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Vedic Mathematics

Class 10th Mathematics Rajasthan Board Solution
Exercise 1.3
  1. 2116 Find the square root by Vedic methods.
  2. 4225 Find the square root by Vedic methods.
  3. 6889 Find the square root by Vedic methods.
  4. 59049 Find the square root by Vedic methods.
  5. 125316 Find the square root by Vedic methods.
  6. 169744 Find the square root by Vedic methods.
  7. 1265625 Find the square root by Vedic methods.
  8. 1522756 Find the square root by Vedic methods.
  9. 68921 Find the cube root of the perfect cube number by Vedic methods.…
  10. 636056 Find the cube root of the perfect cube number by Vedic methods.…
  11. 314432 Find the cube root of the perfect cube number by Vedic methods.…
  12. 493039 Find the cube root of the perfect cube number by Vedic methods.…
  13. 8365427 Find the cube root of the perfect cube number by Vedic methods.…
  14. 1061208 Find the cube root of the perfect cube number by Vedic methods.…
  15. 8489664 Find the cube root of the perfect cube number by Vedic methods.…
  16. 200201625 Find the cube root of the perfect cube number by Vedic methods.…
  17. 258474853 Find the cube root of the perfect cube number by Vedic methods.…
  18. 22665187 Find the cube root of the perfect cube number by Vedic methods.…
  19. 8615125 Find the cube root of the perfect cube number by Vedic methods.…
  20. 660776311 Find the cube root of the perfect cube number by Vedic methods.…
Exercise 1.4
  1. 13 x – 14 = 9x + 10 Find the oral solution of the equation by the formula…
  2. 3y + 4 = 5y – 4 Find the oral solution of the equation by the formula paravartya…
  3. {2x+1}/{3x+4} = frac {1}/{3} Find the oral solution of the equation by the…
  4. {5x-3}/{2} = frac {2x+1}/{5} Find the oral solution of the equation by the…
  5. (x + 7) (x + 9) = (x – 8) (x – 11) Find the oral solution of the equation by the…
  6. (x + 5) (x + 1) = (x + 3) (x + 2) Find the oral solution of the equation by the…
  7. {1}/{x+1} - frac {2}/{x-1} = 0 Find the oral solution of the equation by the…
  8. {5}/{2x-1} - frac {9}/{3x-2} = 0 Find the oral solution of the equation by the…
  9. (2x + 1) + (x + 3) = 5x + 4 Solve the equation by formula Sunyam
  10. a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1) Solve the equation by formula Sunyam…
  11. (x + 1) (x + 9) = (x + 3) (x + 3) Solve the equation by formula Sunyam…
  12. {x}/{2} + frac {x}/{3} = frac {x}/{4} + frac {x}/{1} Solve the equation by…
  13. {1}/{x+4} + frac {1}/{x-6} = 0 Solve the equation by formula Sunyam…
  14. {5}/{3x+2} + frac {5}/{2x+8} = 0 Solve the equation by formula Sunyam…
  15. {2x+4}/{2x+1} = frac {2x+1}/{2x+4} Solve the equation by formula Sunyam…
  16. {3x+2}/{5x+7} = frac {x+1}/{3x+5} Solve the equation by formula Sunyam…
  17. {5x+7}/{2x+1} = frac {x+1}/{3x+5} Solve the equation by formula Sunyam…
  18. {3x+6}/{6x+3} = frac {5x+4}/{2x+7} Solve the equation by formula Sunyam…
  19. {1}/{x+2} + frac {1}/{x+6} = frac {1}/{x+1} + frac {1}/{x+7} Solve the equation…
  20. {1}/{x-4} + frac {1}/{x-6} = frac {1}/{x-2} + frac {1}/{x-8} Solve the equation…

Exercise 1.3
Question 1.

Find the square root by Vedic methods.

2116


Answer:

So finding the square root by Vedic method,


four digits are in the number, so the digits in it’s square root will be 2.



the quotient is 46, so the square root is 46.


