RBSE Solutions Class 8 Maths Chapter 4 Mental Exercise 4.2

Question 1. Find the value of alphabets and show the causes of that process.
(i) 5A + 34 = B2
(ii) 5A + 79 = CB3
(iii) AB + 37 = 6A
(v) 12A + 6AB = A09
(vi) 1A x A = 9A
(vii) AB x B = CAB
(viii) AB x 6 = BBB
Answer:
(i) We need to find the digits A and B. In the units column, \( A + 4 \) must end in 2. This means \( A + 4 = 12 \). So, \( A = 12 - 4 = 8 \). When \( A=8 \), we have a carry-over of 1 to the tens column. In the tens column, \( 5 + 3 + 1 \) (carry-over) gives B. So, \( 5 + 3 + 1 = 9 \). Therefore, \( B = 9 \). The value for A is 8 and for B is 9.
In simple words: For the sum to work, A has to be 8, which makes the first column add up to 12. This leaves a 2 and carries 1 over. Then B has to be 9, completing the sum correctly.

🎯 Exam Tip: Always start solving cryptarithmetic problems from the units column, as carries propagate from right to left.

(ii) We need to find the digits A, B, and C. In the units column, \( A + 9 \) must end in 3. This means \( A + 9 = 13 \). So, \( A = 13 - 9 = 4 \). When \( A=4 \), there is a carry-over of 1 to the tens column. In the tens column, \( 5 + 7 + 1 \) (carry-over) gives a sum of 13. So, the digit B is 3 and there is a carry-over of 1 to the hundreds column. In the hundreds column, the carry-over 1 becomes the digit C. Therefore, \( C = 1 \). The value for A is 4, B is 3, and C is 1.
In simple words: We figure out A first from the units column, which is 4. This makes us carry 1. Then we add the tens column, and B becomes 3 with another carry of 1. That last carry gives us C as 1.

🎯 Exam Tip: Remember to account for carry-overs in each column when solving addition problems with unknown digits.

(iii) We need to find the digits A and B. In the units column, \( B + 7 \) must end in A. Let's assume there is a carry-over. In the tens column, \( A + 3 + \text{carry} = 6 \). If no carry from units column, \( B+7=A \). If there is a carry from units column, \( B+7=10+A \). Considering the sums for single digits, let's work backward from the known result. If \( A+3 = 6 \) (assuming no carry from units), then \( A=3 \). Then \( B+7 = 3 \) (or 13). \( B+7=3 \) is impossible. So \( B+7=13 \), then \( B=6 \). So \( A=3, B=6 \). Let's check: 36 + 37 ---- 73 This does not match `6A`, which would be 63. So \( A=3 \) is wrong. Therefore, there must be a carry from the units column. So, \( B+7=10+A \). And for the tens column, \( A+3+1 = 6 \). This gives \( A+4 = 6 \), so \( A=2 \). Now substitute \( A=2 \) into \( B+7=10+A \): \( B+7=10+2 \) \( B+7=12 \) \( B=12-7 \) \( B=5 \) So, \( A=2 \) and \( B=5 \). Let's verify: 25 + 37 ---- 62 This matches `6A` where `A=2`. The value for A is 2 and for B is 5.
In simple words: By checking the columns, we find that B must be 5 to make the units column add up correctly with a carry. Then A must be 2 so that the tens column, including the carry, adds up to 6.

🎯 Exam Tip: For these types of problems, consider possible carry-overs between columns to find the correct digits. Sometimes testing different possibilities for carries helps narrow down choices.

(v) We need to find the digits A and B in `12A + 6AB = A09`. Let's interpret `12A` as a number where A is the units digit (i.e., \( 120 + A \)) and `6AB` as a number where A is the tens digit and B is the units digit (i.e., \( 600 + 10A + B \)). The sum `A09` is \( 100A + 9 \). So the equation is: \( (120 + A) + (600 + 10A + B) = 100A + 9 \) \( 720 + 11A + B = 100A + 9 \) Now, let's rearrange to solve for A and B: \( 720 - 9 = 100A - 11A - B \) \( 711 = 89A - B \) Since A and B are single digits (0-9), we can test values for A. If \( A=8 \): \( 711 = 89(8) - B \) \( 711 = 712 - B \) \( B = 712 - 711 \) \( B = 1 \) This gives valid digits \( A=8 \) and \( B=1 \). Let's verify: 128 + 681 ----- 809 This matches `A09` where \( A=8 \). The values are \( A=8 \) and \( B=1 \).
In simple words: By writing out the sum in terms of A and B, we get an equation. Solving it shows that A is 8 and B is 1, which makes the addition work correctly.

