Question 4 - Page 143 - Digits in Disguise - Chapter 6 Class 7 - Number Play - Ganita Prakash

Last updated at October 17, 2025 by

[Ganita Prakash Class 7] Solve this alphametic: KP + KP = PRR - Digits in Disguise

part 2 - Question 4 - Page 143 - Digits in Disguise - Chapter 6 Class 7 - Number Play - Ganita Prakash - Class 7 (Ganita Prakash 1, 2 & old NCERT)
part 3 - Question 4 - 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 4 - Page 143 Solve this alphametic: Here, Adding two 2-digit numbers gives 3 digit number Since biggest number can be 99 + 99 = 196 This means Hundreds digit, P, will always be 1 ∴ P = 1 Now, let’s try different values of K - starting from biggest number So, we start from biggest number For K = 9 91 + 91 = 182 But, 182 ≠ PRR as last 2 digits are not same ∴ This is not possible For K = 8 81 + 81 = 162 But, 182 ≠ PRR as last 2 digits are not same ∴ This is not possible For K = 7 71 + 71 = 142 But, 182 ≠ PRR as last 2 digits are not same ∴ This is not possible For K = 6 61 + 61 = 122 Now, 122 = PRR Thus, K = 6, P = 1, R = 2

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