Hint –


I. Four digits in the number so two digits in the square root.


II. First square root digit = 4


III. 21 - = 5 = Remainder


IV. Pair of digits 16 was descended before 5. So new dividend = 516.


V. Divisor = 4 2 = 8, i.e., double of square root digit.


VI. 86 divides 516 by 6, so write 6 ahead digit 4 in the quotient.


VII. 516 – 86 6 = 516 – 516 = 0


Remainder = 0, Square Root = Quotient = 46.



Question 2.

Find the square root by Vedic methods.

4225


Answer:

So finding the square root by vedic method,


four digits are in the number, so the digits in it’s square root will be 2.



the quotient is 65, so the square root is 65.


Hint –


I. Four digits in the number so two digits in the square root.


II. First square root digit = 6


III. 42 - = 6 = Remainder


IV. Pair of digits 25 was descended before 6. So new dividend = 625.


V. Divisor = 6 2 = 12, i.e., double of square root digit.


VI. 125 divides 625 by 5, so write 5 ahead digit 6 in the quotient.


VII. 625 – 125 5 = 625 – 625 = 0


Remainder = 0, Square Root = Quotient = 65.



Question 3.

Find the square root by Vedic methods.

6889


Answer:

So finding the square root by vedic method,


four digits are in the number, so the digits in it’s square root will be 2.



the quotient is 83, so the square root is 83.


Hint –


I. Four digits in the number so two digits in the square root.


II. First square root digit = 4


III. 68 - = 4 = Remainder


IV. Pair of digits 89 was descended before 4. So new dividend = 489.


V. Divisor = 8 2 = 16, i.e., double of square root digit.


VI. 163 divides 489 by 3, so write 3 ahead digit 8 in the quotient.


VII. 489 – 163 3 = 489 – 489 = 0


Remainder = 0, Square Root = Quotient = 83.



Question 4.

Find the square root by Vedic methods.

59049


Answer:

So finding the square root by Vedic method,


five digits are in the number, so the digits in it’s square root will be 3.



the quotient is 243, so the square root is 24.


Hint –


I. Five digits in the number so three digits in the square root.


II. First square root digit = 2


III. 5 - = 1 = Remainder


IV. Pair of digits 90 was descended before 1. So new dividend = 190.


V. Pair of digits 49 was descended before 14. So new dividend = 1449.


VI. Divisor = 2 2 = 4 & 44 + 4 = 48 i.e., double of square root digit respectively.


VII. 483 divides 1449 by 3, so write 3 ahead digit 24 in the quotient.


VIII. 1449 – 483 3 = 1449 – 1449 = 0


Remainder = 0, Square Root = Quotient = 243.



Question 5.

Find the square root by Vedic methods.

125316


Answer:

So finding the square root by Vedic method,


six digits are in the number, so the digits in its square root will be 3.



the quotient is 354, so the square root is 354.


Hint –


I. Six digits in the number so three digits in the square root.


II. First square root digit = 3


III. 12 - = 3 = Remainder


IV. Pair of digits 53 was descended before 3. So new dividend = 353.


V. Pair of digits 16 was descended before 28. So new dividend = 2816.


VI. Divisor = 3 2 = 6 & 65 + 5 = 70 i.e., double of square root digit respectively.


VII. 704 divides 2816 by 4, so write 4 ahead digit 35 in the quotient.


VIII. 2816 – 704 4 = 2816 – 2816 = 0



Question 6.

Find the square root by Vedic methods.

169744


Answer:

So finding the square root by Vedic method,


six digits are in the number, so the digits in it’s square root will be 3.



the quotient is 412, so the square root is 412.


Hint –


I. Six digits in the number so three digits in the square root.


II. First square root digit = 4


III. 16 - = 0 = Remainder


IV. Pair of digits 97 was descended before 0. So new dividend = 97.


V. Pair of digits 44 was descended before 16. So new dividend = 1644.


VI. Divisor = 4 2 = 8 & 81 + 1 = 82 i.e., double of square root digit respectively.


VII. 822 divides 1644 by 2, so write 2 ahead digit 41 in the quotient.


VIII. 1644 – 822 2 = 1644 – 1644 = 0



Question 7.

Find the square root by Vedic methods.

1265625


Answer:

So finding the square root by Vedic method,


seven digits are in the number, so the digits in it’s square root will be 3.



the quotient is 1125, so the square root is 1125.