🎯 Exam Tip: When letters represent digits in numbers, treat them as place values (e.g., AB = 10A + B) and form an algebraic equation to solve.

(vi) We need to find the digit A in `1A x A = 9A`. Here, `1A` means the number \( 10 + A \). `9A` means the number \( 90 + A \). The equation is \( (10 + A) \times A = 90 + A \). Let's expand the left side: \( 10A + A^2 = 90 + A \) Subtract A from both sides: \( 9A + A^2 = 90 \) Rearrange into a quadratic equation: \( A^2 + 9A - 90 = 0 \) We can solve this by factoring or using the quadratic formula. Let's look for two numbers that multiply to -90 and add to 9. These are 15 and -6. So, \( (A + 15)(A - 6) = 0 \) This gives two possible values for A: \( A = -15 \) or \( A = 6 \). Since A must be a single digit (0-9), \( A = 6 \). Let's verify: If \( A=6 \), the problem is `16 x 6 = 96`. \( 16 \times 6 = 96 \). This matches the form `9A` where \( A=6 \). The value for A is 6.
In simple words: We changed the letter problem into a number equation. Solving the equation gave us two possible answers for A, but only one (6) is a single digit. When we check 16 times 6, it really is 96, so A is 6.

🎯 Exam Tip: For multiplication problems involving unknown digits, forming and solving an algebraic equation is often the most direct method to find the correct digit.

(vii) We need to find the digits A, B, and C in `AB x B = CAB`. Here `AB` means \( 10A + B \), and `CAB` means \( 100C + 10A + B \). The product of `AB` multiplied by `B` results in `CAB`. Let's look at the units digit: \( B \times B \) must end in B. Possible values for B are 0, 1, 5, 6 (since \( 0 \times 0 = 0 \), \( 1 \times 1 = 1 \), \( 5 \times 5 = 25 \), \( 6 \times 6 = 36 \)). If \( B=0 \), then `A0 x 0 = C A 0`, which means `0 = C A 0`. This implies A, C are 0, which is not a useful solution. If \( B=1 \), then `A1 x 1 = C A 1`. This means `A1 = C A 1`. So `A=0, C=0` which makes `1 x 1 = 1`. Not a two-digit number `AB`. If \( B=5 \): The problem is `A5 x 5 = C A 5`. Units digit: \( 5 \times 5 = 25 \). So the units digit is 5 (which is B), and there's a carry of 2 to the tens column. Tens digit: \( A \times 5 + 2 \) (carry) must end in A. Let's try values for A: If \( A=1 \), \( 1 \times 5 + 2 = 7 \). This ends in 7, not 1. If \( A=2 \), \( 2 \times 5 + 2 = 12 \). This ends in 2 (which is A!). This works. So, if \( A=2, B=5 \). Let's check `25 x 5`: \( 25 \times 5 = 125 \). This matches the form `CAB`, where \( C=1, A=2, B=5 \). So, \( A=2, B=5, C=1 \). If \( B=6 \): The problem is `A6 x 6 = C A 6`. Units digit: \( 6 \times 6 = 36 \). So the units digit is 6 (which is B), and there's a carry of 3 to the tens column. Tens digit: \( A \times 6 + 3 \) (carry) must end in A. If \( A=1 \), \( 1 \times 6 + 3 = 9 \). Ends in 9, not 1. If \( A=2 \), \( 2 \times 6 + 3 = 15 \). Ends in 5, not 2. So \( A=2, B=5 \) is the correct solution. The values are \( A=2, B=5, C=1 \).
In simple words: We tested possible digits for B that make \( B \times B \) end in B. Using B=5, we then found A=2 by checking the tens column. When we multiply 25 by 5, we get 125, which perfectly fits the `CAB` pattern.

🎯 Exam Tip: In multiplication puzzles like this, often a good starting point is to analyze the units column. Digits that remain themselves when squared (0, 1, 5, 6) are good candidates for the multiplying digit.

