RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions

RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 1
Chapter Name	Vedic Mathematics
Exercise	Additional Questions
Number of Questions Solved	20
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions

Question 1.
Add: 112 kg 065 gm + 360 kg 085 gm + 289 kg 872 gm + 156 kg (RBSESolutions.com) 345 gm
Solution

Hint:

1. Write 65 gm as 065 gm and 85 gm as 085 gm
2. Start adding from top of unit coloumn.
3. 5 + 5 = 10 So ekadhik mark on 8. preccecding digit of 51 remainder = 10 – 10 = 0
4. Write remainder 0 + 2 + 5 = 7 at place of answer.
5. Keep it up.

Question 2.
Add: 7534 + 2459 + 1932 + 6547
Solution

Hint:

1. 34 + 59 = 33+ 1 + 59 = 33 + 60 = 93
2. Remaining 93 + 32 = 93 + 7 + 25 – 100 + 25 = 125
   So ekadhiken make on 9.
3. Write remainng 25 + 47 = 22 + 3 + 47 = 22 + 50 = 72 at the place of answer.
4. Remaining (RBSESolutions.com) addition process is an above.

Question 3.
Subtract by Vedic method:
28 km 375 km 46 cm from 37 km 467 km 35 cm.
Solution

Hint:

1. Arrange the column numbers in m-cm.
2. cm. column; 6 cannot be subtracted from 5, thus add 4 the param Mitra digit of 6 to 5.
3. Write sum = 9 at the answer place and mark ekadhk sign on 4 the pre-dissociator digit.
4. 4 = 5, cannot be subtracted from 3, thus add 5 the param Mitra digit to 3, the pre-dissociator digit.
5. Write sum = 8 below and mark the sign of ekadhik on the pre-dissociator.
6. Write 7 – 5 = 1 below.
7. 7 cannot be (RBSESolutions.com) subtracted from 6, So adding 6 + 3 = 9 write below and mark the sign of ekadhik on the pre-disjunction 3.
8. Write 4 – 3 = 0 below.
9. All the next operations will be done in the same way.
   ∴ Answer = 9 km 91 m 89 cm.

Question 4.
Which of the sutra is the best to solve 842 × 858?
Solution

1. Sutra ekadhiken poorvene is not effective (RBSESolutions.com) because product of R.H.S. i.e., 42 × 58 cannotbe got easily.
2. The sutra Nikhilam Adhara cannot be effective bacause if we consider base 1000 then respectively difference will be -158 and -142. The sutra Nikhilam updhara is not effective here as upadhar (sub-base) = 800.
   We get the difference 42 and 58.
3. The sutra Ekanyeune Poorvene is not effective here.
4. The sutra Urdhava-triyagbhayam is effective and the best here.
5. The new option: To get the product of 842 × 858 first we use sutra ekadhikene poorvene and then sutra urdhva-tiryagbhayam.
   

Question 5.
13579 ÷ 975 (dhwahjank method)
Solution

Hint:

1. 13 ÷ 9, first digit (RBSESolutions.com) of quotient = 1, remainder = 4
2. New dividend = 45, corrected dividend = 45 – 1 × 7 = 38
3. 38 ÷ 9, second digit of quotient = 4, remainder = 2
4. New dividend = 27,
   Corrected dividend = 27 – (4 × 7 + 1 × 5) = 27 – 33 = – 6
   Since we get the corrected dividend as negative so the second digit of quotient must be 3 instead of 4. So (iii) and (iv) steps are rejelable
5. Again 38 ÷ 9, Quotuiet second digit = 3, Remainder = 11.
6. New dividend = 1179 So corrected dividend and Last remainder
   = 1179 – (3 × 7 + 1 × 5) × 10 – 3 × 5
   = 1179 – 260 – 15
   = 904

Question 6.
Find a (RBSESolutions.com) square of 41.
Solution

Hint:

1. Make three columns for answer.
2. In first column, square of ten’s digit = 16
3. In third column, square of unit digit = 1
4. In middle column, product of both digit = 1 × 4 = 4
5. In the middle part write the above product once again.
6. The sum is the square of the given number. In the middle and third column only one digit must be written.

Question 7.
Find the (RBSESolutions.com) square of 17 by upsutra Yavaunam.
Solution
17² = 17 + 7/7²
Base = 10, deviation = +7 = 24/₄9 = 289

Question 8.
Find the square of 354 by Dwandwa yoga.
Solution
Five digit groups can be formed of the number 354 – 3, 35, 354, 54 and 4.
Written the Dwandwa yog of these five digit groups, we get

Question 9.
Find a (RBSESolutions.com) square of 12 by Ishta Sankhya Method.
Solution
12² = (12 + 2)(12 – 2) + 2²
= 14 × 10 + 4
= 144

Question 10.
Find the cube of 15.
Solution
15³ = 15 + 2 × 5/3 × 5²/5³
= 25 / ₇5 / ₁₂5
= 3375
Hint:

1. Base = 10, deviation = +5
2. Only single digit in middle and third part.

Question 11.
Find the (RBSESolutions.com) cube of 24.
Solution

Hint:

1. Base = 10, Sub-base = 10 × 2
2. Sub-base digit = 2, deviation = +4

Question 12.
Find the cube of 43 using formula.
Solution

Hint:

1. Base = 10,deviation = 3 × 3 – 10 = -1
2. Value of 2 in first part 20 of second part

Question 13.
Find the (RBSESolutions.com) square root of perfect square number 10329796.

