Vedic maths questions with solutions and elaborated explanations are provided here for practice. Students must practice these questions based on the Vedic maths sutras to develop mental maths skills, which is essential for any examinations.
There are sixteen sutras and their sub-sutras (corollaries) in Vedic mathematics which are used for almost every type of mathematical calculation.
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1. Ekadhikena Purvena (also a corollary) |
1. Anurupyena |
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2. Nikhilam Navatascaraman Dasatah |
2. Sisyate Sesasam |
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3. Urdhva-tiryagbhyam |
3. Adyamadyenantya-mantyena |
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4. Paravartya Yojayet |
4. Kevalaih Saptakam Gunyit |
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5. Sunyam Samyasamuccaye |
5. Vestanam |
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6. (Anurupye) Sunyamanyat |
6. Yavadunam Tavadunam |
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7. Sankalana-vyavakalanabhyam |
7. Yavadunam Tavadunikrtya Varganca Yojayet |
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8. Puranapuranabhyam |
8. Antyayordasakepi |
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9. Calana-Kalanabhyam |
9. Antyayoreva |
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10. Yavadunam |
10. Samuccayagunitah |
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11. Vyastisamastih |
11. Lopanasthapanabhyam |
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12. Sesanyankena Caramena |
12. Vilokanam |
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13. Sopantyadvayamamtyam |
13. Gunitasamuccayah Samuccayagunitah |
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14. Ekanyunena Purvena |
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15. Gunitasamuccayah |
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16. Gunakasamuccayah |
Multiplication by Nikhilam Sutra:
The Vedic sutra – “Nikhilam Navatascaraman Dasatah” means “all from nine and the last from ten ”
Let us take an example of 1-digit number 7 × 9, both the numbers are close to 10. 7 is 3 less than 10 and 9 1 less than 10.
- 7 | – 3, 7 is 3 less than 10
- 9 | –1, 9 is 1 less than 10
- Take either 7 – 1 = 6 or 9 – 3= 6, this will make the first part of the product
- Multiply (-3) by (-1), –3 × ( –1) = 3
∴ 7 × 9 = 63
Any number can be multiplied by this sutra with few modifications, let us take another example 13 × 7
- 13 | + 3, 13 is 3 more than 10
- 7 | – 3, 7 is 3 less than 10
- Take either 13 – 3 = 10 or 7 + 3 = 10 as the first half of the product
- Now 3 × ( –3) = –9, since the other half is negative number
- Multiply the half of the product by 10, 10 × 10 = 100
- Subtract 9 from it, 100 – 9 = 91
∴ 13 × 7 = 91
If numbers are bigger and closer to 100, for example 98 × 91
- 98 | –02, 98 is two less than 100
- 91 | –09, 91 is 9 less than 100
- Take 98 – 09 = 89 or 91 – 02 = 89, as the first half of the number
- (–02) × (–09) = 18. This is the other half of the number
∴ 98 × 91 = 8918.
We shall use more sutras while solving problems.
Vedic Maths Questions With Solutions
Question 1:
Solve the following using Nikhilam sutra:
(i) 12 × 13
(ii) 9 × 8
(iii) 16 × 11
Solution:
(i) 12 × 13
- 12 | +2
- 13 | +3
- Taking 12 + 3 = 15
- 3 × 2 = 6
∴ 12 × 13 = 156.
(ii) 9 × 8
- 9 | –1
- 8 | –2
- Taking 9 – 2 = 7
- –1 × –2 = 2
∴ 9 × 8 = 72.
(iii) 16 × 11
- 16 | +6
- 11 | +1
- Taking 16 + 1 = 17
- 6 × 1 = 6
∴ 16 × 11 = 176
Question 2:
Solve the following using Nikhilam sutra:
(i) 97 × 89
(ii) 91 × 92
(iii) 99 × 101
Solution:
(i) 97 × 89
- 97 | –03
- 89 | –11
- Taking 89 – 03 = 86
- –03 × –11 = 33
∴ 97 × 89 = 8633
(ii) 91 × 92
- 91 | –09
- 92 | –08
- Taking 91 – 08 = 83
- –9 × –8 = 72
∴ 91 × 92 = 8372
(iii) 99 × 101
- 99 | –01
- 101 | +01
- 101 – 01 = 100
- –1 × 1 = –1
- 100 × 100 = 10000
- 10000 – 1 = 9999
∴ 99 × 101 = 9999.