(viii) We need to find the digits A and B in `AB x 6 = BBB`. Here, `AB` means \( 10A + B \), and `BBB` means \( 100B + 10B + B = 111B \). So the equation is \( (10A + B) \times 6 = 111B \). Divide both sides by 3: \( (10A + B) \times 2 = 37B \) Now, let's analyze the units digit: \( B \times 6 \) must end in B. Possible values for B are: If \( B=0 \), \( 0 \times 6 = 0 \). If \( B=0 \), then \( (10A+0) \times 6 = 000 \), so \( 60A = 0 \), meaning \( A=0 \). `00 x 6 = 000`. This is technically a solution but usually A, B are non-zero if they are the first digits. If \( B=2 \), \( 2 \times 6 = 12 \). Ends in 2. If \( B=4 \), \( 4 \times 6 = 24 \). Ends in 4. If \( B=6 \), \( 6 \times 6 = 36 \). Ends in 6. If \( B=8 \), \( 8 \times 6 = 48 \). Ends in 8. Let's try \( B=4 \) as suggested by the source. Substitute \( B=4 \) into \( (10A + B) \times 2 = 37B \): \( (10A + 4) \times 2 = 37 \times 4 \) \( 20A + 8 = 148 \) \( 20A = 148 - 8 \) \( 20A = 140 \) \( A = \frac{140}{20} \) \( A = 7 \) So, \( A=7 \) and \( B=4 \). Let's verify: If \( A=7, B=4 \), the problem is `74 x 6 = 444`. \( 74 \times 6 = 444 \). This matches the form `BBB` where \( B=4 \). The values are \( A=7 \) and \( B=4 \).
In simple words: We used the given pattern to set up an equation. By looking at which digits B keep their value when multiplied by 6, and then testing them in the equation, we found A is 7 and B is 4. This makes 74 times 6 equal to 444.

🎯 Exam Tip: For multiplication problems where the product is a repeating digit (like BBB), setting up an algebraic equation for \( 10A+B \) and \( 111B \) can simplify the solution process.

Question 2. Find the value of x or (*) in following questions
(i) 2* + *8 + 95 = 167
(ii) 905 + *12 + 88* = 2100
(iii) 7*3 - 281 = 432
(v) 68 x x = 408
(vi) 763 x 3x = 25942
(vii) 216 / (2x) = 8 (remainder 0)
(viii) 907 / (x7) = 24 (remainder 19)
Answer:
(i) Let the first missing digit (in `2*`) be \( x_1 \) and the second missing digit (in `*8`) be \( x_2 \). The addition is: 2x1 x28 + 95 ----- 167 Units column: \( x_1 + 8 + 5 \) must end in 7. \( x_1 + 13 \) ends in 7. For this to happen, \( x_1 \) must be 4 (since \( 4 + 13 = 17 \)). So, \( x_1 = 4 \). We have a carry-over of 1 to the tens column. Tens column: \( 1 \) (carry) \( + 2 + x_2 + 9 = 16 \). \( 12 + x_2 = 16 \) \( x_2 = 16 - 12 \) \( x_2 = 4 \) Both missing digits are 4. So the value of \( * \) is 4.
In simple words: By adding up the units column, we found the first missing digit is 4, with a carry of 1. Then, using that carry and the known numbers in the tens column, we found the second missing digit is also 4.

🎯 Exam Tip: When multiple asterisks or variables are used in one problem, it's crucial to correctly identify which variable corresponds to which missing digit, often by solving column by column.

(ii) Let the first missing digit (in `*12`) be \( x_1 \) and the second missing digit (in `88*`) be \( x_2 \). The addition is: 905 x112 + 88x2 ------- 2100 Units column: \( 5 + 2 + x_2 \) must end in 0. \( 7 + x_2 \) ends in 0. For this to happen, \( x_2 \) must be 3 (since \( 7 + 3 = 10 \)). So, \( x_2 = 3 \). We have a carry-over of 1 to the tens column. Tens column: \( 1 \) (carry) \( + 0 + 1 + 8 \) must end in 0. \( 10 \) ends in 0. This works. We have a carry-over of 1 to the hundreds column. Hundreds column: \( 1 \) (carry) \( + 9 + x_1 + 8 \) must end in 1 (from `2100`). \( 18 + x_1 \) ends in 1. For this to happen, \( 18 + x_1 = 21 \). \( x_1 = 21 - 18 \) \( x_1 = 3 \) Both missing digits are 3. So the value of \( * \) is 3.
In simple words: We solved for the units digit first, finding it to be 3 with a carry. The tens column worked out with another carry. Then, in the hundreds column, we used the carry to find the last missing digit, which was also 3.

🎯 Exam Tip: For problems involving sums that extend to a new digit in the leftmost column (like 2100 here), ensure the carry-over from the hundreds column is accounted for in the thousands place.

