RBSE Solutions Class 8 Maths Chapter 4 Mental Exercises More Ques

 

Question 1. Fill in the blanks
(i) 42 = __________ x 10 + 2
(ii) 60 = __________ x 10 + __________
(iii) 99 = __________ x __________ + __________
(iv) __________ = 7 x 100 + 1 x 10 + 8
Answer:
(i) 42 = 4 × 10 + 2
(ii) 60 = 6 × 10 + 0
(iii) 99 = 9 × 10 + 9
(iv) 718 = 7 × 100 + 1 × 10 + 8
In simple words: To fill in the blanks, we need to think about place values. For example, 42 is 4 tens and 2 ones. This way, we can break down numbers into hundreds, tens, and units.

🎯 Exam Tip: Remember that any two-digit number 'ab' can be written as \( 10a + b \), and a three-digit number 'abc' can be written as \( 100a + 10b + c \).

 

Question 2. Write the following numbers in generalized form
(i) 10 x 5 + 6
(ii) 8 × 100 + 0 × 10 + 5
(iii) 9 × 100 + 9 × 10 + 9
Answer:
(i) \( 10 \times 5 + 6 = 5 \times 10 + 6 \times 1 = 56 \)
(ii) \( 8 \times 100 + 0 \times 10 + 5 = 8 \times 100 + 0 \times 10 + 5 \times 1 = 805 \)
(iii) \( 9 \times 100 + 9 \times 10 + 9 = 9 \times 100 + 9 \times 10 + 9 \times 1 = 999 \)
In simple words: The generalized form shows how a number is made up of its digits multiplied by their place values (hundreds, tens, ones). Adding these parts together gives you the actual number.

🎯 Exam Tip: Always make sure to multiply each digit by its correct place value (1, 10, 100, etc.) before adding them up to get the standard number.

 

Question 3. What will be the result if you think the following numbers?
(i) 27
(ii) 67
(iii) 94
Answer:
(i) For the number 27:
The number thought is 27.
When we reverse its digits, we get 72.
The total of both numbers is \( 27 + 72 = 99 \).
Dividing this total by 11 gives \( \frac{99}{11} = 9 \), with a remainder of 0.
Subtracting the smaller number from the larger number gives \( 72 - 27 = 45 \).
Dividing this difference by 9 gives \( \frac{45}{9} = 5 \), with a remainder of 0.
(ii) For the number 67:
The number thought is 67.
When we reverse its digits, we get 76.
The total of both numbers is \( 67 + 76 = 143 \).
Dividing this total by 11 gives \( \frac{143}{11} = 13 \), with a remainder of 0.
Subtracting the smaller number from the larger number gives \( 76 - 67 = 9 \).
Dividing this difference by 9 gives \( \frac{9}{9} = 1 \), with a remainder of 0.
(iii) For the number 94:
The number thought is 94.
When we reverse its digits, we get 49.
The total of both numbers is \( 94 + 49 = 143 \).
Dividing this total by 11 gives \( \frac{143}{11} = 13 \), with a remainder of 0.
Subtracting the smaller number from the larger number gives \( 94 - 49 = 45 \).
Dividing this difference by 9 gives \( \frac{45}{9} = 5 \), with a remainder of 0.
In simple words: For any two-digit number, if you add it to its reversed version, the sum is always divisible by 11. If you subtract the smaller from the larger, the result is always divisible by 9.

🎯 Exam Tip: These are fun properties of two-digit numbers. Practice with different numbers to see these patterns, which can help in number puzzles.

 

