RBSE Solutions Class 8 Maths Chapter 4 Mental Exercises Important Questions

I. Objective Type Questions

Question 1. The expanded form of 37 is -
(a) 37 = 3 + 7
(b) 37 = 10 x 3 + 7
(c) 37 = 7 x 10 + 3
(d) 37 = 37 + 3 + 7
Answer: (b) 37 = 10 x 3 + 7
In simple words: To expand a number like 37, we break it down into its tens and units parts. 37 is made of 3 tens (30) and 7 units (7), so it is 10 multiplied by 3, plus 7.

🎯 Exam Tip: Remember that in a two-digit number, the first digit represents tens and the second digit represents units. For example, in 37, '3' means 3 tens (30) and '7' means 7 units.

Question 2. A two digit number ab can be written as -
(a) 10a + b
(b) a + 10b
(c) 10 b + a
(d) a + b
Answer: (a) 10a + b
In simple words: When you have a two-digit number written as 'ab', it means 'a' is in the tens place and 'b' is in the units place. So, its value is 'a' multiplied by 10, plus 'b'.

🎯 Exam Tip: Always remember the place value of digits. 'a' in 'ab' holds the tens place value, and 'b' holds the units place value.

Question 3. ab + ba = 11 (a + b) is divisible by -
(a) 4
(b) 5
(c) 11
(d) ab
Answer: (c) 11
In simple words: If you add a two-digit number 'ab' to its reverse 'ba', the sum will always be a multiple of 11. This means the sum is always perfectly divisible by 11 because the sum is 11 times the sum of the digits (a+b).

🎯 Exam Tip: This is a common property of two-digit numbers and their reverses. The sum \( \overline{ab} + \overline{ba} \) is always \( 11(a+b) \), and their difference \( \overline{ab} - \overline{ba} \) is always \( 9(a-b) \).

Question 4. In ab - ba = 9 (a - b), if a = b, the which of the following is true?
(a) ab = 9
(b) ba = 9
(d) None of the options
Answer: (d) None of the options
In simple words: If the two digits 'a' and 'b' are the same, then 'a - b' is zero. This means 'ab - ba' will also be zero, because the numbers are identical. So, neither 'ab = 9' nor 'ba = 9' can be true in this situation.

🎯 Exam Tip: When \( a=b \), any expression involving \( (a-b) \) will simplify to zero. Always substitute values to check such conditions.

Question 6. In a three digit number 24x is divisible by 9, then the value of x will be -
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (a) 3
In simple words: For a number to be divisible by 9, the sum of its digits must also be divisible by 9. Here, the digits are 2, 4, and x. So, 2 + 4 + x must be a multiple of 9. If x is 3, then 2 + 4 + 3 = 9, which is divisible by 9.

🎯 Exam Tip: The divisibility rule for 9 is that the sum of the digits must be a multiple of 9. The smallest multiple of 9 is 0 (for 0 itself), then 9, 18, 27, etc.

Question 7. Which of the following is divisible by 2?
(a) 4438
(b) 5791
(c) 233
(d) 147
Answer: (a) 4438
In simple words: A number is divisible by 2 if its last digit is an even number (0, 2, 4, 6, or 8). Out of the given options, only 4438 ends with an even digit, which is 8.

🎯 Exam Tip: The divisibility test for 2 is the simplest: just check if the last digit is an even number. If it is, the entire number is divisible by 2.

II. Fill In The Blanks

Question 1. If [missing expression]ber, then ab = ______ + b
Answer: 10a
In simple words: When we write a two-digit number like 'ab', it means 'a' in the tens place and 'b' in the units place. So, its full value is ten times 'a' plus 'b'.

🎯 Exam Tip: Always remember that the digit in the tens place is multiplied by 10 and the digit in the units place is multiplied by 1 when writing a number in expanded form.

Question 4. If a three digit number 24x is a multiple of 3, then value of x is _____
Answer: 0, 3, 6 or 9
In simple words: For a number to be a multiple of 3, the sum of its digits must be a multiple of 3. For the number 24x, the sum of digits is 2 + 4 + x, or 6 + x. The possible values for x that make 6 + x a multiple of 3 are 0, 3, 6, and 9.

🎯 Exam Tip: Always list all possible single-digit values for the unknown digit that satisfy the divisibility rule. Remember that 0 is a valid digit.

Question 5. A number is divisible by____if it has zero in its ones place.
Answer: 10
In simple words: A number is divisible by 10 if its last digit is 0. If a number ends in zero, it can always be divided evenly by 10.

🎯 Exam Tip: Divisibility by 10 is very easy to check: simply look at the last digit. If it's a 0, the number is divisible by 10.