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Subtraction by Nikhilam sutra: To subtract any number by 10, 100, 1000, 10000, … etc, we use this sutra – “all from nine and the last from ten ” For example, 10000 – 7456, we only subtract the unit place digit by 10, and rest of all by 9 9 – 7 = 2 9 – 4 = 5 9 – 5 = 4 10 – 6 = 4 ∴ 10000 – 7456 = 2544. |
Also refer:
Question 3:
Solve the following using Nikhilam sutra:
(i) 1000 – 62
(ii) 1 – 0.654
(iii) 80.000 – 0.7312
Solution:
(i) 1000 – 62
We take 1000 – 062
9 – 0 = 9
9 – 6 = 3
10 – 2 = 8
∴ 1000 – 62 = 938
(ii) 1 – 0.654
We take 1.000 – 0.654
9 – 6 = 3
9 – 5 = 4
10 – 4 = 6
∴ 1 – 0.654 = 0. 346
(iii) 80.000 – 0.7312
The main operation here is 10000 – 7312 where one is borrowed from the left number
9 – 7 = 2
9 – 3 = 6
9 – 1 = 8
10 – 2 = 8
Since 1 is borrowed, 80 – 1 = 79
∴ 80.000 – 0.7312 = 79.2682
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Multiplication by Anurupyena: When the numbers given for multiplication are not close to any 10n (n = 1, 2, 3, ..), in that case, we use a corollary sutra, “Anurupyena” means “Proportionately”. For example, 41 × 49, these numbers are closer to 50, which is a multiple of 10. Now doing as previous,
∴ 41 × 49 = 2009 |
Question 4:
Solve the following by the Nikhlam sutra:
(i) 23 × 25
(ii) 46 × 49
(iii) 65 × 69
Solution:
(i) 23 × 25
- 23 | –7
- 25 | –5
- Taking 25 – 7 = 18
- –7 × –5 = 35
- Since 10 × 3 = 30, we multiply just the first part of the number by 3, 3 × 18 = 54
∴ 23 × 25 = 54+ 35 = 575.
(ii) 46 × 49
- 46 | –4
- 49 | –1
- Taking 46 – 1 = 45
- –4 × –1 = 4
- Since 10 × 5 = 50, we multiply just the first part of the number by 5, 5 × 45 = 225
∴ 46 × 49 = 2254
(iii) 65 × 69
- 65 | –5
- 69 | –1
- Taking 65 – 1 = 64
- –5 × –1 = 5
- Since 10 × 7 = 70, we multiply just the first part of the number by 7, 7 × 64 = 448
∴ 65 × 69 = 4485.
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Multiplication by Urdhva-tiryagbhyam: Another Vedic maths sutra, “Urdhva-tiryagbhyam”, means “Vertically and crosswise”. To see the application let us take an example 34 × 12. The Nikhilam sutra cannot solve this.
∴ 34 × 12 = 408. This method could be followed both from right to left and left to right as per convenience. |
Question 5:
Evaluate:
(i) 23 × 33
(ii) 12 × 26
(iii) 34 × 23
Solution:
(i) 23 × 33
- 3 × 3 = 9
- (2 × 3) + (3 × 3) = 6 + 9 = 15; 5 carry 1
- 2 × 3 = 6
∴ 23 × 33 = 759
(ii) 12 × 26
- 2 × 6 = 12; 2 carry 1
- (1 × 6) + (2 × 2) = 6 + 4 = 10; 0 carry 1
- 1 × 2 = 2
∴ 12 × 26 = 312
(iii) 34 × 23
- 4 × 3 = 12; 2 carry 1
- (4 × 2) + (3 × 3) = 8 + 9 = 17; 7 carry 1
- 3 × 2 = 6
∴ 34 × 23 = 782
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Duplex Method of Squaring Numbers: ‘Dwanda’ or ‘Duplex’ is denoted by D.
