RBSE Class 5 Maths Chapter 7 Solutions Magic of Maths

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RBSE Class 5 Maths Chapter 7 Question Answer Magic of Maths

Magic of Maths Class 5 Questions and Answers

Magic of Maths Class 5 Question Answer – InText

Page – 54

Question 1.
Add the following by ekadhiken purven :

Solution :

Question 2.
Subtract by Vedic method :

(i)

Answer:
Hint :

1. Subtract 6 from 6 and we get 0.
2. Param mitra digit of 2 is 8, we add 8 to 1 and write 9.
3. We put eknunen sign below 8 and write 8 i.e. 7.
4. Subtract 3 from 7 and write 4.

(ii)

Answer:
Hint :

1. Param mitra digit of 9 is 1, we add 1 to 5 and write 6.
2. We put eknunen sign below 3 and write 3 i.e. 2.
3. Param mitra digit of 5 is 5, we add 5 to 2 and write 7.
4. Put eknunen sign below 9 and write 9 i.e. 8.
5. Subtract 4 from 8 and write 4.

(iii)

Answer:
Hint :

1. Param mitra digit of 9 is 1, add 1 to 5 and write 6.
2. Put eknunen sign below 5 and write 5 i.e. 4.
3. Param mitra digit of 7 is 3, add 3 to 4 and write 7.
4. Put eknunen sign below 3 and write 3.
5. 3 means 2, subtract 1 from 2 and get 1.

(iv)

Answer:
Hint :

1. Param mitra digit of 4 is 6, we add 6 to 2 and get 8.
2. Put eknunen sign below 3, which is previous to 2. 3 means 2 and param mitra digit of 7 is 3. We add 2 to 3 and get 5.
3. We put eknunen sign below 6 and write 6.6 means 5, subtract 2 from 5 and get 3.

Page – 55

Question 3.
Find deviation :z

Solution:

Page – 58

Question 4.
Multiply using Nikhilam sutra (base 10 or 100) :
(i) 11 × 11
(ii) 12 × 8
(iii) 101 × 105
(iv) 102 × 98
(v) 93 × 97
(vi) 9 × 7
(vii)
(viii)
(ix)
Solution :
(i) 11 × 11

Step :
1. Base 10, deviation + 1 and + 1
2. Product of deviations = 1 × 1 = 1
3. One number + deviation of other number = 11 + 1 = 12
4. 11 + 1 / 1 × 1
5. 121

(ii) 12 × 8

Step :
1. Base 10, deviation + 2 and – 2
2. Product of deviation = (+ 2) × (- 2) = – 4
3. One number + deviation of other number = 12 + (- 2) = 10
4. 10 / (- 4)
5. For changing negative number of right side to positive number we take one carry over from left side to right side.
6. Required solution = (10 – 1) / 10 – 4 = 96

(iii) 101 × 105

Step :

1. Base 100, deviation + 1 and + 5
2. Product of deviation = (+ 1) × (+ 5) = 5 = 05
3. One number + deviation of other number = 101 + 5 = 106
4. 106 / 05
5. 10605

RBSE Class 5th Maths Chapter 7 Question Answer – Practice Work

Question 1.
Write deviation on the base of 10 :
(i) of 12
Solution :
Deviation of 12 = 12 – 10 = 2

(ii) of 18
Solution :
Deviation of 18 = 18 – 10 = 8

(iii) of 6
Solution :
Deviation of 6 = 6 – 10 = – 4

(iv) of 9
Solution :
Deviation of 9 = 9 – 10 = – 1

Question 2.
Write ekadhiken purven :
(i) 89 of 9
Answer:
Ekadhiken purven 89 of 9 = 89 = 99

(ii) 125 of 2
Answer:
Ekadhiken purven 125 of 2 = 125 = 225

(iii) 76 of 7
Answer:
Ekadhiken purven 76 of 7 = 076 = 176

(iv) 703 of 3
Answer:
Ekadhiken purven 703 of 3 = 703 = 713

Question 3.
Write ekadhik.
(i) of 8
(ii) of 12
(iii) of 17
(iv) of 37
Solution :
(i) 8 = 9
(ii) 12 = 13
(iii) 17 = 18
(iv) 37 = 38