III. Very Short Answer Type Questions

Question 1. If a two digit number ab can be written as 10a + b, then write down the value of ba.
Answer: The value of 'ba' is \( 10b + a \).
In simple words: If 'ab' means 'a' tens and 'b' units, then 'ba' means 'b' tens and 'a' units. So, 'b' is multiplied by 10 and 'a' is added to it to find its value.

🎯 Exam Tip: Swapping the digits also swaps their place values. The digit that was in the tens place moves to the units place, and vice versa.

Question 2. Write the expanded form of a three digit number abc.
Answer: The expanded form of a three-digit number 'abc' is \( 100a + 10b + c \).
In simple words: In a three-digit number 'abc', 'a' is in the hundreds place, 'b' is in the tens place, and 'c' is in the units place. So, you multiply 'a' by 100, 'b' by 10, and add 'c'.

🎯 Exam Tip: For any multi-digit number, the expanded form breaks down the number by multiplying each digit by its respective place value (units, tens, hundreds, thousands, etc.) and then adding them together.

Question 3. What is the divisibility test for 5?
Answer: A number is divisible by 5 if its last digit (the digit in the units place) is either 0 or 5. This makes it easy to quickly check if a number can be divided by 5 without doing the full division.
In simple words: To check if a number can be divided by 5, just look at its very last digit. If that digit is a 0 or a 5, then the whole number can be divided by 5.

🎯 Exam Tip: Divisibility rules are quick shortcuts to determine if a number can be divided by another number evenly, without performing long division.

Question 5. In 99a - 99c, if a = c, then what will be the difference?
Answer: If \( a = c \), then the difference \( 99a - 99c \) will be zero. This is because if \( a \) and \( c \) are equal, then \( 99a \) and \( 99c \) are also equal, and subtracting equal numbers always results in zero.
In simple words: If 'a' and 'c' are the same number, then 99 times 'a' will be the same as 99 times 'c'. When you subtract two numbers that are exactly the same, you always get zero.

🎯 Exam Tip: Any expression of the form \( XA - XC \) will be zero if \( A=C \), because \( XA - XC = X(A-C) = X(0) = 0 \).

IV. Short Answer Type Questions

Question 1. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Answer: If the number 31z5 is a multiple of 3, it means the sum of its digits must also be a multiple of 3.
The sum of the digits is \( 3 + 1 + z + 5 = 9 + z \).
Since \( z \) is a digit, its value can be from 0 to 9.
We need \( 9 + z \) to be a multiple of 3.
Possible values for \( 9 + z \) that are multiples of 3 are 9, 12, 15, 18.
If \( 9 + z = 9 \), then \( z = 0 \).
\( \implies \) If \( 9 + z = 12 \), then \( z = 3 \).
\( \implies \) If \( 9 + z = 15 \), then \( z = 6 \).
\( \implies \) If \( 9 + z = 18 \), then \( z = 9 \).
Any other multiple of 3 for \( 9+z \) (like 21) would make \( z \) a two-digit number (e.g., \( z=12 \)), which is not allowed.
So, the possible values for \( z \) are 0, 3, 6, or 9.
In simple words: For a number to be divisible by 3, its digits must add up to a number that can also be divided by 3. For 31z5, the digits 3, 1, z, and 5 add up to 9 + z. We need 9 + z to be a multiple of 3. Since z is a single digit (0-9), the only numbers it can be are 0, 3, 6, or 9.

🎯 Exam Tip: When solving for an unknown digit, always remember that a digit can only be from 0 to 9. This helps narrow down the possible solutions quickly.

Find the values of the letters in each of the following and give reasons for the steps involved.

Question 2. Find the values of A and B in the following addition puzzle:
\( \begin{array}{r} 3 \, A \\ + 2 \, 5 \\ \hline B \, 2 \end{array} \)
Answer: Consider the units column: \( A + 5 \) must end in 2.
This means \( A + 5 \) could be 12 (since A is a single digit, \( A+5 \) cannot be 2).
So, \( A = 12 - 5 \implies A = 7 \).
Now, carry over 1 to the tens column.
Consider the tens column: \( 1 + 3 + 2 = B \).
So, \( B = 6 \).
Thus, the completed addition is:
\( \begin{array}{r} 3 \, 7 \\ + 2 \, 5 \\ \hline 6 \, 2 \end{array} \)
Therefore, \( A = 7 \) and \( B = 6 \).
In simple words: First, look at the last column. A plus 5 should give a number ending in 2. So A must be 7 (because 7 + 5 = 12). We carry over 1 to the next column. Then, in the middle column, 1 (carried over) plus 3 plus 2 equals 6. So B is 6. The completed sum is 37 + 25 = 62.

