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All questions of Algebra Play for Class 8 Exam
In a number pyramid, the bottom row is: a, b, c.
The top number is:- a)a + b + c
- b)a + 2b + c
- c)2a + b + c
- d)a + b + 2c
Correct answer is option 'B'. Can you explain this answer?
In a number pyramid, the bottom row is: a, b, c.
The top number is:
The top number is:
a)
a + b + c
b)
a + 2b + c
c)
2a + b + c
d)
a + b + 2c
 | Nidhi Bhatt answered |
Middle row: a+b , b+c
Top = (a+b)+(b+c) = a+2b+c
Top = (a+b)+(b+c) = a+2b+c
In the "Addition Trick," the sum of a two-digit number and its reverse is always divisible by?- a)7
- b)11
- c)9
- d)3
Correct answer is option 'B'. Can you explain this answer?
a)
7
b)
11
c)
9
d)
3
 | Kritika Chopra answered |
Understanding the Addition Trick
The Addition Trick involves a two-digit number and its reverse. Let's break down why the sum of these numbers is always divisible by 11.
Two-Digit Number Representation
- Let the two-digit number be represented as AB, where A is the tens digit and B is the units digit.
- Mathematically, this can be expressed as: 10A + B.
Reversed Number
- The reverse of the number AB is BA, which can be represented as: 10B + A.
Calculating the Sum
- Now, let's find the sum of the original number and its reversed form:
Sum = (10A + B) + (10B + A)
= 10A + B + 10B + A
= 11A + 11B
= 11(A + B).
Divisibility by 11
- The expression 11(A + B) clearly shows that the sum is a multiple of 11 because it is 11 times the sum of the digits (A + B).
- Since A and B are digits (0-9), A + B can range from 0 to 18, but regardless of its value, 11(A + B) itself is always divisible by 11.
Conclusion
- Therefore, the sum of a two-digit number and its reverse is always divisible by 11.
- The correct answer to the Addition Trick is option 'B' (11).
This fascinating property highlights the unique characteristics of two-digit numbers and their reversals!
The Addition Trick involves a two-digit number and its reverse. Let's break down why the sum of these numbers is always divisible by 11.
Two-Digit Number Representation
- Let the two-digit number be represented as AB, where A is the tens digit and B is the units digit.
- Mathematically, this can be expressed as: 10A + B.
Reversed Number
- The reverse of the number AB is BA, which can be represented as: 10B + A.
Calculating the Sum
- Now, let's find the sum of the original number and its reversed form:
Sum = (10A + B) + (10B + A)
= 10A + B + 10B + A
= 11A + 11B
= 11(A + B).
Divisibility by 11
- The expression 11(A + B) clearly shows that the sum is a multiple of 11 because it is 11 times the sum of the digits (A + B).
- Since A and B are digits (0-9), A + B can range from 0 to 18, but regardless of its value, 11(A + B) itself is always divisible by 11.
Conclusion
- Therefore, the sum of a two-digit number and its reverse is always divisible by 11.
- The correct answer to the Addition Trick is option 'B' (11).
This fascinating property highlights the unique characteristics of two-digit numbers and their reversals!
How does the "3-Digit Cycling Trick" demonstrate properties of number divisibility?- a)It shows a number is divisible by 37
- b)It shows a number is divisible by 9
- c)It shows a number is divisible by 3
- d)It shows a number is divisible by 11
Correct answer is option 'A'. Can you explain this answer?
a)
It shows a number is divisible by 37
b)
It shows a number is divisible by 9
c)
It shows a number is divisible by 3
d)
It shows a number is divisible by 11
| | Nidhi Bhatt answered |
Let the 3-digit number be N = 100a + 10b + c.
The two cyclic permutations are 100b + 10c + a and 100c + 10a + b.
Their sum is (100a+10b+c)+(100b+10c+a)+(100c+10a+b) = 111(a+b+c).
Since 111 = 3 × 37, the sum is a multiple of 37. Hence the trick demonstrates divisibility by 37; option A is correct.
If p < q < r, the largest product formed using pq × r, qp × r, rp × q is:- a)pq × r
- b)qp × r
- c)rp × q
- d)rq × p
Correct answer is option 'B'. Can you explain this answer?