Question 4.
Write deviation on the base of 100 :
(i) of 109
(ii) of 115
(iii) of 112
(iv) of 98
Solution :
(i) of 109 = 109 – 100 = + 9
(ii) of 115 = 115 – 100 = + 15
(iii) of 112 = 112 – 100 = + 12
(iv) of 98 = 98 – 100 = – 2

Question 5.
Write Param mitra digit of the numbers :
(i) 5
(ii) 7
(iii) 8
(iv) 6
Solution :
(i) Param mitra digit of 5 = 5(5 + 5 = 10)
(ii) Param mitra digit of 7 = 3(7 + 3 = 10)
(iii) Param mitra digit of 8 = 2(8 + 2 = 10)
(iv) Param mitra digit of 6 = 4(6 + 4 = 10)

Question 6.
Write one Nuenen purven :
(i) in number 748 of 8
Answer:
One nuenen purven in 748 of 8
= 748 = 738

(ii) in number 68 of 8
Answer:
One nuenen purven in 68 of 8 = 68 = 58

(iii) in number 7432 of 2
Answer:
One nuenen purven in 7432 of 2 = 7432 = 7422

(iv) in number 843 of 4
Answer:
One nuenen purven in 843 of 4 = 843 = 743

Question 7.
Addition with ekadhiken purven :

Solution :
(i)

Hint :

1. Total of unit digit
2. Here, total is more than 10, we write unit digit in sum and mark ekadhik symbol on digit before unit digit.
3. Total of tens digit
4. Total of hundreds digit

(ii)

Hint :

1. Total of unit digit = 4 + 6 = 10
2. Total is 10, we write unit digit in sum and mark ekadhik symbol on digit 6.
3. Total of tens digit = 9 + 6 = 9 + 7 = 16
4. Sum is more than 10, we write unit digit in sum and mark ekadhik symbol on digit 2.
5. Total of hundreds unit = 8 + 2 = 8 + 3 = 11

(iii)

Hint :

1. Total of unit digit = 4 + 7 = 11
2. Here, total is more than 10, we write unit digit in sum and mark ekadhik symbol on digit 2.
3. Total of tens digit = 8 + 2 = 8 + 3 = 11
4. Again, total is more than 10, we write unit digit in sum and mark ekadhik symbol on digit 3.
5. Total of hundreds digit = 7 + 3 = 7 + 4 = 11

(iv)

Hint :
As mentioned earlier.

Question 8.
Substract with the help of nuenen purven and param mitra digit :

Solution :
(i)

Hint :

1. We add param mitra digit of 6 i.e. 4 to 5 and get 9.
2. We mark eknuen sign below 9.
3. 9 means 8, we subtract 3 from 8 and get 5.

(ii)

Hint :

1. We add param mitra digit of 4 i.e. 6 to 3 and get 9.
2. We mark eknuen sign below 4.
3. 4 means 3, we subtract 2 from 3 and get 1.
4. We subtract 6 from 7 and get 1.

(iii)

Hint :

1. Param mitra digit of 5 is 5, we add 5 to 4 and get 9.
2. We mark eknuen sign below 3.
3. 3 means 2 and param mitra digit of 4 is 6, we add 2 to 6 and get 8. Mark Ekuen sign below 6.
4. 6 = 6 – 1 = 5, we subtract 2 from 5 and get 3.

(iv)

Hint :

1. Param mitra digit of 6 is 4, we add 4 to 5 and get 9.
2. We make eknuen sign below 2, 2 = 1.
3. Param mitra digit of 3 is 7, we add 7 to 1 and get 8.
4. Eknuen purven of 2 is ? i.e. we get 2.
5. Subtract 2 from 2 and get 0.
6. Subtract 1 from 4 and get 3.