🎯 Exam Tip: Always start solving cryptarithms from the rightmost column (units place) and systematically move left, accounting for any carry-overs.

Question 3. Find the values of A, B and C in the following addition puzzle:
\( \begin{array}{r} 4 \, A \\ + 9 \, 8 \\ \hline C \, B \, 3 \end{array} \)
Answer: Consider the units column: \( A + 8 \) must end in 3.
This means \( A + 8 \) must be 13 (as A is a single digit).
So, \( A = 13 - 8 \implies A = 5 \).
Carry over 1 to the tens column.
Consider the tens column: \( 1 + 4 + 9 \) must equal B, and potentially carry over.
\( 1 + 4 + 9 = 14 \).
This means B is 4 (the units digit of 14) and we carry over 1 to the hundreds column.
Consider the hundreds column: The carry-over 1 becomes C.
So, \( C = 1 \).
Thus, the completed addition is:
\( \begin{array}{r} 4 \, 5 \\ + 9 \, 8 \\ \hline 1 \, 4 \, 3 \end{array} \)
Therefore, \( A = 5 \), \( B = 4 \), and \( C = 1 \).
In simple words: Starting from the rightmost column, A plus 8 must end in 3. This means A must be 5 (because 5 + 8 = 13). We write 3 and carry over 1. Next, in the middle column, 1 (carried over) plus 4 plus 9 equals 14. So B is 4 and we carry over 1 to the last column. This carried-over 1 becomes C. So C is 1. The full sum is 45 + 98 = 143.

🎯 Exam Tip: In cryptarithmetic, always check if any value for a letter is a two-digit number or negative. Letters represent single digits from 0 to 9.

Question 5. If a three digit number 4y2 is multiple of 3, and then find all possible values of y where y is one digit number.
Answer: For the three-digit number 4y2 to be a multiple of 3, the sum of its digits must also be divisible by 3.
The sum of the digits is \( 4 + y + 2 = 6 + y \).
Since \( y \) is a single digit (0-9), we need \( 6 + y \) to be a multiple of 3.
Possible values for \( y \) that make \( 6 + y \) divisible by 3 are:
If \( y = 0 \), \( 6 + 0 = 6 \) (divisible by 3)
\( \implies \) If \( y = 3 \), \( 6 + 3 = 9 \) (divisible by 3)
\( \implies \) If \( y = 6 \), \( 6 + 6 = 12 \) (divisible by 3)
\( \implies \) If \( y = 9 \), \( 6 + 9 = 15 \) (divisible by 3)
Thus, the possible values for \( y \) are 0, 3, 6, and 9.
In simple words: For the number 4y2 to be a multiple of 3, the digits 4, y, and 2 must add up to a number that can be divided by 3. This sum is 6 + y. Since y is a single digit, the only numbers y can be to make 6 + y divisible by 3 are 0, 3, 6, or 9.

🎯 Exam Tip: When using the divisibility rule for 3, remember to consider all single-digit options for the unknown digit that make the sum a multiple of 3.

Question 6. Solve the Cryptarithm: \( \overline{AB} \times \overline{AB} = \overline{ACB} \)
Answer: We have the cryptarithm: \( \overline{AB} \times \overline{AB} = \overline{ACB} \).
First, consider the units column: \( B \times B \) must end in \( B \).
The digits that satisfy this are 1 (because \( 1 \times 1 = 1 \)), 5 (because \( 5 \times 5 = 25 \), ends in 5), and 6 (because \( 6 \times 6 = 36 \), ends in 6). So \( B \) can be 1, 5, or 6.
The solution indicates that \( B=1 \). Let's explore this case.
If \( B = 1 \), the equation becomes \( \overline{A1} \times \overline{A1} = \overline{AC1} \).
This can be written as \( (10A + 1)^2 = 100A + 10C + 1 \).
Expanding the left side: \( 100A^2 + 20A + 1 = 100A + 10C + 1 \).
Subtract 1 from both sides: \( 100A^2 + 20A = 100A + 10C \).
Divide by 10: \( 10A^2 + 2A = 10A + C \).
Rearrange: \( 10A^2 - 8A = C \).
Since \( A \) and \( C \) are single digits (0-9), let's test values for \( A \).
If \( A=1 \), \( 10(1)^2 - 8(1) = 10 - 8 = 2 \). So \( C=2 \).
This gives \( A=1, B=1, C=2 \).
Let's check: \( \overline{11} \times \overline{11} = 11 \times 11 = 121 \).
And \( \overline{AC1} \) with these values is \( \overline{121} \). This matches.
If \( A=2 \), \( 10(2)^2 - 8(2) = 40 - 16 = 24 \). \( C \) would be 24, which is not a single digit.
Therefore, the only possible values are \( A=1, B=1, C=2 \).
In simple words: First, look at the last digit: B multiplied by B must end in B. So B can be 1, 5, or 6. The answer tells us B is 1. If B is 1, the puzzle is A1 x A1 = AC1. Let's try A=1. Then 11 x 11 = 121. This fits the pattern where A is 1, C is 2, and B is 1. This is the only single-digit solution that works.