If p < q < r, the largest product formed using pq × r, qp × r, rp × q is:
a)
pq × r
b)
qp × r
c)
rp × q
d)
rq × p
| | Nidhi Bhatt answered |
Largest digit should be multiplier and others in decreasing order → qp × r
In a number pyramid, if a + b = 60 and 12 + c = a, what is the value of c when these equations are solved?- a)10
- b)40
- c)30
- d)20
Correct answer is option 'D'. Can you explain this answer?
In a number pyramid, if a + b = 60 and 12 + c = a, what is the value of c when these equations are solved?
a)
10
b)
40
c)
30
d)
20
| | Nidhi Bhatt answered |
To find the value of c, we use the equations given: a + b = 60, 12 + c = a, and c + 8 = b. By substituting b from the first equation into the second, we find c = 20. This highlights how algebra can simplify complex relationships into solvable equations.
If you take a 3-digit number, cycle its digits, and add all three resulting numbers, what is the sum always divisible by?- a)111
- b)9
- c)37
- d)3
Correct answer is option 'C'. Can you explain this answer?
If you take a 3-digit number, cycle its digits, and add all three resulting numbers, what is the sum always divisible by?
a)
111
b)
9
c)
37
d)
3
| | Nidhi Bhatt answered |
The sum of the three numbers formed by cycling the digits of a 3-digit number is always divisible by 37. This occurs because the sum can be expressed as 111(a + b + c), where a, b, and c are the digits, and since 111 is divisible by 37, the entire sum will also be divisible by 37, showcasing interesting properties of number manipulation.
What is the final result of the classic "Think of a Number" trick when you follow these steps: think of a number, double it, add four, divide by two, and subtract the original number?- a)2
- b)1
- c)0
- d)3
Correct answer is option 'A'. Can you explain this answer?
What is the final result of the classic "Think of a Number" trick when you follow these steps: think of a number, double it, add four, divide by two, and subtract the original number?
a)
2
b)
1
c)
0
d)
3
| | Nidhi Bhatt answered |
Let the original number be x.
After doubling: 2x.
After adding 4: 2x + 4.
After dividing by 2: (2x + 4)/2 = x + 2.
After subtracting the original number: (x + 2) - x = 2.
Final result: 2.
What is the largest product you can create using the digits 2, 3, and 5 in the format __ __ × __?- a)106
- b)160
- c)156
- d)115
Correct answer is option 'B'. Can you explain this answer?
What is the largest product you can create using the digits 2, 3, and 5 in the format __ __ × __?
a)
106
b)
160
c)
156
d)
115
| | Nidhi Bhatt answered |
Form a two-digit number from two of the digits and use the remaining digit as the multiplier; evaluate all possibilities.
53 × 2 = 106
35 × 2 = 70
52 × 3 = 156
25 × 3 = 75
32 × 5 = 160
23 × 5 = 115
The largest product is 160, achieved by 32 × 5.
In the date trick, the final expression obtained is:
100M + 165 + D.
Why can the date be extracted correctly?- a)Because 165 is a multiple of 31
- b)Because D always has two digits and M has one or two digits
- c)Because 100M separates month and day digits
- d)Because M and D are prime numbers
Correct answer is option 'C'. Can you explain this answer?
In the date trick, the final expression obtained is:
100M + 165 + D.
Why can the date be extracted correctly?
100M + 165 + D.
Why can the date be extracted correctly?
a)
Because 165 is a multiple of 31
b)
Because D always has two digits and M has one or two digits
c)
Because 100M separates month and day digits
d)
Because M and D are prime numbers
| | Nidhi Bhatt answered |
100M shifts the month to the hundreds place and D occupies last two digits.
In the calendar magic trick, if your friend tells you the sum of a 2x2 grid is 44, what is the formula to express the sum of the numbers in the grid in terms of the top-left number, a?- a)3a + 12
- b)4a + 16
- c)4a + 12
- d)5a + 10
Correct answer is option 'B'. Can you explain this answer?
a)
3a + 12
b)
4a + 16
c)
4a + 12
d)
5a + 10
| | Nidhi Bhatt answered |
On a calendar, moving one square right adds 1 and moving one row down adds 7, so a 2×2 block with top-left entry a contains: a, a+1, a+7, a+8.
Sum = a + (a+1) + (a+7) + (a+8).
Sum = 4a + 16.
As a check, if the sum is 44 then a = (44 - 16)/4 = 7.
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