Let us see how use this to find square of a number: (ab)2 = D(a)/D(ab)/D(b) (abc)2 = D(a)/D(ab)/D(abc)/D(bc)/D(c) For example, we have to find 352 352 = D(3)/D(35)/D(5) = 32/2.3.5/52 = 9/30/25 = 9/32/5….keep 5 and carry over 2 = 12/2/5….keep 2 and carry over 3 ∴ 352 = 1225 |
Question 6:
Find the square of the following:
(i) 27
(ii) 78
(iii) 45
Solution:
(i) 27
272 = D(2)/D(27)/D(7)
= 4/28/49
Passing on the double digits, we get,
= 4/32/9
= 7/2/9
∴ 272 = 729.
(ii) 78
782 = D(7)/D(78)/D(8)
= 49/112/64
Passing on the double digits, we get,
= 49/118/4
= 60/8/4
∴ 782 = 6084.
(iii) 45
452 = D(4)/D(45)/D(5)
= 16/40/25
Passing on the double digits, we get,
= 16/42/5
= 20/2/5
∴ 452 = 2025.
Question 7:
Find the squares of the following numbers:
(i) 234
(ii) 123
(iii) 168
Solution:
(i) 234
2342 = D(2)/D(23)/D(234)/D(34)/D(4)
= 4/12/25/24/16
Passing on the double digits, we get,
= 4/12/25/25/6
= 4/12/27/5/6
= 4/14/7/5/6
= 5/4/7/5/6
∴ 2342 = 54756
(ii) 123
1232 = D(1)/D(12)/D(123)/D(23)/D(3)
= 1/4/10/12/9
Passing on the double digits, we get,
= 1/5/1/2/9
∴ 1232 = 15129
(iii) 168
1682 = D(1)/D(16)/D(168)/D(68)/D(8)
= 1/12/52/96/64
Passing on the double digits, we get
= 1/12/52/102/4
= 1/12/62/2/4
= 1/18/2/2/4
= 2/8/2/2/4
∴ 1682 = 28224.
Also Read:
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Factorisation by Anurupyena: Let us take an example to understand this sutra: 2x2 + 5x + 2
Here 5 = 4 + 1, such that 2 : 4 = 1 : 2 and 1 : 2 = 1 : 2
2x2 + 2 = 2x + 1 x 2 ∴ 2x2 + 5x + 2 = (x + 2) (2x + 1) |
Question 8:
Factorise the following:
(i) 2x2 + 5x – 3
(ii) 3x2 + x – 14
(iii) 7x2 – 6x – 1
Solution:
(i) 2x2 + 5x – 3
5 = 6 – 1
Such that 2 : 6 = 1 : 3 and –1 : –3 = 1 : 3
∴ the first factor is (x + 3)
For the second factor,
2x2 – 3 = (2x – 1)
x 3
∴ 2x2 + 5x – 3 = (x + 3)(2x – 1)
(ii) 3x2 + x – 14
x = 7 – 6
Such that 3 : (–6) = 1: –2 and 7: –14 = 1: –2
∴ the first factor is x – 2
For the second factor,
3x2 – 14 = (3x + 7)
x –2
∴ 3x2 + x – 14 = (x – 2)(3x + 7)
(iii) 7x2 – 6x – 1
–6x = 1 – 7
Such that 7 : –7 = 1: (–1) and 1 : ( –1)
∴ x – 1 is the first factor
For the second factor,
7x2 – 1 = 7x + 1
x –1
∴ 7x2 – 6x – 1 = (x – 1)(7x + 1)
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Division by Ekadhikena Purvena: “Ekadhikena purvena” simply means “by one more than the previous one”. We will use this technique to perform a division whose divisor ends with 9. Steps are
For example 6/19
6/2 → 0.3 3/2 → 0.311 (sub 1 is the remainder) 11/2 → 0.3115 (sub 1 is the remainder) 15/2 → 0.31517 (sub 1 is the remainder) 17/2 → 0.315718 (sub 1 is the remainder) 18/2 → 0.315789 9/2 → 0.31578914 (sub 1 is the remainder) 14/2 → 0.31578947 ∴ 6/19 = 0.31578947 |
Question 9:
Evaluate using the sutra Ekadhikena Purvena:
(i) 22/39
(ii) 12/29
(iii) 3/9
Solution:
(i) 22/39
- Adding 1 to the denominator, 39 + 1 = 40
- We have to divide 22 by 40
- Removing zero and placing decimal, 2.2/4
- Carrying out step-by-step division
22/4 → 0.25 (sub 2 is the remainder)
