Problems on Numbers: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

Let the tens digit be a and the units digit be b. The original number equals 10a + b. Interchanging the digits gives the number 10b + a. The difference between these is (10b+a) – (10a+b) = 9b – 9a = 9(b-a). So the value always changes by 9 times the difference between the digits. Check: for example, if a=2 and b=5, the numbers are 25 and 52, and 52-25=27=9×(5-2)=9×3, which matches.

Let the larger number be a and the smaller be b. Fact 1: a + b = S, their sum. Fact 2: a – b = D, their difference. Subtracting the second equation from the first: (a+b) – (a-b) = S – D. The a terms cancel, leaving 2b = S – D. Dividing both sides by 2 gives b = (S-D)/2. Check: then a = S – b = (S+D)/2, and a – b = D, confirming b is the smaller number.

Let the larger number be a and the smaller be b. Fact 1: a + b = S, their sum. Fact 2: a – b = D, their difference. Adding these two equations: (a+b) + (a-b) = S + D. The b terms cancel, leaving 2a = S + D. Dividing both sides by 2 gives a = (S+D)/2. Check: then b = S – a = (S-D)/2, and a – b = D, confirming a is the larger number.

Let the two digits be p and q, with p the larger one. Statement I gives p + q = 8. Statement II gives p – q = 6. Adding these equations: 2p = 14, so p = 7. Subtracting them: 2q = 2, so q = 1. So the digits are 7 and 1, but we don’t know which is tens and which is units. If 7 is the tens digit, the number is 71. If 1 is the tens digit, the number is 17. Both numbers fit both statements, so together they still don’t give a unique number. Check: 71 and 17 both have digits 7 and 1, summing to 8 and differing by 6.

Let the two digits be p and q, with p the larger one. Statement I gives p + q = 9. Statement II gives p – q = 5. Adding these equations: 2p = 14, so p = 7. Subtracting them: 2q = 4, so q = 2. So the digits are 7 and 2, but we don’t know which is tens and which is units. If 7 is the tens digit, the number is 72. If 2 is the tens digit, the number is 27. Both numbers fit both statements, so together they still don’t give a unique number. Check: 72 and 27 both have digits 7 and 2, summing to 9 and differing by 5.

Let the two digits be p and q, with p the larger one. Statement I gives p + q = 9. Statement II gives p – q = 3. Adding these equations: 2p = 12, so p = 6. Subtracting them: 2q = 6, so q = 3. So the digits are 6 and 3, but we don’t know which is tens and which is units. If 6 is the tens digit, the number is 63. If 3 is the tens digit, the number is 36. Both numbers fit both statements, so together they still don’t give a unique number. Check: 63 and 36 both have digits 6 and 3, summing to 9 and differing by 3.

Let the two digits be p and q, with p the larger one. Statement I gives p + q = 10. Statement II gives p – q = 6. Adding these equations: 2p = 16, so p = 8. Subtracting them: 2q = 4, so q = 2. So the digits are 8 and 2, but we don’t know which is tens and which is units. If 8 is the tens digit, the number is 82. If 2 is the tens digit, the number is 28. Both numbers fit both statements, so together they still don’t give a unique number. Check: 82 and 28 both have digits 8 and 2, summing to 10 and differing by 6.

Let the two digits be p and q, with p the larger one. Statement I gives p + q = 11. Statement II gives p – q = 5. Adding these equations: 2p = 16, so p = 8. Subtracting them: 2q = 6, so q = 3. So the digits are 8 and 3, but we don’t know which is tens and which is units. If 8 is the tens digit, the number is 83. If 3 is the tens digit, the number is 38. Both numbers fit both statements, so together they still don’t give a unique number. Check: 83 and 38 both have digits 8 and 3, summing to 11 and differing by 5.