Question 9.
Multiplication with nuenen purven mehod :
(i) 11 × 99
(ii) 57 × 99
(iii) 325 × 999
(iv) २३ × ९९
(v) ३४२ × ९९९
(vi)
Solution:
(i) 11 × 99
Solution :
= 11 × 99
= 10 / 99
= 10 / 99 – 10 by 99.
= 1089
Step :

1. Eknuen of 11 will 10.
2. 10 which is eknuen of 11 will be subtracted
3. When join both the number we get product.

(ii) 57 × 99
= 57 × 99
= 56 / 99
= 56 / 99 – 56

Step :

1. Eknuen of 57 will be 56.
2. 56 will be subtracted
3. When join both the number we get product.

(iii) 325 × 999
= 325 × 999
= 324 / 999 – 324
= 324/64
= 324675

Step :

1. Eknuen of 325 will be
2. 324 will be subtracted
3. When join both the number we get product.

(iv)

Step :

1. Eknuen of २३ is २२.
2. २२ will be subtracted by ९९, we will get ७७.
3. When join both the number we get product.

(v)

Step :

1. Eknuen of ३४२ will be
2. 389 will be subtracted
3. When join both the

Step :

1. Base = १०० deviation = ७ and ९
2. Product of the deviation = ७ × ९ = ६ ३
3. On left side in १०७ + ९ or १०९ + ७ = ११६
4. On left side we get two + ६३, the required product is १९६६३.

(vi)

Step :

1. Eknuen of ५খ will be ५७
2. ५७ will be subtracted by ९९.
3. When join both the number we get product.

Question 10.
Multiply with nikhilam formula (sutra) :
(i) 117 × 103
(ii) 104 × 92
(iii) 102 × 109
(iv) 95 × 105
(v)
(vi)
Sol. :
(i) 117 × 103

Step :

1. Nearest base =100 Deviation = + 17, + 3
2. Product of deviations = 17 × 3 = 51, we will write it on the right side.
3. 117 + 3 or 103 + 17 = 120, we will write it on the left side.
4. Thus we get multiplication.

(ii) 104 × 92

Step :

1. Base =100, deviation +4 and -8.
2. Product of deviations = 4(-8) = -32
3. As -32 , we will take 1 as carry over from left side to right side number and make it positive left side 96 – 1 = 95, right side 100 – 32 = 68
4. Now combining both part we get 9568 as product.

(iii) 102 × 109

Step :

1. Base = 100, deviations +2,+9
2. Product of the deviations = 2 × 9 = 18, we will write it on the right side.
3. On left side in 102+9 or 109 + 2 = 111.
4. Base 100 has 2 zeros, so on right side we will write 2 digits. And product has 3 digits, we will write 2 digits at righ side.

(iv) 95 × 105

Step :

1. Base =100, dèviation -5 and +5.
2. Product of the deviations = (-5)(5) = -25
3. On left side in 95 + 5 or 105 – 5 = 100
4. Right side is negative, so we will take 1 as carry over from left side then we will get 100 – 25 = 75 there.
5. On left side, we write 100 – 1 = 99.

(v)

Step :

(vi)

Step :

RBSE Class 5 Maths Chapter 7 Magic of Maths Important Questions

Multiple Choice Questions :

Question 1.
Ekadhik of 6 i.e. 6 will be :
(A) 5
(B) 7
(C) 6
(D) 8
Answer:
(D) 8

Question 2.
Eknuenen of 7 i.e. ? will be :
(A) 8
(B) 7
(C) 5
(D) 6
Answer:
(B) 7

Question 3.
Param mitra digit of 8 is :
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(A) 2

Question 4.
Ekadhiken purven in 63 of 6 will be :
(A) 62
(B) 64
(C) 0
(D) 163
Answer:
(D) 163

Question 5.
Eknuen purven in 523 of 2 will be :
(A) 524
(B) 423
(C) 422
(D) 623
Answer:
(B) 423

Question 6.
Param mitra digit of number 1 is :
(A) 9
(B) 0
(C) 8
(D) 3
Answer:
(A) 9

Question 7.
Charam Ank in number 623 is :
(A) 6
(B) 2
(C) 3
(D) 0
Answer:
(C) 3

Fill in the blanks :

(i) The _____ digit of any number is called Charam digit.
Answer: unit

(ii) Ekadhik of 5 is represented as _____.
Answer: 5

(iii) Eknuen of 8 is _____.
Answer: 7

(iv) Ekadhiken purven of digit 3 in number 23 is _____.
Answer: 33

(v) _____ is a method of multiplication in Vedic maths.
Answer: Nikhilam method.