25/4 → 0.516 (sub 1 is the remainder)
16/4 → 0.564
4/4 → 0.5641
¼ → 0.564110 (sub 1 is the remainder)
10/4 → 0.5641022 (sub 2 is the remainder)
22/4 → 0.56410225 (sub 2 is the remainder)
25/4 → 0.564102516 (sub 1 is the remainder)
16/4 → 0.56410256
∴ 22/39 = 0.56410256
(ii) 12/29
- Adding 1 to the denominator, 29 + 1 = 30
- We have to divide 12 by 30
- Removing zero and placing decimal, 1.2/3
- Carrying out step-by-step division
12/3 → 0.4
4/3 → 0.411 (sub 1 is the remainder)
11/3 → 0.4123 (sub 2 is the remainder)
23/3 → 0.41327 (sub 2 is the remainder)
27/3 → 0.41379
9/3 → 0.413793
3/3 → 0.4137931
∴ 12/29 = 0.4137931
(iii) 3/9
- Adding 1 to the denominator, 9 + 1 = 10
- We have to divide 3 by 10
- Removing zero and placing decimal, 0.3/1
- Carrying out step-by-step division
3/1 → 0.3
3/9 = 0.333333…
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Divisors ending with 1: Steps of division:
For example, 4/21
3/2 → 0.11 subtract quotient 1 from 9, we get 0.18 18/2 → 0.19 subtract quotient 9 from 9, we get 0.0 0/2 → 0.190 subtract quotient 0 from 9, we get 0.9 9/2 → 0.19014 subtract quotient 4 from 9, we get 0.15 15/2 → 0.190417 subtract quotient 7 from 9, we get 0.12 12/2 → 0.190476 subtract quotient 6 from 9, we get 0.3 3/2 → 0.19047611 subtract quotient 1 from 9, we get 0.18 and so on. ∴ 4/21 = 0.1904761 |
Question 10:
Evaluate the following (up to 4 decimal point):
(i) 12/21
(ii) 7/11
(iii) 22/31
Solution:
(i) 12/21
- Subtract 1 from numerator and denominator
- We have 11/20
- Removing the zero and placing the decimal, 1.1/2
- Carrying out step-by-step division
11/2 → 0.15 subtract quotient 5 from 9, we get 0.14
14/2 → 0.57 subtract quotient 7 from 9, we get 0.2
2/2 → 0.571 subtract quotient 1 from 9, we get 0.8
8/2 → 0.5714
∴ 12/21 = 0.5714.
(ii) 7/11
- Subtract 1 from the numerator and denominator
- We have 6/10
- Removing the zero and placing the decimal, 0.6/1
- Carrying out step-by-step division
6/1 → 0. 6 subtract quotient 6 from 9, we get 0.3
3/1 → 0.63 subtract quotient 3 from 9, we get 0.6
6/1 → 0.636 subtract quotient 6 from 9, we get 0.3
3/1 → 0.6363
∴ 7/11= 0.6363…
(iii) 22/31
- Subtract 1 from the numerator and denominator
- We have 21/30
- Removing the zero and placing the decimal, 2.1/3
- Carrying out step-by-step division
21/3 → 0.7 subtract quotient 7 from 9, we get 0.2
2/3 → 0.720 subtract quotient 0 from 9, we get 0.29
29/3 → 0.7029 subtract quotient 9 from 9, we get 0.20
20/3 → 0.70926
∴ 22/31 = 0.7096
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Practice Questions on Vedic Maths
1. Evaluate the following:
(i) 18 × 11
(ii) 98 × 97
(iii) 15 × 14
(iv) 25 × 31
(v) 46 × 48
(vi) 88 × 87
2. Evaluate the following:
(i) 23 × 18
(ii) 28 × 17
(iii) 18 × 21
(iv) 34 × 96
3. Evaluate the following:
(i) 100 – 78
(ii) 1000 – 378
(iii) 10000 – 7786
(iv) 8 – 0.562
4. Evaluate the following:
(i) 12/19
(ii) 23/39
(iii) 8/21
(iv) 14/31
(v) 23/41
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