Let the two digits be p and q, with p the larger one. Statement I gives p + q = 9. Statement II gives p – q = 1. Adding these equations: 2p = 10, so p = 5. Subtracting them: 2q = 8, so q = 4. So the digits are 5 and 4, but we don’t know which is tens and which is units. If 5 is the tens digit, the number is 54. If 4 is the tens digit, the number is 45. Both numbers fit both statements, so together they still don’t give a unique number. Check: 54 and 45 both have digits 5 and 4, summing to 9 and differing by 1.

Let the tens digit be a = 1 and the units digit be b = 9. The original number is 10a + b = 19. Interchanging the digits gives the new number 10b + a = 91. The increase is (10b + a) – (10a + b) = 9(b – a). Here b – a = 8, so the increase is 9 × 8 = 72. Check: 19 + 72 = 91, and the reversed number is 91, which matches.

Let the tens digit be a = 4 and the units digit be b = 5. The original number is 10a + b = 45. Interchanging the digits gives the new number 10b + a = 54. The increase is (10b + a) – (10a + b) = 9(b – a). Here b – a = 1, so the increase is 9 × 1 = 9. Check: 45 + 9 = 54, and the reversed number is 54, which matches.

Let the tens digit be a = 8 and the units digit be b = 9. The original number is 10a + b = 89. Interchanging the digits gives the new number 10b + a = 98. The increase is (10b + a) – (10a + b) = 9(b – a). Here b – a = 1, so the increase is 9 × 1 = 9. Check: 89 + 9 = 98, and the reversed number is 98, which matches.

Let the tens digit be a = 4 and the units digit be b = 9. The original number is 10a + b = 49. Interchanging the digits gives the new number 10b + a = 94. The increase is (10b + a) – (10a + b) = 9(b – a). Here b – a = 5, so the increase is 9 × 5 = 45. Check: 49 + 45 = 94, and the reversed number is 94, which matches.

Let the tens digit be a = 5 and the units digit be b = 9. The original number is 10a + b = 59. Interchanging the digits gives the new number 10b + a = 95. The increase is (10b + a) – (10a + b) = 9(b – a). Here b – a = 4, so the increase is 9 × 4 = 36. Check: 59 + 36 = 95, and the reversed number is 95, which matches.

Let the tens digit be a = 6 and the units digit be b = 9. The original number is 10a + b = 69. Interchanging the digits gives the new number 10b + a = 96. The increase is (10b + a) – (10a + b) = 9(b – a). Here b – a = 3, so the increase is 9 × 3 = 27. Check: 69 + 27 = 96, and the reversed number is 96, which matches.

Let the tens digit be a = 6 and the units digit be b = 8. The original number is 10a + b = 68. Interchanging the digits gives the new number 10b + a = 86. The increase is (10b + a) – (10a + b) = 9(b – a). Here b – a = 2, so the increase is 9 × 2 = 18. Check: 68 + 18 = 86, and the reversed number is 86, which matches.

Let the tens digit be a = 4 and the units digit be b = 6. The original number is 10a + b = 46. Interchanging the digits gives the new number 10b + a = 64. The increase is (10b + a) – (10a + b) = 9(b – a). Here b – a = 2, so the increase is 9 × 2 = 18. Check: 46 + 18 = 64, and the reversed number is 64, which matches.

Let the tens digit be a and the units digit be b. The original number is 10a + b, and the reversed number is 10b + a. Fact 1: the digits sum to 6, so a + b = 6. Fact 2: the number exceeds its reverse by 18, so (10a+b) – (10b+a) = 18. That simplifies to 9(a – b) = 18, so a – b = 2. Adding a + b = 6 and a – b = 2 gives 2a = 8, so a = 4. Then b = 6 – 4 = 2. So the original number is 10×4 + 2 = 42. Check: reversing gives 10×2 + 4 = 24, and 42 – 24 = 18, which matches.

