# 10th Maths Book Back Algebra Ex 3.6

## Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 3 – Algebra Ex 3.6 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 3 Algebra consists of 19 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Algebra Ex 3.6 Answers in English.

### 10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Unit 3 Chapter 6 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,

### 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 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 quantity of apples = 30 kg
The quantity of bananas = 20 kg