Question 4. Test if Chhotu thinks the following numbers then what would be the result?
(i) 237
(ii) 119
(iii) 397<
(iv) 435
Answer:
(i) For the number 237:
Number thought by Chhotu is 237.
The number obtained after reversing its digits is 732.
When we interchange the digits to form other numbers using 2, 3, and 7, we can get 723 and 372. These are permutations of the original digits.
The sum of these three numbers is \( 237 + 723 + 372 = 1332 \).
Dividing the sum by 37 gives \( \frac{1332}{37} = 36 \). The result has a remainder of 0.
We can also observe that \( 36 = 3 \times 12 = 3 \times (2 + 3 + 7) \), which means the quotient is 3 times the sum of the original digits.
(ii) For the number 119:
Result 1 (Subtraction property):
Number thought by Chhotu is 119.
The number obtained after reversing its digits is 911.
Subtracting the smaller number from the larger number gives \( 911 - 119 = 792 \).
Dividing the result by 99 gives \( \frac{792}{99} = 8 \). The result has a remainder of 0.
Result 2 (Sum of permutations property):
Number thought by Chhotu is 119.
After interchanging the digits, we can get numbers like 911 and 191 (using digits 1, 1, 9).
The sum of these three numbers is \( 119 + 911 + 191 = 1221 \).
Dividing the sum by 37 gives \( \frac{1221}{37} = 33 \). The result has a remainder of 0.
The quotient \( 33 = 3 \times 11 = 3 \times (1 + 1 + 9) \), which is 3 times the sum of the digits.
(iii) For the number 397:
Result 1 (Subtraction property):
Number thought by Chhotu is 397.
The number obtained after reversing its digits is 793.
Subtracting the smaller number from the larger number gives \( 793 - 397 = 396 \).
Dividing the result by 99 gives \( \frac{396}{99} = 4 \). The result has a remainder of 0.
Result 2 (Sum of permutations property):
Number thought by Chhotu is 397.
After interchanging the digits, we can form numbers like 739 and 973 (using digits 3, 9, 7).
The total of these three numbers is \( 397 + 739 + 973 = 2109 \).
Dividing the sum by 37 gives \( \frac{2109}{37} = 57 \). The result has a remainder of 0.
The quotient \( 57 = 3 \times 19 = 3 \times (3 + 9 + 7) \), which is 3 times the sum of the digits.
(iv) For the number 435:
Result 1 (Subtraction property):
Number thought by Chhotu is 435.
The number obtained after reversing its digits is 534.
Subtracting the smaller number from the larger number gives \( 534 - 435 = 99 \).
Dividing the result by 99 gives \( \frac{99}{99} = 1 \). The result has a remainder of 0.
Result 2 (Sum of permutations property):
Number thought by Chhotu is 435.
After interchanging the digits, we can form numbers like 543 and 354 (using digits 4, 3, 5).
The total of these three numbers is \( 435 + 543 + 354 = 1332 \).
Dividing the sum by 37 gives \( \frac{1332}{37} = 36 \). The result has a remainder of 0.
The quotient \( 36 = 3 \times 12 = 3 \times (4 + 3 + 5) \), which is 3 times the sum of the digits.
In simple words: For three-digit numbers, if you form all possible numbers by mixing up the digits and add them, the sum will always be divisible by 37. Also, if you subtract a number from its reverse, the difference is often divisible by 99 or 495, depending on the digits.

🎯 Exam Tip: Remember these two common number puzzles: for a two-digit number 'ab', \( (10a+b) + (10b+a) = 11(a+b) \) and \( |(10a+b) - (10b+a)| = 9|a-b| \). For a three-digit number 'abc' with distinct digits, the sum of all permutations (abc, acb, bac, bca, cab, cba) is \( 222(a+b+c) \), and the sum of circular permutations (abc, bca, cab) is \( 111(a+b+c) \). Remember that \( 111 = 3 \times 37 \).

 

Question 6. If 79y is divisible by 9, then is it possible more than one value of y?
Answer:
The given number is 79y.
For a number to be divisible by 9, the sum of its digits must be divisible by 9.
The sum of the digits is \( 7 + 9 + y = 16 + y \).
This sum \( (16 + y) \) must be a multiple of 9.
Possible multiples of 9 are 18, 27, 36, and so on.
Since \( y \) is a single digit (meaning \( y \) can be 0, 1, 2, ..., 9), we check the possibilities:
If \( 16 + y = 18 \), then \( y = 18 - 16 = 2 \). This is a valid single digit.
If \( 16 + y = 27 \), then \( y = 27 - 16 = 11 \). This is not a single digit, so it's not possible.
Therefore, only one value of \( y \) is possible, which is 2.
In simple words: For a number to be divided by 9 without anything left over, its digits must add up to a number that can also be divided by 9. Since 'y' has to be a single number, only 2 works in this case.

