Unit Test (Solutions): Quadratic Equations | Mathematics (Maths) Class 10 PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: Equation of (x+1)²-x²=0 has number of real roots equal to: (1 Mark)
(a) 1
(b) 2
(c) 3
(d) 4
Ans: (a)
(x+1)²-x²=0
x²+2x+1-x² = 0
2x+1=0
x=-½
Hence, there is one real root.

Q2: The sum of two numbers is 27 and product is 182. The numbers are: (1 Mark)
(a) 12 and 13
(b) 13 and 14
(c) 12 and 15
(d) 13 and 24
Ans: (b)
Let x is one number
Another number = 27 – x
Product of two numbers = 182
x(27 – x) = 182
⇒ x² – 27x – 182 = 0
⇒ x² – 13x – 14x + 182 = 0
⇒ x(x – 13) -14(x – 13) = 0
⇒ (x – 13)(x -14) = 0
⇒ x = 13 or x = 14

Q3: 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. (1 Mark)
(a) 30 km/hr
(b) 40 km/hr
(c) 50 km/hr
(d) 60 km/hr

Ans: (b)

Let x km/hr be the speed of train.
Time required to cover 360 km = 360/x hr.
As per the question given,
⇒ (x + 5)(360-1/x) = 360
⇒ 360 – x + 1800-5/x = 360
⇒ x² + 5x + 10x – 1800 = 0
⇒ x(x + 45) -40(x + 45) = 0
⇒ (x + 45)(x – 40) = 0
⇒ x = 40, -45
Negative value is not considered for speed hence the answer is 40km/hr.

Q4: A quadratic equation ax2 + bx + c = 0 has no real roots, if _______________________________. (1 Mark)

Ans: b2 – 4ac < 0

A quadratic equation ax2 + bx + c = 0 has no real roots, if b2 – 4ac < 0. That means, the quadratic equation contains imaginary roots.

Q5: The maximum number of roots for a quadratic equation is equal to (1 Mark)
(a) 1
(b) 2
(c) 3
(d) 4

Ans: (b)

The maximum number of roots for a quadratic equation is equal to 2 since the degree of a quadratic equation is 2.

Q6: The sum of areas of two squares is 468m2. If the difference of their perimeters is 24cm, find the sides of the two squares. (2 Marks)
Ans: 
Let, the side of the larger square be x. Let, the side of the smaller square be y.
x²+y²=468
Cond. II 4x-4y = 24
⇒ xy = 6
⇒ x = 6+y
⇒ x² + y² = 468
⇒(6+y)² + y² = 468
on solving we get
y = 12
⇒ x = (12+6) = 18 m
∴ The length of the sides of the two squares are 18m and 12m.

Q7: If x =1, is a solution of the quadratic equation 3x² + 2kx – 3 = 0, find the value of k. (2 Marks)
Ans:
The given quadratic equation can be written as, 3x² + 2kx – 3 = 0 

Q8: 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. (2 Marks)
Ans:
Let us say, the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
Diagonal of the rectangle = √[x²+(x+30)²] As given, the length of the diagonal is = x + 60 m
⇒ x² + (x + 30)² = (x + 60)²
⇒ x² + x² + 900 + 60x = x² + 3600 + 120x
⇒ x² – 60x – 2700 = 0
⇒ x² – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30(x -90)
⇒ (x – 90)(x + 30) = 0
⇒ x = 90, -30

Q9: Solve the quadratic equation 2×2 – 7x + 3 = 0 by using quadratic formula. (3 Marks)
Ans:
2×2 – 7x + 3 = 0
On comparing the given equation with ax2 + bx + c = 0, we get,
a = 2, b = -7 and c = 3
By using quadratic formula, we get,
x = [-b±√(b² – 4ac)]/2a
⇒ x = [7±√(49 – 24)]/4
⇒ x = [7±√25]/4
⇒ x = [7±5]/4
Therefore,
⇒ x = 7+5/4 or x = 7-5/4
⇒ x = 12/4 or 2/4
∴ x = 3 or ½

Q10: Find two numbers whose sum is 30 and product is 200. (3 Marks)

Ans: Let the two numbers be $xx$x and $yy$y. We are given:
$x+y=\mathrm{30 }$(sum of the numbers)
$xy=200(product of the numbers)xy\; =\; 200\; \backslash quad\; \backslash text\{(product\; of\; the\; numbers)\}$ (product of the numbers)