Write whether True or False :

(i) Eknuen purven of digit 7 in 372 is 371.
Answer: False

(ii) Eknuen purven of digit 0 in 5210 is 5200.
Answer: True

(iii) Deviation of 7 on base 10 is – 3.
Answer: True

(iv) Ekadhik of digit 5 is 4.
Answer: False

Very Short Answer Type Questions :

Question 1.
What are the param mitra digits?
Solution :
The two digits whose sum is 10 are called pram mitra digits to each other.

Question 2.
Write Eknuen purven of 0 in number 2710.
Solution :
2710 = 2710 = 2700

Question 3.
What is Charam digit?
Solution :
The unit digit of any number is called Charam digit.

Question 4.
Write deviation of 13 on base 10.
Solution :
Deviation = + 3

Question 5.
What is Nikhilam digit?
Solution :
The digits other than Param digit are called Nikhilam digit.

Question 6.
Write deviation of 8 on base 10.
Solution :
Deviation = – 2

Question 7.
Write ekadhik of 99.
Solution :
Ekadhik of 99 = 9 9 = 99 + 1 = 100

Short Answer Type Questions :

Question 1.
Multiply 11 × 6 by Nikhilam method taking base 10.
Solution :
Finding 11 × 6 by Nikhilam method :

Question 2.
Multiply. by Nikhilam method :
14 × 13
Solution :
14 × 13

Question 3.
Multiply by Vedic maths method :
11 × 15
Solution :
11 × 15

Hint :

1. Nearest base = 10 deviation = + 1 and + 5
2. Product of deviations = 1 × 5 = 5
3. On left side in 11 + 5 = 16 or 15 + 1 = 16

Question 4.
Multiply 9 × 11.
Solution :
9 × 11

Hint :

1. Nearest base = 10
   Deviation = + 1 and – 1
2. Product of deviations = – 1 × 1 = – 1
3. On left side in 9 + 1 = 10 or 11 – 1 = 10
4. To make right side positive we take 1 as carry over and in left side write 1 × 10 = 10
5. Write 10 – 1 = 9 on left side.

Question 5.
Multiply 14 × 17.
Solution :
14 × 17

Hint :

1. Nearest base = 10
   Deviation = + 4 and + 7
2. Product of deviation = 7 × 4 = 28
3. On left side in 17 + 4 = 21 or 14 + 7 = 21
4. There are two digits at left side so one digit will be transferred to right side.
5. We will write 21 + 2 = 23.

Long Answer Type and Essay Type Questions :

Question 1.
Subtract by Eknuen purven method and param digit.
(i) 753 – 584
Solution :

Hint :

1. 4 is not subtracted by 3, param mitra digit of 4 is 6, add 6 to 3 and get sum 9.
2. We write eknuen sign below 5 ((?) which is previous to 3.
3. 8 is not subtracted from 5 = 4, so, we add param mitra digit of 8 i.e. 2 to 4 and get sum 6.
4. We put eknuen sign below 7 (?)
5. 7 = 6 – 5 = 1

(ii) 8321 – 7654
Solution:

Hint :

1. 4 is not subtracted from 1. Param mitra digit of 4 is 6, we add 6 to 1 and write 7. Write eknuen sign below 2.
2. 2 = 1, 5 is not subtracted from 1. Param mitra digit of 5, we add 5 to 1 and write eknuen sign below 3.
3. 3 = 2, 6 is not subtracted from 2. Param mitra digit of 6 is 4. Add 4 to 2 and write eknuen sign below 8.
4. 8 = 7 and 7 – 7 = 0.

Question 2.
Subtract with Vedic method Ekadhiken purven + param mitra digit :
700 – 432
Solution:

Hint :

1. 2 is not subtracted from 0. Param mitra digit of 2 is 8, we add 8 to 0 and get sum 8. Write ekadhik on 3.
2. 3 = 4 is not subtracted from 0. Param mitra digit of 4 is 6, we add 6 to 0 and get 6.
3. Write ekadhik on 4.
4. 4 = 5, subtracting 5 from 7 we get 2.