🎯 Exam Tip: Always remember the divisibility rule for 9: "A number is divisible by 9 if the sum of its digits is divisible by 9." Also, when solving for a digit, remember that a digit must be between 0 and 9.

 

Question 7. Test the divisibility of 5629003 by 11
Answer:
To test the divisibility of 5629003 by 11, we use the alternating sum of digits rule.
Starting from the rightmost digit, we alternately add and subtract the digits:
\( 3 - 0 + 0 - 9 + 2 - 6 + 5 \)
\( = (3 + 0 + 2 + 5) - (0 + 9 + 6) \)
\( = 10 - 15 \)
\( = -5 \)
Since the result, -5, is not divisible by 11, the number 5629003 is not divisible by 11.
In simple words: To check if a number can be divided by 11, we add the digits in the odd places and subtract the digits in the even places. If the final answer can be divided by 11 (or is zero), then the whole number can be divided by 11.

🎯 Exam Tip: The divisibility rule for 11 is crucial: Calculate the alternating sum of the digits from right to left. If this sum is 0 or a multiple of 11, then the number is divisible by 11. Be careful with the order of operations and signs.

 

Question 8. Find the value of * in the following arithmetic-operations
(i)
\( \quad 3* \)
\( + 57 \)
\( + 34 \)
\( ----- \)
\( \quad 127 \)
(ii)
\( \quad 56 \)
\( + 77 \)
\( + *3 \)
\( ----- \)
\( \quad 216 \)
(iii)
\( \quad 443 \)
\( + *57 \)
\( + 128 \)
\( ----- \)
\( \quad 928 \)
(iv)
\( \quad 803 \)
\( - 2*6 \)
\( ----- \)
\( \quad 567 \)
Answer:
(i) For \( 3* + 57 + 34 = 127 \):
Let's look at the unit place digits: \( * + 7 + 4 \). The unit digit of the sum is 7.
So, \( * + 11 \) must end in 7. This means \( * + 1 \) must have a unit digit of 7.
Therefore, \( * = 6 \).
Let's check: \( 36 + 57 + 34 = 127 \). This is correct.
(ii) For \( 56 + 77 + *3 = 216 \):
Let's look at the unit place digits: \( 6 + 7 + 3 = 16 \). So, we write 6 and carry over 1 to the tens place.
Now, let's look at the tens place digits, including the carry-over: \( 1 + 5 + 7 + * \). The tens digit of the sum is 1.
So, \( 13 + * \) must end in 1. This means \( 13 + * = 21 \) (since the hundreds digit is 2).
Therefore, \( * = 21 - 13 = 8 \).
Let's check: \( 56 + 77 + 83 = 216 \). This is correct.
(iii) For \( 443 + *57 + 128 = 928 \):
Let's look at the unit place digits: \( 3 + 7 + 8 = 18 \). So, we write 8 and carry over 1 to the tens place.
Now, let's look at the tens place digits, including the carry-over: \( 1 + 4 + 5 + 2 = 12 \). So, we write 2 and carry over 1 to the hundreds place.
Now, let's look at the hundreds place digits, including the carry-over: \( 1 + 4 + * + 1 \). The hundreds digit of the sum is 9.
So, \( 6 + * = 9 \).
Therefore, \( * = 9 - 6 = 3 \).
Let's check: \( 443 + 357 + 128 = 928 \). This is correct.
(iv) For \( 803 - 2*6 = 567 \):
Let's look at the unit place digits: \( 3 - 6 \). Since 3 is smaller than 6, we borrow 10 from the tens place. So, \( (10 + 3) - 6 = 13 - 6 = 7 \). The unit digit of the difference is 7. This matches.
Now, let's look at the tens place digits. The original 0 in the tens place became 9 after borrowing (since it borrowed from the hundreds place first). So, \( 9 - * \) must equal 6 (the tens digit of the difference).
Therefore, \( * = 9 - 6 = 3 \).
Let's check the hundreds place: The original 8 in the hundreds place became 7 after lending to the tens place. So, \( 7 - 2 = 5 \). This matches.
So, \( * = 3 \).
Let's check: \( 803 - 236 = 567 \). This is correct.
In simple words: To find the missing star in these math problems, we work from right to left (units, then tens, then hundreds). We use what we know about adding or subtracting digits and remembering to carry over or borrow when needed.