We can express $yy$ in terms of $xx$ from the first equation:
$y=30-xy\; =\; 30\; -\; x$

Substitute this into the second equation:
$x(30-x)=200$ 
$30x-x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}=20030x\; -\; x^2\; =\; 200$
$x2-30x+200=0$

Now, solve the quadratic equation using the quadratic formula:

Q11: Find the roots of the following quadratic equations by factorization: (3 Marks)
(i) x² + 7x + 10 = 0
(ii) 3x² − 5x − 2 = 0$3x^2\; -\; 5x\; -\; 2\; =\; 0$

Ans:

(i) Solve x² + 7x + 10 = 0

We need two numbers whose product is 10 and sum is 7. The numbers are 5 and 2 since $5\times 2=105\; \backslash times\; 2\; =\; 10$ and $5+2=7$.
Rewrite the middle term:
$x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}+5x+2x+10=0$
Factorize by grouping:
$x(x+5)+2(x+5)=0$

Factor out the common term:
$(x+5)(x+2)=0$

Thus, the roots are:
$x+5=0\Rightarrow x=-5x\; +\; 5\; =\; 0\; \backslash Rightarrow\; x\; =\; -5$
$x+2=0\Rightarrow x=-2x\; +\; 2\; =\; 0\; \backslash Rightarrow\; x\; =\; -2$

(ii) Solve 3x²−5x−2=0:

We need two numbers whose product is $3\times (-2)=-63\; \backslash times\; (-2)\; =\; -6$3×(−2)=−6 and sum is -5. The numbers are -6 and 1 since $-6\times 1=-6-6\; \backslash times\; 1\; =\; -6$ and −6+1=−5.
Rewrite the middle term:
$3x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}-6x+x-2=03x^2\; -\; 6x\; +\; x\; -\; 2\; =\; 0$

Factorize by grouping:
$3x(x-2)+1(x-2)=03x(x\; -\; 2)\; +\; 1(x\; -\; 2)\; =\; 0$

Factor out the common term:
$(3x+1)(x-2)=0(3x\; +\; 1)(x\; -\; 2)\; =\; 0$

Thus, the roots are:

Q.12: 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. (5 Marks)
Ans:
Let the number of articles produced on that day be $xx$x. The cost of production of each article is $2x+32x\; +\; 3$2x+3.
The total cost of production is given as:
$x\times (2x+3)=90x\; \backslash times\; (2x\; +\; 3)\; =\; 90$ 
$2x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}+3x-90=02x^2\; +\; 3x\; -\; 90\; =\; 0$
Using the quadratic formula:

The two possible values of $x$x are:

$x=6(since\; the\; number\; of\; articles\; cannot\; be\; negative,\; discard$ (since the number of articles cannot be negative, discard x=−7.5)

The cost of production for each article is:

$2\left(6\right)+3=152(6)\; +\; 3\; =\; 15$

Thus, the number of articles produced is 6, and the cost of each article is Rs. 15.

Q13: The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field. (5 Marks)

Ans: Let the shorter side of the rectangle be $xx$x meters.
Then, the diagonal of the rectangle is $x+60$ meters, and the longer side is $x+30$ meters.
According to the Pythagoras theorem:

$(Diagon$(Diagonal)² = (Shorter side)² + (Longer side)²
Substituting the values:
$(x+60)\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}=x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}+(x+30)\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}$

Expanding both sides:
$x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}+120x+3600=x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}+x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}+60x+900x^2\; +\; 120x\; +\; 3600\; =\; x^2\; +\; x^2\; +\; 60x\; +\; 900$

Simplifying:
$x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}+120x+3600=2x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}+60x+900x^2\; +\; 120x\; +\; 3600\; =\; 2x^2\; +\; 60x\; +\; 900$

Bringing all terms to one side:

$0=2x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}+60x+900-x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}-120x-3600$ 
$0=x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}-60x-2700$

Solve the quadratic equation:

The two possible values for x are:

Since the length cannot be negative, we discard $x=-30.x\; =\; -30$

Thus, $x=90$ meters is the shorter side.

The longer side is:

$x+30=90+30=120\hspace{0.17em}metersx\; +\; 30\; =\; 90\; +\; 30\; =\; 120\; \backslash ,\; \backslash text\{meters\}$meters

Thus, the shorter side is 90meters and the longer side is 120meters.