RBSE Solutions for Class 6 Maths Chapter 7 Vedic Mathematics Ex 7.1

RBSE Solutions for Class 6 Maths Chapter 7 Vedic Mathematics Ex 7.1 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 7 Vedic Mathematics Exercise 7.1.

Rajasthan Board RBSE Class 6 Maths Chapter 7 Vedic Mathematics Ex 7.1

Question 1.
Find out the addition by formula (RBSESolutions.com) of Ekadhiken poorven.


Solution.
Finding the sum, using (RBSESolutions.com) Ekadhiken poorven,

Hint :
(a) Addition of digits at unit’s place 6 + 8 = 14. Thus, we will mark the 16 4 one more sign on poorven digit of 8 i.e., 6, and we will put 4 below whole sum (at the place of unit).
(b) In the addition of digit at tens place 9 + \(\overset { \bullet }{ 6 } \) = 16. (where \(\overset { \bullet }{ 6 } \) = 7)
(c) Therefore, mark one more sign on poorven digit of 6 i.e., 0.
(d) Write the remainder .6 at the place of addition (tens place).
(e) \(\overset { \bullet }{ 0 } \) = 1, shall be written at hundreds place.

Hint :
(a) Addition of digits at the units place 8 + 9 = 17, therefore (RBSESolutions.com) mark one + 35 more sign at poorven digit of 9 i.e, 4.
(b) Left 7 + 5 = 12, therefore mark one more sign would be at poorven digit of 5, i.e., 3.
(c) Remainder 2, shall be written at the place of addition (in unit place).
(d) In the addition of tens place 9 + \(\overset { \bullet }{ 4 } \) = 14, therefore mark one more sign at the poorven digit of 4 i.e., 0.
(e) Remainder 4 + 3 = 8, shall be written at the place of addition (at tens place).
(f) \(\overset { \bullet }{ 0 } \) = 1, at the place of hundreds.

Hint :
(a) Addition of digits at the units place 7 + 6 = 13, therefore mark (RBSESolutions.com) one more sign at poorven digit of 6 i.e., 9.
(b) Left 3 + 8 = 11, therefore put one more sign at poorven digit of 8 i.e., 2.
(c) Remainder 1, shall be written at the place of addition (in unit place).
(d) In the addition of tens place 2 + \(\overset { \cdot }{ 9 } \) = 12, therefore put one more sign at poorven digit of 9 i.e., 4.
(e) Remainder 2 + \(\overset { \cdot }{ 2 } \) = 5 shall be written at the place of addition (at tens place).
(f) In the addition of hundreds place 3 + \(\overset { \cdot }{ 4 } \) = 8 and 8 + 5 = 13, therefore put one more sign at the poorven digit of 5 i.e., 0 and remainder 3, shall be written at the place of addition (at hundreds place).
(g) 0 = 1, the place of thousands.

Hint :
(a) 5 + 6 = 11, therefore put sign at poorven digit of 6 i.e., 3, and (RBSESolutions.com) write 1 at hundredth place (in paisa).
(b) 7 + \(\overset { \cdot }{ 3 } \) = 11, therefore put one more sign at poorven digit of \(\overset { \cdot }{ 3 } \) i.e., 5, at write 1 at tenth place (in paisa).
(c) 8 + \(\overset { \cdot }{ 5 } \) = 14, therefore put one more sign at poorven digit of \(\overset { \cdot }{ 5 } \) i.e., 9, and write 4 at unit place (in ₹).
(d) 1 + \(\overset { \cdot }{ 9 } \) = 11, therefore put one more sign at poorven digit of \(\overset { \cdot }{ 9 } \) i.e., 3, and write 1 at tens place (in ₹).
(e) 4 + \(\overset { \cdot }{ 3 } \) = 8, written at hundreds place (in ₹).

Hints :
(a) 6 + 5 = 11, therefore put sign at poorven digit of 5 i.e., 4, and (RBSESolutions.com) write 1 at thousand place (in m).
(b) 8 + \(\overset { \cdot }{ 4 } \) = 13, therefore put one more sign at poorven digit of 4 i.e., 3, and write 3 at hundredth place (in m).
(c) 7 + \(\overset { \cdot }{ 3 } \) = 11, therefore put one more sign at poorven digit of 3 i.e., 5, and write 1 at tenth place (in m.).
(d) 6 + \(\overset { \cdot }{ 5 } \) = 12 , therefore put one more sign at poorven digit of 5 i.e., 7, and write 2 at units place(in km).
(e) 8 + \(\overset { \cdot }{ 7 } \) = 16, therefore put one more sign at poorven digit of 7 i.e., 0, and write 6 at tens place (in km.)
(f) \(\overset { \cdot }{ 0 } \) = 1 at the place of hundreds place (in km).

Hint :
(a) 5 + 3 = 8, remainder 8, written at (RBSESolutions.com) hundredth place (in gm).
(b) 6 + 8 = 14, therefore put one more sign at poorven digit of 8 i.e., 7, and remainder 4 written at tenth place (in gm).
(c) 9 + \(\overset { \cdot }{ 7 } \) = 17 , therefore put one more sign at poorven digit of 7 i.e., 8, and remainder 7 written at units place (in kg).
(d) 3 + \(\overset { \cdot }{ 8 } \) = 12 , therefore put one more sign at poorven digit of 8 i.e., 0, and remainder 2 written at tens place (in kg).
(e) 1 + \(\overset { \cdot }{ 0 } \) = 2, remainder 2 written at hundreds place (in kg).

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