Hint –


I. Seven digits in the number so four digits in the square root.


II. First square root digit = 1


III. 1 -1 = 0 = Remainder


IV. Pair of digits 26 was descended before 0. So new dividend = 26.


V. Pair of digits 56 was descended before 5. So new dividend = 556.


VI. Pair of digits 112 was descended before 25. So new dividend = 11225.


VII. Divisor = 1 1 = 1 & 222 + 2 = 224 i.e., double of square root digit respectively.


VIII. 2245 divides 11225 by 5, so write 5 ahead digit 112 in the quotient.


IX. 11225 – 2245 5 = 11225 – 11225 = 0



Question 8.

Find the square root by Vedic methods.

1522756


Answer:

So finding the square root by Vedic method,


seven digits are in the number, so the digits in it’s square root will be 3.



the quotient is 1234, so the square root is 1234.


Hint –


I. Seven digits in the number so four digits in the square root.


II. First square root digit = 1


III. 1 -1 = 0 = Remainder


IV. Pair of digits 52 was descended before 0. So new dividend = 52.


V. Pair of digits 27 was descended before 8. So new dividend = 827.


VI. Pair of digits 56 was descended before 98. So new dividend = 9856.


VII. Divisor = 1 1 = 1 & 243 + 3 = 246 i.e., double of square root digit respectively.


VIII. 2464 divides 9856 by 4, so write 4 ahead digit 123 in the quotient.


IX. 9856 – 2464 4 = 9856 – 9856 = 0



Question 9.

Find the cube root of the perfect cube number by Vedic methods.

68921


Answer:

So finding the cube root by Vedic method,


five digits are in the number, so the digits in its cube root will be 2.



the quotient is 41, so the cube root is 41.


I. 68 - = 4


II. New dividend = 49.


III. Divide the new dividend by 3 4 = 48


IV. Write above quotient digit = 1. Subtract 3 4 1. Remainder = 12. New Dividend = 12.


V. Subtract 3 4 12 = 864. Subtract 12 and the remainder is 12.


VI. Again new dividend = = 1.


VII. New dividend = 1 - = 0


VIII. Therefore cube root = 41.



Question 10.

Find the cube root of the perfect cube number by Vedic methods.

636056


Answer:

So finding the cube root by Vedic method,


six digits are in the number, so the digits in its cube root will be 2.



the quotient is 86, so the cube root is 86.


I. 636 - = 124


II. New dividend = 1240.


III. Divide the new dividend by 3 8 = 192


IV. Write above quotient digit = 6. Subtract 3 8 6. Remainder = 88. New Dividend = 885.


V. Subtract 3 8 62 = 864. Subtract 864 and the remainder is 21.


VI. Again new dividend = = 216.


VII. New dividend = 216 - = 0


VIII. Therefore cube root = 86.



Question 11.

Find the cube root of the perfect cube number by Vedic methods.

314432


Answer:

So finding the cube root by Vedic method,


six digits are in the number, so the digits in its cube root will be 2



the quotient is 68, so the cube root is 68.


I. 314 - = 98


II. New dividend = 984.


III. Divide the new dividend by 3 6 = 108


IV. Write above quotient digit = 8. Subtract 3 6 8. Remainder = 120. New Dividend = 1203.


V. Subtract 3 6 82 = 1152. Subtract 1152 and the remainder is 51.


VI. Again new dividend = = 512.


VII. New dividend = 512 - = 0


VIII. Therefore cube root = 68.



Question 12.

Find the cube root of the perfect cube number by Vedic methods.

493039


Answer:

So finding the cube root by Vedic method,


three digits are in the number, so the digits in its cube root will be 2



the quotient is 79, so the cube root is 79.


I. 493 - = 15


II. New dividend = 1500.


III. Divide the new dividend by 3 7 = 147


IV. Write above quotient digit = 7. Subtract 3 7 9. Remainder = 177. New Dividend = 1773.


V. Subtract 3 7 92 = 1701. Subtract 1701 and the remainder is 72.


VI. Again new dividend = = 729.


VII. New dividend = 729 - = 0


VIII. Therefore cube root = 79.



Question 13.

