Question 3 - Page 143 - Digits in Disguise - Chapter 6 Class 7 - Number Play - Ganita Prakash

Last updated at October 17, 2025 by

[Chapter 6 Class 7] Solve this alphametic: B5 + 3D = ED5 - Digits in Disguise

part 2 - Question 3 - Page 143 - Digits in Disguise - Chapter 6 Class 7 - Number Play - Ganita Prakash - Class 7 (Ganita Prakash 1, 2 & old NCERT)
part 3 - Question 3 - Page 143 - Digits in Disguise - Chapter 6 Class 7 - Number Play - Ganita Prakash - Class 7 (Ganita Prakash 1, 2 & old NCERT)

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Question 3 - Page 143 Solve this alphametic: Here, Adding two 2-digit numbers gives 3 digit number Since 3 digit number has unit digit 5 And we are adding B5 + 3D – so D = 0 Now, let’s try different values of B - starting from biggest number For B = 9 95 + 30 = 125 But, 125 ≠ EO5 as last middle digit is not same ∴ This is not possible For B = 8 85 + 30 = 115 But, 115 ≠ EO5 as last middle digit is not same ∴ This is not possible For B = 7 75 + 30 = 105 Now, 105 = EO5 Thus, B = 7, D = 0, E = 1

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