence, the roots of the equation are real and unequal.
5. Find the sum of the first 200 natural numbers.
Solution:
First 200 natural numbers: 1, 2, 3, 4,...,200
n = 200
We know that,
Sum of first n natural numbers =
n(n + 1)/2
Sum of the first 200 natural numbers = 200(200 + 1)/2
= 100 × 201
= 20100
6. For what values of m, the pair of equations 3x + my = 10 and 9x + 12y = 30 have a unique solution.
Solution:
Given.
3x + my = 10
9x + 12y = 30
Comparing with the standard form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,
a1 = 3, b1 = m, c1 = -10
a2 = 9, b2 = 12, c2 = -30
Condition for unique solution
a1/a2 ≠ b1/b2
3/9 ≠ m/12
⇒ m ≠ 12/3
⇒ m ≠ 4
Hence, m takes all the real values except 4.
7. Find the midpoint of the line segment joining the points (-5, 5) and (5, -5).
Solution:
Let the given points be:
(x1, y1) = (-5, 5)
(x2, y2) = (5, -5)
Midpoint = [(x1 + x2)/2, (y1 + y2)/2]
= [(-5 + 5)/2, (5 - 5)/2]
= (0/2, 0/2)
= (0, 0)
SECTION - II
8. If x2 + y2 = 7xy, then show that 2 log(x + y) = log x + log y + 2 log 3.
Solution:
Given,
x²+ y² = 7xy
Adding 2xy on both the sides,
x²+ y²+ 2xy = 7xy + 2xy
(x + y)² = 9xy
(x + y)²= (3)²xy
Taking log on both sides,
log (x + y)² = log (3)²xy
2 log (x + y) = log 32 + log x + log y
2 log (x + y) = 2 log 3 + log x + log y
9. Length of a rectangle is 5 units more than its breadth. Express its perimeter in polynomial form.
Solution:
Let x be the breadth of a rectangle.
Length = (x + 5) units
Perimeter of rectangle = 2 (Length + Breadth)
= 2(x + 5 + x)
= 2(2x + 5)
= 4x + 10
10. Measures of sides of a triangle are in Arithmetic Progression. Its perimeter is 30 cm and the difference between the longest and shortest side is 4 cm, then find the measures of the sides.
Solution:
Let a - d, a, a + d be the measures of three sides of a triangle.
According to the given,
Perimeter = 30 cm
a - d + a + a + d = 30
3a = 30
a = 30/3
a = 10
Also,
a + d - (a - d) = 4
2d = 4
d = 4/2
d = 2
Thus, a - d = 10 - 2 = 8
a + d = 10 + 2 = 12
Hence, the measures of the triangle are 8 cm, 10 am and 12 cm.
11. Show that the points A(-3, 3), B(0, 0), C(3, -3) are collinear.
Solution:
If (x1, y1), (x2, y2) and (x3, y3) are collinear, then the area of the triangle formed by these vertices is 0.
That means ½ [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] = 0
Given,
A(-3, 3), B(0, 0), C(3, -3)
Area of triangle ABC = ½ [-3(0 + 3) + 0(-3 - 3) + 3(3 - 0)]
= ½ [-9 + 0 + 9]
= ½ (0)
= 0
Hence, the given points are collinear.
12. Solve the following pair of linear equations by substitution method.
2x - 3y = 19
3x - 2y = 21
Solution:
Given,
2x - 3y = 19….(i)
3x - 2y = 21….(ii)
From (i),
2x - 3y = 19
2x = 3y + 19
x = (3y + 19)/2….(iii)
Substituting (iii) in (ii),
3[(3y + 19)/2] - 2y = 21
9y + 57 - 4y = 42
5y = 42 - 57
5y = -15
y = -15/5
y = -3
Substituting y = -3 in (iii),
x = [3(-3) + 19)]/2
= (-9 + 19)/2
= 10/2
= 5
Hence, the solution of the given pair of linear equations is x = 5 and y = -3.
13. If 9x2 + kx + 1 = 0 has equal roots, find the value of k.
Solution:
Given,
9x2 + kx + 1 = 0
Comparing with the standard form ax
2 + bx + c = 0,
a = 9, b = k, c = 1
Condition for equal roots:
b
2
- 4ac = 0
k
2
- 4(9)(1) = 0
k
2
- 36 = 0
k
2 = 36
k = √36
k = ±6
SECTION - III
14. Use Euclid's division lemma to show that the cube of any positive integer is of the form 7m or 7m + 1 or 7m +6.
Solution:
Let a be any positive integer and b = 7.
By Euclid's division lemma,
a = bq + r, 0 ≤ r < b
a = 7q + r; r = 0, 1, 2, 3, 4, 5, 6
When r = 0,
a = 7q
a³ = (7q)³
a³ = 343q³
a³= 7(49q³)
a³ = 7m, where m = 49q³
Also,in (7q + r)³
consider r³
and divide by 7. The remainder will give the result in each case.
When r = 1,
1³ = 1 and a³ = 7m + 1
When r = 2,
2³ = 8, divided by 7, the remainder is 1.
Therefore, a³= 7m + 1
When r = 3,
3³ = 27 divided by 7, the remainder is 6.
Therefore, a³ = 7m + 6
When r = 4,
3³ = 64 divided by 7, the remainder is
1.
Therefore, a³ = 7m + 1
When r = 5,
3⁵ = 125 divided by 7, the remainder is 6.
Therefore, a³ = 7m + 6
When r = 6,
3⁶ = 216 divided by 7, the remainder is 6.
Therefore, a³ = 7m + 6
Hence, the cube of any positive integer is of the form 7m or 7m + 1 or 7m + 6
OR
Prove that √2 - 3√5 is an irrational number
Solution:
Let √2 - 3√5 be a rational number.
√2 - 3√5 = a, where a is an integer.
Squaring on both sides,
(√2 - 3√5)²= a²
(√2)² + (3√5)²- 2(√2)(3√5) = a²
2 + 45 - 6√10 = a²
47 - 6√10 = a²
6√10 = a²- 47
(47 - a²)/6 is a rational number since a is an integer.
Therefore, √10 is also an integer.
We know that integers are not rational numbers.
Thus, our assumption that √2 - 3√5 is a rational number is wrong.
Hence, √2 - 3√5 is an irrational number.
15. Draw the graph for the polynomial p(x) = x²- 3x + 2 and find the zeroes from the graph.
Solution:
Given,
p(x) = x²- 3x + 2
Let y = x²- 3x + 2