Find the cube root of the perfect cube number by Vedic methods.

8365427


Answer:

So finding the cube root by Vedic method,


seven digits are in the number, so the digits in its cube root will be 3.



the quotient is 203, so the cube root is 203.


I. 8 - = 0


II. New dividend = 36.


III. Divide the new dividend by 3 2 = 12


IV. Write above quotient digit = 0. Subtract 3 2 0. Remainder = 3. New Dividend = 36.


V. Subtract 3 2 02 = 0. Subtract 0 and the remainder is 36.


VI. Again new dividend = = 27.


VII. New dividend = 27 - = 0


VIII. Therefore cube root = 203.



Question 14.

Find the cube root of the perfect cube number by Vedic methods.

1061208


Answer:

So finding the cube root by Vedic method,


seven digits are in the number, so the digits in its cube root will be 3.



the quotient is 102, so the cube root is 102.


I. 1 -1 = 0


II. New dividend = 00.


III. Divide the new dividend by 3 1 = 3


IV. Write above quotient digit = 1. Subtract 3 1 0. Remainder = 0. New Dividend = 6.


V. Subtract 3 1 02 = 0. Subtract 0 and the remainder is 6.


VI. Again new dividend = = 8


VII. New dividend = 512 - = 0


VIII. Therefore cube root = 102.



Question 15.

Find the cube root of the perfect cube number by Vedic methods.

8489664


Answer:

So finding the cube root by Vedic method,


seven digits are in the number, so the digits in its cube root will be 3.



the quotient is 204, so the cube root is 204.


I. 8 - = 0


II. New dividend = 4.


III. Divide the new dividend by 3 2 = 12


IV. Write above quotient digit = 2. Subtract 3 2 0. Remainder = 4. New Dividend = 48.


V. Subtract 3 2 02 = 0. Subtract 0 and the remainder is 48.


VI. Again new dividend = = 64.


VII. New dividend = 64 - = 0


VIII. Therefore cube root = 204.



Question 16.

Find the cube root of the perfect cube number by Vedic methods.

200201625


Answer:

So finding the cube root by Vedic method,


nine digits are in the number, so the digits in its cube root will be 3.



the quotient is 585 so the cube root is 585.


I. 200 - = 75


II. New dividend = 752.


III. Divide the new dividend by 3 5 = 75


IV. Write above quotient digit = 5. Subtract 3 5 8. Remainder = 152. New Dividend = 1520.


V. Subtract 3 5 82 = 960. Subtract 560 and the remainder is 5601.


VI. Again new dividend = = 125.


VII. New dividend = 125 - = 0


VIII. Therefore cube root = 585.



Question 17.

Find the cube root of the perfect cube number by Vedic methods.

258474853


Answer:

So finding the cube root by Vedic method,


nine digits are in the number, so the digits in its cube root will be 3.



the quotient is 637 so the cube root is 637.


I. 258 - = 42


II. New dividend = 424.


III. Divide the new dividend by 3 6 = 108


IV. Write above quotient digit = 6. Subtract 3 6 3. Remainder = 100. New Dividend = 1007.


V. Subtract 3 6 32 = 162. Subtract 162 and the remainder is 845.


VI. Again new dividend = = 343.


VII. New dividend = 343 - = 0


VIII. Therefore cube root = 637.



Question 18.

Find the cube root of the perfect cube number by Vedic methods.

22665187


Answer:

So finding the cube root by Vedic method,


eight digits are in the number, so the digits in its cube root will be 3.



the quotient is 283 so the cube root is 283.


I. 22 - = 14


II. New dividend = 146.


III. Divide the new dividend by 3 2 = 12


IV. Write above quotient digit = 2. Subtract 3 2 8. Remainder =50. New Dividend = 506.


V. Subtract 3 2 82 = 384. Subtract 384 and the remainder is 122.


VI. Again new dividend = = 27.


VII. New dividend = 27 - = 0


VIII. Therefore cube root = 283.



Question 19.

Find the cube root of the perfect cube number by Vedic methods.