Solution
Hint:

1. 4 pairs of digits in number, so 4 digits in the square root.
2. first square root digit = 3
3. Remainder = 10 – 3² = 1, Take next pair 32, besides 1, so new divident = 132.
4. Divisor = double of 3 = 6
5. 13 of the number 132, can be divided by 6, 2 times so write quotient 2 next to 3 in quotient column.
6. Write 2 next to divisor 6. So, the corrected divisor = 62
7. 132 – 62 × 2 = 132 – 124 = 8 = Remainder
8. Take 97 next to remainder 8, So new dividend = 897 and new divisor = 32 × 2 = 64
9. 89 ÷ 64 = 1 quotient, So write 1 next to quotient 32.
10. Write next to divisor 64, So the corrected divisor = 641.
11. 897 – 641 × 1 = 897 – 641 = 256, take 96 to ths remainder 256.
12. So, the new dividend = 25696 and new divisor = 321 × 2 = 642.
13. 2569 ÷ 642 = 4 quotient, So write 4 next to the quotient 321.
14. Write 4 next to divisor 642, So the corrected divisor = 6424
15. 25696 – 6424 × 4 = 0 remainder.
    Hence, remainder = 0
    Square root = quotient = 3214

Question 14.
Find the (RBSESolutions.com) square root of 41254929 by Dwandwa Yoga method.
Solution

Hint:

1. 4 digit in the square root.
2. 41 – 6² = 5, write before 2 just below of it.
3. New dividend = 52, corrected dividend = 52
4. 52 ÷ 12, quotient digit = 4. Write 4 between 2 and 5 and just below of them.
5. New dividend = 45, corrected dividend = 45 – 4² = 29.
6. 29 ÷ 12,quotient = 2, remainder digit = 5
7. Write 5 between 5 and 4 just below them.
8. New dividend = 54, corrected dividend = 54 – 4 × 2 × 2 = 38
9. 38 ÷ 12, quotient digit = 3 and remainder = 2.
   Write 2 between 4 and 9 just below. Then we have obtained 4 digits in the square root so we are to find the final remainder. After finding complete square root the decimal point and the zeros can be placed. To find the final remainder, as many zeros are placed after the decimal point as the new dividend are to be corrected.
10. Remainder = 29 – 4 × 3 × 2 – 2² = 1. Write 1 between 9 and 2, just below of them.
11. New dividend = 12, remainder = 12 – 2 × 3 × 2 = 0. Write 0 before and just below 9.
12. New dividend = 9, final remainder = 0
    Hence square root = 6423

Question 15.
Find the (RBSESolutions.com) cube root of 355045312441 perfect cube number, by division method.
Solution

Cube root = 7081
Hint:

1. No. of steps = cube root digit no. × 3 – 2
2. Don’t calculate the last three steps to obtain the cube root of a perfect cube number. As soon as we see the unit digit of a number we get the unit digit of its cube root.
3. Using the Dwandwa Yog Method. We can obtain the cube root of a number having so many digits in it.
4. If the cube root has four digits, find the second and third digit by division method. Unit and last digits can be calculated orally.

Question 16.
Simplify 2(x + 1) = 7(x + 1).
Solution
(x + 1) is (RBSESolutions.com) common factor in each term, so x + 1 = 0 ⇒ x = – 1

Question 17.
Simplify (x + 1)(x + 9) = (x + 3) (x + 3)
Solution
Constant terms in both sides are same, so, x = 0

Question 18.
Simplify: \(\frac { m }{ 2x+1 }\) + \(\frac { m }{ 3x+4 }\)
Solution
Both the numerators of fractions due same = m.
By formula 2x + 1 + 3x + 4 = 0
or 5x + 5 = 0
or x = – 1

Question 19.
Simplify : \(\frac { 3x+4 }{ 6x+7 }\) = \(\frac { x+1 }{ 2x+3 }\)
Solution
Sum of numerators (RBSESolutions.com) of both sides = 3x + 4 + x + 1 = 4x + 5 …(i)
Sum of denominators of both sides = 6x + 7 + 2x + 3 = 8x + 10 …(ii)
Ratio of (i) by (ii) = 1 : 2
According to question any sum equation to zero.
From 4x + 5 = 0
or 8x + 10 = 0
⇒ x = \(\frac { -5 }{ 4 }\)

Question 20.
Simplify the equation:

Solution
Sum of the denominator of (RBSESolutions.com) both sides are equal = 2x – 17
According to formula
2x – 17 = 0
⇒ x = \(\frac { 17 }{ 2 }\) = 8\(\frac { 1 }{ 2 }\)

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