Let the tens digit be a and the units digit be b. The original number is 10a + b, and the reversed number is 10b + a. Fact 1: the digits sum to 13, so a + b = 13. Fact 2: the number exceeds its reverse by 27, so (10a+b) – (10b+a) = 27. That simplifies to 9(a – b) = 27, so a – b = 3. Adding a + b = 13 and a – b = 3 gives 2a = 16, so a = 8. Then b = 13 – 8 = 5. So the original number is 10×8 + 5 = 85. Check: reversing gives 10×5 + 8 = 58, and 85 – 58 = 27, which matches.

Let the tens digit be a and the units digit be b. The original number is 10a + b, and the reversed number is 10b + a. Fact 1: the digits sum to 12, so a + b = 12. Fact 2: the number exceeds its reverse by 54, so (10a+b) – (10b+a) = 54. That simplifies to 9(a – b) = 54, so a – b = 6. Adding a + b = 12 and a – b = 6 gives 2a = 18, so a = 9. Then b = 12 – 9 = 3. So the original number is 10×9 + 3 = 93. Check: reversing gives 10×3 + 9 = 39, and 93 – 39 = 54, which matches.

Let the tens digit be a and the units digit be b. The original number is 10a + b, and the reversed number is 10b + a. Fact 1: the digits sum to 12, so a + b = 12. Fact 2: the number exceeds its reverse by 18, so (10a+b) – (10b+a) = 18. That simplifies to 9(a – b) = 18, so a – b = 2. Adding a + b = 12 and a – b = 2 gives 2a = 14, so a = 7. Then b = 12 – 7 = 5. So the original number is 10×7 + 5 = 75. Check: reversing gives 10×5 + 7 = 57, and 75 – 57 = 18, which matches.

Let the tens digit be a and the units digit be b. The original number is 10a + b, and the reversed number is 10b + a. Fact 1: the digits sum to 8, so a + b = 8. Fact 2: the number exceeds its reverse by 54, so (10a+b) – (10b+a) = 54. That simplifies to 9(a – b) = 54, so a – b = 6. Adding a + b = 8 and a – b = 6 gives 2a = 14, so a = 7. Then b = 8 – 7 = 1. So the original number is 10×7 + 1 = 71. Check: reversing gives 10×1 + 7 = 17, and 71 – 17 = 54, which matches.

Let the tens digit be a and the units digit be b. The original number is 10a + b, and the reversed number is 10b + a. Fact 1: the digits sum to 5, so a + b = 5. Fact 2: the number exceeds its reverse by 9, so (10a+b) – (10b+a) = 9. That simplifies to 9(a – b) = 9, so a – b = 1. Adding a + b = 5 and a – b = 1 gives 2a = 6, so a = 3. Then b = 5 – 3 = 2. So the original number is 10×3 + 2 = 32. Check: reversing gives 10×2 + 3 = 23, and 32 – 23 = 9, which matches.

Let the tens digit be a and the units digit be b. The original number is 10a + b, and the reversed number is 10b + a. Fact 1: the digits sum to 5, so a + b = 5. Fact 2: the number exceeds its reverse by 27, so (10a+b) – (10b+a) = 27. That simplifies to 9(a – b) = 27, so a – b = 3. Adding a + b = 5 and a – b = 3 gives 2a = 8, so a = 4. Then b = 5 – 4 = 1. So the original number is 10×4 + 1 = 41. Check: reversing gives 10×1 + 4 = 14, and 41 – 14 = 27, which matches.

Let the tens digit be a and the units digit be b. The original number is 10a + b, and the reversed number is 10b + a. Fact 1: the digits sum to 6, so a + b = 6. Fact 2: the number exceeds its reverse by 36, so (10a+b) – (10b+a) = 36. That simplifies to 9(a – b) = 36, so a – b = 4. Adding a + b = 6 and a – b = 4 gives 2a = 10, so a = 5. Then b = 6 – 5 = 1. So the original number is 10×5 + 1 = 51. Check: reversing gives 10×1 + 5 = 15, and 51 – 15 = 36, which matches.