8615125


Answer:

So finding the cube root by Vedic method,


seven digits are in the number, so the digits in its cube root will be 3.



the quotient is 205 so the cube root is 205.


I. 8 - = 0


II. New dividend = 6.


III. Divide the new dividend by 3 2 = 12


IV. Write above quotient digit = 2. Subtract 3 2 0. Remainder = 0. New Dividend = 615.


V. Subtract 3 2 02 = 0. Subtract 0 and the remainder is 615.


VI. Again new dividend = = 125.


VII. New dividend = 125 - = 0


VIII. Therefore cube root = 205.



Question 20.

Find the cube root of the perfect cube number by Vedic methods.

660776311


Answer:

So finding the cube root by Vedic method,


nine digits are in the number, so the digits in its cube root will be 3.



the quotient is 871 so the cube root is 871.


I. 660 - = 148


II. New dividend = 1487.


III. Divide the new dividend by 3 8 = 72


IV. Write above quotient digit = 8. Subtract 3 8 7. Remainder = 143. New Dividend = 1437.


V. Subtract 3 8 72 = 1176. Subtract 1176 and the remainder is 261.


VI. Again new dividend = = 1.


VII. New dividend = 1 - = 0


VIII. Therefore cube root = 871.




Exercise 1.4
Question 1.

Find the oral solution of the equation by the formula paravartya Yojayet.

13 x – 14 = 9x + 10


Answer:

As per the formula of Parvartya Yojayet, its application is used as –


ax + b = cx + d



So comparing the equations,


a = 13, b = -14, c = 9 & d = 10



x = 6



Question 2.

Find the oral solution of the equation by the formula paravartya Yojayet.

3y + 4 = 5y – 4


Answer:

As per the formula of Parvartya Yojayet, its application is used as –


ax + b = cx + d



So comparing the equations,


a = 3, b = 4, c = 5 & d = -4



x = 4



Question 3.

Find the oral solution of the equation by the formula paravartya Yojayet.




Answer:

As per the formula of Parvartya Yojayet, its application is used as –





So comparing the equations,


a = 2, b = 1, c = 3, d = 4, p = 1 & q = 3





Question 4.

Find the oral solution of the equation by the formula paravartya Yojayet.




Answer:

As per the formula of Parvartya Yojayet, its application is used as –




So comparing the equations,


a=5, b=-3, c=2, d=1, p=2 & q=5





Question 5.

Find the oral solution of the equation by the formula paravartya Yojayet.

(x + 7) (x + 9) = (x – 8) (x – 11)


Answer:

As per the formula of Parvartya Yojayet, its application is used as –


(x + a) (x + b) = (x + c) (x + d)



So comparing the equations,


a = 7, b = 9, c = -8 & d = -11





Question 6.

Find the oral solution of the equation by the formula paravartya Yojayet.

(x + 5) (x + 1) = (x + 3) (x + 2)


Answer:

As per the formula of Parvartya Yojayet, its application is used as –


(x + a) (x + b) = (x + c) (x + d)



So comparing the equations,


a=5, b=1, c=3 & d=2



x = 1



Question 7.

Find the oral solution of the equation by the formula paravartya Yojayet.




Answer:

As per the formula of Parvartya Yojayet, its application is used as –




So comparing the equations,


a = 1, b = -1, m = 1 & n = -2



x = -3



Question 8.

Find the oral solution of the equation by the formula paravartya Yojayet.




Answer:

As per the formula of Parvartya Yojayet, its application is used as –



Dividing the part and second part of the equation by 2 & 3 respectively.




So comparing the equations,


, , m = 2.5 & n = -3





Question 9.

Solve the equation by formula Sunyam Samyasamuchchaye.

(2x + 1) + (x + 3) = 5x + 4


Answer:

As per the formula of Sunyam Samaysamuchchaye, its second application is used –


If the independent terms of both the sides of the equation are equal, then the value of variable quantity is zero.


the value of independent term is 4.


x = 0



Question 10.

Solve the equation by formula Sunyam Samyasamuchchaye.

a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1)


Answer:

As per the formula of Sunyam Samaysamuchchaye, its second application is used –


If the independent terms of both the sides of the equation are equal, then the value of variable quantity is zero.


x - 1 = 0


x = 1



Question 11.

