## 10th Maths Book Back Question and Answers – Chapter 3 Exercise 3.6:

Samacheer Kalvi 10th Standard Maths Book Back Questions with Answers PDF uploaded and the same given below. Class-tenth candidates and those preparing for TNPSC exams can check the Maths Book Back Answers PDF below. Samacheer Kalvi Class 10th Std Maths Book Back Answers Chapter 3 Exercise 3.6 Solutions are available below. Check the complete Samacheer Kalvi 10th Maths – Algebra Ex 3.6 Book Back Answers below:

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### Samacheer Kalvi 10th Maths Book Back Answers – Ex 3.6 Algebra

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#### Exercise 3.6 Algebra

1. Simplify

Solution:

2. Simplify

Solution:

3. Subtract 1×2+2 from 2×3+x2+3(x2+2)2
Solution:

4. Which rational expression should be subtracted from x2+6x+8×3+8 to get 3×2−2x+4
Solution:

5.

Solution:

6. If A = xx+1, B = 1x+1, prove that

Solution:

7. Pari needs 4 hours to complete a work. His friend Yuvan needs 6 hours to complete the same work. How long will be take to complete if they work together?
Let the work done by Pari and Yuvan together be x
Work done by part = 14
Work done by Yuvan = 16
By the given condition
14 + 16 = 1x ⇒ 3+212 = 1x
512 = 1x
5x = 12 ⇒ x = 125
x = 2 25 hours (or) 2 hours 24 minutes

8. Iniya bought 50 kg of fruits consisting of apples and bananas. She paid twice as much per kg for the apple as she did for the banana. If Iniya bought Rs. 1800 worth of apples and Rs. 600 worth of bananas, then how many kgs of each fruit did she buy?
Let the quantity of apples and bananas purchased be ‘x’ and ‘y’
By the given condition
x + y = 50 ………(1)
Cost of one kg of apple = 1800x
Cost of one kg of banana = 600y
By the given condition
One kg of apple = 2(600)y
Total cost of fruits purchased = 1800 + 600
x × 2 (600)y + y (600)y = 2400
1200xy = 2400 – 600
1200xy = 1800
1200 x = 1800 × y
x = 1800×1200 = 3y2
Substitute the value of x in (1)
3y2 + y = 50
5y2 = 50
5y = 100 ⇒ y = 1005 = 20
x = 3y2 = 3×202
= 30
The number of apples = 30 kg
The number of bananas = 20 kg