🎯 Exam Tip: When testing possible values for digits in cryptarithms, use logical reasoning and substitution to quickly narrow down the options and find the correct solution.

Question 7. Test the divisibility of following numbers by 3 and 9.
(i) 1839
(ii) 236637
Answer:
(i) For the number 1839:
**Divisibility by 3:** Sum of digits \( = 1 + 8 + 3 + 9 = 21 \). Since 21 is divisible by 3, the number 1839 is also divisible by 3.
**Divisibility by 9:** Sum of digits \( = 1 + 8 + 3 + 9 = 21 \). Since 21 is not divisible by 9, the number 1839 is not divisible by 9.
(ii) For the number 236637:
**Divisibility by 3:** Sum of digits \( = 2 + 3 + 6 + 6 + 3 + 7 = 27 \). Since 27 is divisible by 3, the number 236637 is also divisible by 3.
**Divisibility by 9:** Sum of digits \( = 2 + 3 + 6 + 6 + 3 + 7 = 27 \). Since 27 is divisible by 9, the number 236637 is also divisible by 9.
In simple words: To check if a number can be divided by 3 or 9, we add up all its digits. If this sum can be divided by 3, the number is divisible by 3. If this sum can be divided by 9, the number is divisible by 9. For 1839, the sum is 21, so it is divisible by 3 but not 9. For 236637, the sum is 27, so it is divisible by both 3 and 9.

🎯 Exam Tip: Remember that if a number is divisible by 9, it is automatically divisible by 3. However, if a number is divisible by 3, it is not necessarily divisible by 9.

Question 8. Find A and B. Where A and B are digits.
\( \begin{array}{r} A \, 2 \\ + 6 \, 3 \, 7 \\ \hline B \, 2 \, A \end{array} \)
Answer: Consider the units column: \( 2 + 7 = A \).
This means \( A = 9 \).
Now consider the tens column: \( A + 3 \) must end in 2.
Substituting \( A = 9 \), we get \( 9 + 3 = 12 \).
So, we write 2 in the tens place and carry over 1 to the hundreds column.
Consider the hundreds column: \( 1 \text{ (carry)} + 6 = B \).
This means \( B = 7 \).
Thus, the completed addition is:
\( \begin{array}{r} 9 \, 2 \\ + 6 \, 3 \, 7 \\ \hline 7 \, 2 \, 9 \end{array} \)
Therefore, \( A = 9 \) and \( B = 7 \).
In simple words: Start from the rightmost column. 2 plus 7 equals A, so A is 9. In the middle column, A plus 3 must give a number ending in 2. Since A is 9, 9 plus 3 is 12. We write 2 and carry over 1. In the first column, 1 (carried over) plus 6 equals B. So B is 7. The sum is 92 + 637 = 729.

🎯 Exam Tip: In addition cryptarithms, carry-overs are crucial. Always remember to add the carry-over from the previous column to the current column's sum.

Question 9. If a 4-digit number 72x6 is divisible by 3. Find all values of x, where x is a digit.
Answer: For the four-digit number 72x6 to be divisible by 3, the sum of its digits must also be divisible by 3.
The sum of the digits is \( 7 + 2 + x + 6 = 15 + x \).
Since \( x \) is a single digit (0-9), we need \( 15 + x \) to be a multiple of 3.
Possible values for \( 15 + x \) that are multiples of 3 are 15, 18, 21, and 24.
If \( 15 + x = 15 \), then \( x = 0 \).
\( \implies \) If \( 15 + x = 18 \), then \( x = 3 \).
\( \implies \) If \( 15 + x = 21 \), then \( x = 6 \).
\( \implies \) If \( 15 + x = 24 \), then \( x = 9 \).
If \( 15 + x \) were 27, \( x \) would be 12, which is not a single digit.
Thus, the possible values for \( x \) are 0, 3, 6, or 9.
In simple words: For the number 72x6 to be divisible by 3, the sum of its digits must be divisible by 3. The digits are 7, 2, x, and 6, which add up to 15 + x. Since x can be any digit from 0 to 9, we find the values of x that make 15 + x a multiple of 3. These values are 0, 3, 6, and 9.

🎯 Exam Tip: Always remember that a digit (like 'x' here) must be a whole number from 0 to 9. Any result for 'x' outside this range means it's not a valid digit.