🎯 Exam Tip: Always start solving these digit puzzle questions from the units column and move left. Don't forget to account for any carries in addition or borrows in subtraction, as they affect the next column.

 

Question 9. Find the value of * in the following subtraction operations
(i)
\( \quad 76 \)
\( - 5* \)
\( ----- \)
\( \quad 25 \)
(ii)
\( \quad 54 \)
\( - 2* \)
\( ----- \)
\( \quad 28 \)
(iii)
\( \quad 84 \)
\( - *8 \)
\( ----- \)
\( \quad 16 \)
(iv)
\( \quad 803 \)
\( - 2*6 \)
\( ----- \)
\( \quad 567 \)
(v)
\( \quad 782 \)
\( - *73 \)
\( ----- \)
\( \quad 209 \)
Answer:
(i) For \( 76 - 5* = 25 \):
Look at the unit place: \( 6 - * = 5 \). This means \( * = 6 - 5 = 1 \).
Look at the tens place: \( 7 - 5 = 2 \). This matches the given result.
So, the blank digit \( * \) is 1.
(ii) For \( 54 - 2* = 28 \):
Look at the unit place: \( 4 - * = 8 \). Since 4 is smaller than 8, we need to borrow 10 from the tens place.
So, \( (10 + 4) - * = 8 \), which means \( 14 - * = 8 \). Therefore, \( * = 14 - 8 = 6 \).
Look at the tens place: The 5 in the tens place became 4 after lending. So, \( 4 - 2 = 2 \). This matches.
So, the blank digit \( * \) is 6.
(iii) For \( 84 - *8 = 16 \):
Look at the unit place: \( 4 - 8 \). Since 4 is smaller than 8, we borrow 10 from the tens place.
So, \( (10 + 4) - 8 = 14 - 8 = 6 \). This matches the unit digit of the result.
Look at the tens place: The 8 in the tens place became 7 after lending. So, \( 7 - * = 1 \). This means \( * = 7 - 1 = 6 \).
So, the blank digit \( * \) is 6.
(iv) For \( 803 - 2*6 = 567 \):
Look at the unit place: \( 3 - 6 \). We borrow 10 from the tens place. So, \( (10 + 3) - 6 = 7 \). This matches.
Look at the tens place: The 0 in the tens place became 9 (after the 8 in hundreds lent 1, and then 0 borrowed 10). So, \( 9 - * = 6 \). This means \( * = 9 - 6 = 3 \).
Look at the hundreds place: The 8 in the hundreds place became 7 after lending. So, \( 7 - 2 = 5 \). This matches.
So, the blank digit \( * \) is 3.
(v) For \( 782 - *73 = 209 \):
Look at the unit place: \( 2 - 3 \). We borrow 10 from the tens place. So, \( (10 + 2) - 3 = 9 \). This matches.
Look at the tens place: The 8 in the tens place became 7 after lending. So, \( 7 - 7 = 0 \). This matches.
Look at the hundreds place: \( 7 - * = 2 \). This means \( * = 7 - 2 = 5 \).
So, the blank digit \( * \) is 5.
In simple words: When finding missing digits in subtraction, always start with the rightmost column (the ones place). If a top digit is smaller than the bottom digit, remember to borrow from the next column to the left, which changes its value.

🎯 Exam Tip: Pay close attention to borrowing in subtraction. A digit that lends 10 to the right decreases by 1, and a 0 in the middle becomes a 9 if it borrowed from the left, before lending to the right.

 

