Tag: Maths Class 10th Exercise 4.2 Solution

Q1. Find the roots by factorisation

(i) $x^2-3x-10=0$
Factor: $(x-5)(x+2)=0$
Roots: $x=5,-2$

(ii) $2x^2+x-6=0$
Factor: $(2x-3)(x+2)=0$
Roots: $x={\displaystyle \frac{3}{2}},-2$

(iii) $2x^2+7x+5=0$ (interpreting the printed form as $2x^2+7x+5=0$)
Factor: $(2x+5)(x+1)=0$
Roots: $x=-{\displaystyle \frac{5}{2}},-1$

(iv) $2x^2-x+{\displaystyle \frac{1}{8}}=0$
Multiply by 8: $16x^2-8x+1=0$. This is $(4x-1{)}^2=0$.
Double root: $x={\displaystyle \frac{1}{4}}$

(v) $100x^2-20x+1=0$
This is $(10x-1{)}^2=0$
Double root: $x={\displaystyle \frac{1}{10}}$

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Q2. Solve the problems given in Example 1 (from the book).

(i) John and Jivanti have 45 marbles. After each loses 5, product = 124.
Let John = $x$. Then Jivanti = $45-x$. Equation: $(x-5)(40-x)=124$⇒ $x^2-45x+324=0$
Discriminant $D={45}^2-4\cdot 1\cdot 324=2025-1296=729$ 

$\sqrt{D}=27$

$x={\displaystyle \frac{45\pm 27}{2}}\Rightarrow x=36$ or $x=9$
So John could have had 36 (then Jivanti 9) or John 9 (then Jivanti 36). Both satisfy the condition.

(ii) Toys: let number $=x$. Cost per toy $=55-x$. Total $x(55-x)=750$ ⇒ $x^2-55x+750=0$
$D={55}^2-4\cdot 1\cdot 750=3025-3000=25,\sqrt{D}=5.$
$x={\displaystyle \frac{55\pm 5}{2}}\Rightarrow x=30$ or $x=25$
Cost per toy: if $x=30$→ cost = $25$₹; if $x=25$ → cost = $30$.
So number produced = 30 (cost ₹25) or 25 (cost ₹30).

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Q3. Find two numbers whose sum is 27 and product is 182.

Answer – Two numbers sum 27 and product 182

Let $x$ and $27-x$: $x(27-x)=182$ ⇒ $x^2-27x+182=0$
$D={27}^2-4\cdot 182=729-728=1$
$x={\displaystyle \frac{27\pm 1}{2}}\Rightarrow x=14$ or $13$.
Numbers: 14 and 13.

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Q4. Find two consecutive positive integers, sum of whose squares is 365.

Answer – Two consecutive positive integers whose sum of squares is 365

Let integers $n$ and $n+1$: $n^2+(n+1{)}^2=365$

⇒ $2n^2+2n+1=365$ ⇒ $n^2+n-182=0$
$D=1+728=729,\sqrt{D}=27$

$n={\displaystyle \frac{-1\pm 27}{2}}\Rightarrow n=13$ (positive).
Integers: 13 and 14.

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Q5.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 – Right triangle: altitude = base − 7, hypotenuse = 13. Find the other two sides

Let base $=b$, altitude $=b-7$. Then $b^2+(b-7{)}^2={13}^2=169$
So $2b^2-14b-120=0$ ⇒ $b^2-7b-60=\mathrm{}$ $(b-12)(b+5)=0$
Positive solution $b=12$. Altitude $=12-7=5$
Sides: base $12$ cm, altitude $5$ cm, hypotenuse $13$ cm.

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Q6.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 ` 90, find the number of articles produced and the cost of each article.

Answer – Cottage industry: cost per article = $2\times (\text{no.})+3$, total cost = 90

Let number $=x$. Total: $x(2x+3)=90$ ⇒ $2x^2+3x-90=0$.
$D=3^2-4\cdot 2\cdot (-90)=9+720=729,\sqrt{D}=27.$
$x={\displaystyle \frac{-3\pm 27}{4}}\Rightarrow x={\displaystyle \frac{24}{4}}=6$ (positive). Cost per article $=2\cdot 6+3=15$
Answer: 6 articles; cost ₹15 each.