(iii) We need to find the missing digit in `7*3 - 281 = 432`. Let the missing digit be \( x \). The subtraction is: 7x3 - 281 ----- 432 Units column: \( 3 - 1 = 2 \). This is correct and no borrowing is needed. Tens column: \( x - 8 \) must result in 3. Since \( x \) is a single digit, \( x - 8 \) cannot be 3 without borrowing. So, we must borrow from the hundreds column. This means we have \( (x + 10) - 8 = 3 \). \( x + 2 = 3 \) \( x = 3 - 2 \) \( x = 1 \) So, the missing digit is 1. Hundreds column: Since we borrowed 1 from 7, it becomes 6. Now, \( 6 - 2 = 4 \). This matches the result. The value of \( * \) is 1.
In simple words: We looked at the units column, which was simple. For the tens column, we realized we needed to borrow from the hundreds to make the subtraction work, which showed the missing digit is 1. The hundreds column then also matched up.

🎯 Exam Tip: In subtraction problems with missing digits, always consider if borrowing from the next column is necessary. This is a common point where errors occur if overlooked.

(v) We need to find the value of \( x \) in `68 x x = 408`. This is a simple multiplication problem where we need to find the multiplier. We can find \( x \) by dividing the product by the known factor. \( x = \frac{408}{68} \) Performing the division: \( 408 \div 68 = 6 \) So, \( x = 6 \). Let's verify: \( 68 \times 6 = 408 \). This is correct. The value of \( x \) is 6.
In simple words: We simply divided the total product (408) by the number we already knew (68) to find the missing number (x), which turned out to be 6.

🎯 Exam Tip: For simple multiplication problems with one unknown, use division as the inverse operation to quickly find the missing value.

(vi) We need to find the value of \( x \) in `763 x 3x = 25942`. Here `3x` means a two-digit number where 3 is the tens digit and \( x \) is the units digit (i.e., \( 30 + x \)). The equation is \( 763 \times (30 + x) = 25942 \). To find \( (30 + x) \), we divide the product by 763: \( 30 + x = \frac{25942}{763} \) \( 30 + x = 34 \) Now, solve for \( x \): \( x = 34 - 30 \) \( x = 4 \) So, the value of \( x \) is 4. Let's verify: \( 763 \times 34 = 25942 \). This is correct.
In simple words: We first divided the large total by 763 to find the value of the `3x` number. Then, by subtracting 30, we found that x must be 4.

🎯 Exam Tip: When a variable is part of a two-digit number like `3x`, interpret it as \( 30+x \) to set up the correct algebraic equation. Don't multiply 3 by x.

(vii) We need to find the value of \( x \) in `216 / (2x) = 8` with remainder 0. This is a division problem. We know that `Dividend = Divisor x Quotient + Remainder`. Here, Dividend = 216, Quotient = 8, Remainder = 0. The divisor is `2x`. So, \( 216 = (2x) \times 8 + 0 \) \( 216 = (20 + x) \times 8 \) (interpreting `2x` as the number \( 20 + x \)) Divide both sides by 8: \( \frac{216}{8} = 20 + x \) \( 27 = 20 + x \) Now, solve for \( x \): \( x = 27 - 20 \) \( x = 7 \) So, the value of \( x \) is 7. Let's verify: \( 216 \div 27 = 8 \). This is correct.
In simple words: We used the rule that dividend equals divisor times quotient to set up an equation. Solving this equation, we found that the missing digit x is 7, making the divisor 27.

🎯 Exam Tip: Always remember the basic formula for division (Dividend = Divisor x Quotient + Remainder) when solving problems involving missing numbers in division.

(viii) We need to find the value of \( x \) in `907 / (x7) = 24` with remainder 19. Using the formula `Dividend = Divisor x Quotient + Remainder`: Here, Dividend = 907, Quotient = 24, Remainder = 19. The divisor is `x7`. So, \( 907 = (x7) \times 24 + 19 \) First, subtract the remainder from the dividend: \( 907 - 19 = (x7) \times 24 \) \( 888 = (x7) \times 24 \) Now, interpret `x7` as the number \( 10x + 7 \). \( 888 = (10x + 7) \times 24 \) Divide both sides by 24: \( \frac{888}{24} = 10x + 7 \) \( 37 = 10x + 7 \) Now, solve for \( x \): \( 10x = 37 - 7 \) \( 10x = 30 \) \( x = \frac{30}{10} \) \( x = 3 \) So, the value of \( x \) is 3. Let's verify: \( (37 \times 24) + 19 = 888 + 19 = 907 \). This is correct.
In simple words: We set up the division problem as an equation using the remainder rule. After some simple math, we found that the number `x7` must be 37, which means the missing digit x is 3.

🎯 Exam Tip: When a remainder is involved, subtract it from the dividend first to simplify the problem before finding the unknown divisor or quotient.