Question 10. Find the value of x in the following multiplication operations.
(i) \( 56 \times x5 = 1400 \)
(ii) \( 4x \times 37 = 1554 \)
(iii) \( 23 \times 3x = 736 \)
Answer:
(i) Given multiplication: \( 56 \times x5 = 1400 \)
Here, \( x \) is the digit in the tens place of the second number. So, \( x5 \) can be written as \( 10x + 5 \).
Thus, the equation becomes: \( 56 \times (10x + 5) = 1400 \).
Now, we expand the multiplication:
\( 56 \times 10x + 56 \times 5 = 1400 \)
\( 560x + 280 = 1400 \)
Subtract 280 from both sides:
\( 560x = 1400 - 280 \)
\( 560x = 1120 \)
Divide by 560:
\( x = \frac{1120}{560} \)
\( x = 2 \).
So, the value of \( x \) is 2. The operation is \( 56 \times 25 = 1400 \).
(ii) Given multiplication: \( 4x \times 37 = 1554 \)
Here, \( x \) is the digit in the unit place of the first number. So, \( 4x \) can be written as \( 40 + x \).
Thus, the equation becomes: \( (40 + x) \times 37 = 1554 \).
Now, we expand the multiplication:
\( 40 \times 37 + x \times 37 = 1554 \)
\( 1480 + 37x = 1554 \)
Subtract 1480 from both sides:
\( 37x = 1554 - 1480 \)
\( 37x = 74 \)
Divide by 37:
\( x = \frac{74}{37} \)
\( x = 2 \).
So, the value of \( x \) is 2. The operation is \( 42 \times 37 = 1554 \).
(iii) Given multiplication: \( 23 \times 3x = 736 \)
Here, \( x \) is the digit in the unit place of the second number. So, \( 3x \) can be written as \( 30 + x \).
Thus, the equation becomes: \( 23 \times (30 + x) = 736 \).
Now, we expand the multiplication:
\( 23 \times 30 + 23 \times x = 736 \)
\( 690 + 23x = 736 \)
Subtract 690 from both sides:
\( 23x = 736 - 690 \)
\( 23x = 46 \)
Divide by 23:
\( x = \frac{46}{23} \)
\( x = 2 \).
So, the value of \( x \) is 2. The operation is \( 23 \times 32 = 736 \).
In simple words: To find the missing number 'x' in multiplication, we first write the number with 'x' using place values (like \( 10x + 5 \)). Then, we solve the equation by doing the opposite of the operations shown.

🎯 Exam Tip: When a digit 'x' is part of a number (like \( x5 \) or \( 4x \)), remember to express it correctly as \( 10x + 5 \) or \( 40 + x \) before solving the equation. This is key for algebra in number puzzles.

 

Question 11. Find the value of x in following division-operation.
(i) \( 27) 217 (x \) with remainder \( 1 \)
(ii) \( x6) 100 (6 \) with remainder \( 4 \)
(iii) \( 1x) 120 (9 \) with remainder \( 3 \)
Answer:
(i) For the division \( 217 \div 27 \), the quotient is \( x \) and the remainder is 1.
We use the formula: Dividend = Divisor \( \times \) Quotient + Remainder.
So, \( 217 = 27 \times x + 1 \).
Subtract 1 from both sides:
\( 217 - 1 = 27x \)
\( 216 = 27x \)
Divide by 27:
\( x = \frac{216}{27} \)
\( x = 8 \).
So, the value of \( x \) is 8. The operation is \( 217 = 27 \times 8 + 1 \).
(ii) For the division \( 100 \div x6 \), the quotient is 6 and the remainder is 4.
Here, \( x6 \) represents the divisor \( 10x + 6 \).
Using the formula: Dividend = Divisor \( \times \) Quotient + Remainder.
So, \( 100 = (10x + 6) \times 6 + 4 \).
Expand the right side:
\( 100 = 60x + 36 + 4 \)
\( 100 = 60x + 40 \)
Subtract 40 from both sides:
\( 100 - 40 = 60x \)
\( 60 = 60x \)
Divide by 60:
\( x = \frac{60}{60} \)
\( x = 1 \).
So, the value of \( x \) is 1. The operation is \( 100 = 16 \times 6 + 4 \).
(iii) For the division \( 120 \div 1x \), the quotient is 9 and the remainder is 3.
Here, \( 1x \) represents the divisor \( 10 + x \).
Using the formula: Dividend = Divisor \( \times \) Quotient + Remainder.
So, \( 120 = (10 + x) \times 9 + 3 \).
Subtract 3 from both sides:
\( 120 - 3 = (10 + x) \times 9 \)
\( 117 = (10 + x) \times 9 \)
Divide by 9:
\( \frac{117}{9} = 10 + x \)
\( 13 = 10 + x \)
Subtract 10 from both sides:
\( x = 13 - 10 \)
\( x = 3 \).
So, the value of \( x \) is 3. The operation is \( 120 = 13 \times 9 + 3 \).
In simple words: For division problems, we use the rule that the big number being divided (dividend) equals the number doing the dividing (divisor) times the answer (quotient), plus any leftover (remainder). By putting the numbers into this rule, we can find the missing 'x'.