Solve the equation by formula Sunyam Samyasamuchchaye.

(x + 1) (x + 9) = (x + 3) (x + 3)


Answer:

As per the formula of Sunyam Samaysamuchchaye, its second application is used –


If the independent terms of both the sides of the equation are equal, then the value of variable quantity is zero.


the value of independent term is 4.


x = 0



Question 12.

Solve the equation by formula Sunyam Samyasamuchchaye.




Answer:

As per the formula of Sunyam Samaysamuchchaye, its first application is used –


If in each term’x’ is the common factor, then the value of ;x’ is zero.


the value of independent term is 4.


x = 0



Question 13.

Solve the equation by formula Sunyam Samyasamuchchaye.




Answer:

As per the formula of Sunyam Samaysamuchchaye, its third application is used –


If the numerators of two fractions in the equation are equal i.e., 1.




So comparing the equations,


a = 4 & b = -6



x = 1



Question 14.

Solve the equation by formula Sunyam Samyasamuchchaye.




Answer:

As per the formula of Sunyam Samaysamuchchaye, its third application is used –


If the numerators of two fractions in the equation are equal i.e., 5.



x + a + x + b = 0


So adding the denominators,


3x + 2 + 2x + 8 = 0


5x = -10


x = -2



Question 15.

Solve the equation by formula Sunyam Samyasamuchchaye.




Answer:

As per the formula of Sunyam Samaysamuchchaye, its fourth application is used –


If the sum of the numerators & the sum of the denominators of the equation are mutually equal.


So adding the denominators,


2x + 4 + 2x + 1 = 0


4x = -5




Question 16.

Solve the equation by formula Sunyam Samyasamuchchaye.




Answer:

As per the formula of Sunyam Samaysamuchchaye, its fourth application is used –


The question given is incorrect as the correct question is , (GIVEN IN REFERNEC MATERIAL)


If the sum of the numerators & the sum of the denominators of the equation are mutually equal.


So adding the denominators,


3x + 2 + x + 1 = 0 5x + 7 + 3x – 1 = 0


4x + 3 = 0 8x + 6 = 0




Question 17.

Solve the equation by formula Sunyam Samyasamuchchaye.




Answer:

As per the formula of Sunyam Samaysamuchchaye, its fifth application is used –


If the difference of the numerators & denominators of both the part of the equation are equal.


So adding the denominators,


5x + 7 - 2x - 1 x + 1 - 3x - 5


3x + 6 -2x - 4


Equating both the equations to zero,


3x + 6 = 0 -2x – 4 = 0


x = -2 x = -2



Question 18.

Solve the equation by formula Sunyam Samyasamuchchaye.




Answer:

As per the formula of Sunyam Samaysamuchchaye, its fifth application is used –


If the difference of the numerators & denominators of both the part of the equation are equal.


So adding the denominators,


3x + 6 - 6x - 3 5x + 4 - 2x - 7


-3x + 3 3x - 3


Equating both the equations to zero,


x = 1



Question 19.

Solve the equation by formula Sunyam Samyasamuchchaye.




Answer:

As per the formula of Sunyam Samaysamuchchaye, its sixth application is used –


If there are two terms in each side of an equation and each of the numerators of the terms are mutually equal and the sum of denominators on the left side is equal to the sum of denominators on the right side, then putting this sum equal to zero, the value of variable quantity is obtained.


So adding the denominators on LHS & RHS,


x + 2 + x + 6 x + 1 + x + 7


2x + 8 2x + 8


2x + 8 = 0


x = -4



Question 20.

Solve the equation by formula Sunyam Samyasamuchchaye.




Answer:

As per the formula of Sunyam Samaysamuchchaye, its sixth application is used –


If there are two terms in each side of an equation and each of the numerators of the terms are mutually equal and the sum of denominators on the left side is equal to the sum of denominators on the right side, then putting this sum equal to zero, the value of variable quantity is obtained.


So adding the denominators on LHS & RHS,


x – 4 + x - 6 x – 2 + x - 8


2x - 10 2x - 10


2x – 10 = 0


x = 5