[4 marks sum]

Question 14 Marks

If $a = c$, show that $cba – abc = 0.$

Answer

Given: $a = c$
To show: $cba – abc = 0$
Proof:
$cba = 100c + 106 + a …………(i)$
$($By using property $3)$
$abc = 100a + 106 + c …………(ii)$
$($By using property $3)$
Since, $a = c,$
Substitute the value of $a = c$ in equation $(i)$ and $(ii)$, we get
$cba = 100c + 10b + c ……….(iii)$
$abc = 100c + 10b + c …………(iv)$
Subtracting $(iv)$ from $(iii)$, we get
$cba – abc – 100c + 106 + c – 100c – 106 – c$
$\Rightarrow cba – abc = 0$
$\Rightarrow cba = abc$
Hence proved.

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Question 24 Marks

If $a > c;$ show that $abc – cba = 99(a – c).$

Answer

Given, $a > c$
To show: $abc – cba = 99(a – c)$
Proof: $abc = 100a + 10b + c ……….(i)$
$($By using property $3)$
$cba = 100c + 10b + a ………..(ii)$
$($By using property $3)$
Subtracting, equation $(ii)$ from $(i)$, we get
$abc – cba = 100a + c – 100c – a$
$abc – cba = 99a – 99c$
$abc – cba = 99(a – c)$
Hence proved.

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Question 34 Marks

Show that $527 + 752 + 275$ is exactly divisible by $14.$

Answer

Property:
$abc = 100a + 106 + c ………(i)$
$bca = 1006 + 10c + a ……..(ii)$
and $cab = 100c + 10a + b ……….(iii)$
Adding,$ (i), (ii)$ and $(iii)$, we get $abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) = 3 \times 37(a + b + c)$
Now, let us try this method on
$527 + 752 + 275$ to check is it exactly divisible by $14$
Here, $a = 5, 6 = 2, c = 7$
$527 + 752 + 275 = 3 \times 37(5 + 2 + 7) = 3 \times 37 \times 14$
Hence, it shows that $527 + 752 + 275$ is exactly divisible by $14$.

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Question 44 Marks

If $a = 6$, show that $abc = bac.$

Answer

Given: $a = b$
To show: $abc = bac$
Proof:
Let $a = b = k ......(i)$
$k =$ any constant value
then $\text{LHS} = abc$
$abc = 100a + 10b + c$
From equation $(i)$
$abc = 100(k) + 10k + c$
$abc = 100k + 10k + c$
$abc = 110k + c$
$\text{RHS} = bac$
$\text{RHS} = 100b + 10a + c$
From equation $(i)$
$\text{RHS} = 100k + 10k + c$
$\text{RHS} = 110k + c$
Hence,
$\text{LHS = RHS}$
$abc = bac.$

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Question 54 Marks

Show that $954 – 459$ is exactly divisible by $99$.

Answer

To show: $954 – 459$ is exactly divisible by $3 99,$ where $a = 9, b = 5, c = 4$
$abc = 100a + 10b + c$
$\Rightarrow 954 = 100 \times 9 + 10 \times 5 + 4$
$\Rightarrow 954 = 900 + 50 + 4 ………(i)$
and $459 = 100 \times 4+ 10 \times 5 + 9$
$\Rightarrow 459 = 400 + 50 + 9 ……..(ii)$
Subtracting $(ii)$ from $(i),$ we get
$954 – 459 = 900 + 50 + 4 – 400 – 50 – 9$
$\Rightarrow 954 – 459 = 500 – 5$
$\Rightarrow 954 – 459 = 495$
$\Rightarrow 954 – 459 = 99 \times 5$
Hence, $954 – 459$ is exactly divisible by $99$
Hence proved.

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MATHS [4 marks sum]

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