🎯 Exam Tip: Always remember the basic relationship: Dividend = Divisor \( \times \) Quotient + Remainder. This formula is essential for solving division problems with missing values.

 

Question 11. Find the value of x in following division-operation.
(iii) The division is represented as \( 1x ) 120 ( 9 \) with a remainder of \( 3 \).
Answer: We know that in any division, the relationship is: Dividend = Divisor \( \times \) Quotient + Remainder.
\( \implies \) Here, the Dividend is 120, the Divisor is \( 1x \) (which means \( 10 + x \)), the Quotient is 9, and the Remainder is 3.
\( \implies 120 = (10 + x) \times 9 + 3 \)
\( \implies 120 = 90 + 9x + 3 \)
\( \implies 120 = 93 + 9x \)
\( \implies 120 - 93 = 9x \)
\( \implies 27 = 9x \)
\( \implies x = \frac { 27 }{ 9 } \)
\( \implies x = 3 \) Therefore, the value of x is 3. This makes the divisor 13.
In simple words: To find 'x', we use the rule that the dividend (120) equals the divisor (1x) multiplied by the quotient (9) plus the remainder (3). By solving this simple equation, we find the number that 'x' stands for.

🎯 Exam Tip: Always double-check your answer by plugging the value of x back into the original division problem to ensure the equation holds true.

 

Question 12. What is the sum of digits of vertical and horizontal digits? Is this total same?
Answer: Let's find the sum of digits in each vertical column and horizontal row of the given number grid.

Sum of digits in vertical columns:
(i) Column 1: \( 2 + 9 + 4 = 15 \)
(ii) Column 2: \( 7 + 5 + 3 = 15 \)
(iii) Column 3: \( 6 + 1 + 8 = 15 \)
Sum of digits in horizontal rows:
(i) Row 1: \( 2 + 7 + 6 = 15 \)
(ii) Row 2: \( 9 + 5 + 1 = 15 \)
(iii) Row 3: \( 4 + 3 + 8 = 15 \) Yes, the total sum of digits is the same for all vertical columns and horizontal rows in this puzzle. This is a common property found in magic squares.
In simple words: When we add up the numbers in each row and each column, the answer is always 15. So, yes, the totals are the same.

🎯 Exam Tip: For puzzles like these, clearly show each sum calculation to avoid mistakes and make your answer easy to follow.

 

Question 13. In a square puzzle of 4 x 4, we change diagonally the digits written in symbols to \( \triangle \) to \( \bigcirc \) to \( \diamondsuit \) to \( \square \) and to. Find sum of digits of the vertical column and horizontal total of squares.
Answer: Let's consider the standard 4x4 square puzzle where numbers are arranged such that sums across rows, columns, and diagonals are equal. The example grid provided in the solution is:

For this puzzle, the sum of digits in each horizontal row and vertical column is 34. This type of grid, where all row, column, and main diagonal sums are equal, is known as a magic square.
In simple words: For this 4x4 puzzle, if you add numbers across any row or down any column, the total sum is always 34. This makes it a special kind of number puzzle called a magic square.

🎯 Exam Tip: When dealing with number puzzles, confirm sums for all rows, columns, and at least the main diagonals to verify its properties like being a magic square.

 

Question 14. If 2 to 17 numbers are entered in a 4 x 4 square puzzle, what will be the sum?
Answer: When numbers from 2 to 17 are entered into a 4x4 square puzzle in a specific order, as shown below, the sum of each horizontal row, vertical column, and both main diagonal lines is 38.

This arrangement forms a specific type of numerical sequence within the square. The sum across any row, down any column, or along the main diagonals is 38.
In simple words: When you place numbers from 2 to 17 into a 4x4 box in order, if you add the numbers in any row, column, or main corner-to-corner line, the total will always be 38.

🎯 Exam Tip: When a problem involves a sequence of numbers in a grid, try to identify the pattern or mathematical properties of the arrangement to quickly find the required sums.