16 Divisibility Rules
Which of the following numbers is divisible by 11?
Concept:
1
Divisibility by 11: A number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is either 0 or a multiple of 11.
Hint:
1
Test Option A: The digits in odd positions are 5, 2, and 9. The digits in even positions are 7 and 9. Add them up separately and find the difference!
Mistakes to Avoid:
!
Summing All Digits: Do not just add all the digits together. That is the rule for divisibility by 3 or 9, NOT 11. You must alternate them.
👉 Scroll horizontally to read the full handwritten solution.
Let's check the divisibility rule of 11 for option (A) $57299$:
Sum of digits at odd places:
$S_1 = 5 + 2 + 9 = 16$ [½ M]
Sum of digits at even places:
$S_2 = 7 + 9 = 16$ [½ M]
Difference = $S_1 - S_2$
Difference = $16 - 16 = 0$
Since the difference is 0, the number 57299 is divisible by 11.
Correct Option: (A)
17 Remainders
Which number will leave the same remainder as 7245 when divided by 9?
Concept:
1
Remainder Theorem for 9: The remainder when a number is divided by 9 is exactly the same as the remainder when the *sum of its digits* is divided by 9.
Hint:
1
Find the sum of the digits of 7245. What is the remainder when that sum is divided by 9? Then check which option gives that exact same remainder.
Mistakes to Avoid:
!
Long Division: Do not waste time doing long division for 7245 divided by 9. Use the digit-sum shortcut!
👉 Scroll horizontally to read the full handwritten solution.
Let's find the remainder of $7245 \div 9$.
Using the divisibility rule for 9, we sum the digits:
Sum = $7 + 2 + 4 + 5 = 18$ [½ M]
Since 18 is perfectly divisible by 9, the remainder is 0.
Now we look for an option that also leaves a remainder of 0
when divided by 9.
Option (A) is 9. And $9 \div 9$ leaves a remainder of 0. [½ M]
Correct Option: (A)
18 Algebraic Identities
Using identity, the value of $87 \times 113$ is:
Concept:
1
Difference of Squares: We can rewrite difficult multiplications using the identity $(a - b)(a + b) = a^2 - b^2$. Notice that 87 and 113 are equidistant from 100.
Hint:
1
Express 87 as $(100 - 13)$ and 113 as $(100 + 13)$. Now calculate $100^2 - 13^2$.
Mistakes to Avoid:
!
Ignoring the Instructions: If you just do long multiplication, you might get the right answer, but you are more prone to calculation errors. Use the identity to save time!
👉 Scroll horizontally to read the full handwritten solution.
We need to evaluate: $87 \times 113$
This can be written as: $(100 - 13)(100 + 13)$ [½ M]
Using the identity: $(a - b)(a + b) = a^2 - b^2$
Here, $a = 100$ and $b = 13$.
Substitute the values:
$= 100^2 - 13^2$
$= 10000 - 169$ [½ M]
$= 9831$
Correct Option: (A)
19 Divisibility Rules
Which of the following numbers is divisible by both 3 and 11?
Concept:
1
Dual Divisibility: The number must pass BOTH tests. For 3: The sum of the digits must be a multiple of 3. For 11: The difference of alternating sums must be 0 or 11.
Hint:
1
Check the rule for 3 first, as it is faster. Eliminate any options that fail. Then test the remaining options for 11!
Mistakes to Avoid:
!
Partial Checking: Do not select an answer just because it works for 3, without checking if it also works for 11.
👉 Scroll horizontally to read the full handwritten solution.
Let's test the options for divisibility by 3 first:
(A) $124 \rightarrow 1+2+4 = 7$ (Not divisible)
(B) $165 \rightarrow 1+6+5 = 12$ (Divisible!) [½ M]
(C) $184 \rightarrow 1+8+4 = 13$ (Not divisible)
(D) $245 \rightarrow 2+4+5 = 11$ (Not divisible)
Since only 165 passed the test for 3, let's verify it for 11:
Alternating sum for $165 \rightarrow (1 + 5) - 6 = 6 - 6 = 0$.
Since 0 is divisible by 11, it passes both tests! [½ M]
Correct Option: (B)
20 Angles in Quadrilateral
In a quadrilateral, three angles are $120^\circ$, $100^\circ$, and $80^\circ$. The value of the fourth angle "$x$" is:
Concept:
1
Angle Sum Property of Quadrilaterals: Regardless of the shape, the sum of all four interior angles of a quadrilateral is exactly $360^\circ$.
Hint:
1
Add up the three given angles: $120 + 100 + 80$. Subtract this total from 360 to find the missing angle $x$.
Mistakes to Avoid:
!
Wrong Total Sum: Some students confuse quadrilaterals with triangles and try to subtract from $180^\circ$. Always use $360^\circ$ for 4-sided figures!
👉 Scroll horizontally to read the full handwritten solution.
By the angle sum property of a quadrilateral:
Sum of all four angles = $360^\circ$
Therefore, we can write:
$120^\circ + 100^\circ + 80^\circ + x = 360^\circ$ [½ M]
$300^\circ + x = 360^\circ$
$x = 360^\circ - 300^\circ$
$x = 60^\circ$ [½ M]
Correct Option: (C)
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