Question 14 Marks

Tanya had some sweets which she distributed among her five friends $A, B, C, D$ and $E$. She gave $x$ sweets to $A$. To $B$, she gave 10 sweets less than twice of those she gave to $A$. To $C$, she gave 4 sweets more than 4 times of those she gave to $A$. To $D$, she gave $(x+12)$ sweets more than those she gave to $B$. To $E$, she gave $(11-x)$ sweets less than those she gave to $C$. Tanya still had 16 sweets left.
(1) If $C$ got 28 sweets, how many sweets did Tanya have in all, in the beginning?
(a) 79$\quad$(b) 85$\quad$(c) 91$\quad$(d) 95
(2) How many more sweets does $E$ have than $D$ ?
(a) $(-2 x+9)$$\quad$(b) $(2 x-9)$$\quad$(c) $(2 x-5)$$\quad$(d) $(-2 x+5)$
(3) $B$ and $C$ mixed their sweets together and then distributed them equally between themselves. Which of the following algebraic expressions denotes the number of sweets that each of them got?
(a) $3(x-1)$$\quad$(b) $3(x-2)$$\quad$(c) $3(x+1)$$\quad$(d) $3(x+2)$
(4) Had Tanya distributed all the sweets equally amongst her five friends then how many toffees would each friends get?
(a) $\left(5 x-\frac{11}{3}\right)$$\quad$(b) $(3 x+1)$$\quad$(c) $\left(3 x-\frac{11}{3}\right)$$\quad$(d) $(3 x-1)$

Answer

$\begin{array}{l}\text { Number of sweets } A \text { got }=x \\ \text { Number of sweets } B \text { got }=(2 x-10) \\ \text { Number of sweets } C \text { got }=(4 x+4) \\ \text { Number of sweets } D \text { got }=(2 x-10)+(x+12)=(3 x+2) \\ \text { Number of sweets } E \text { got }=(4 x+4)-(11-x)=4 x+4-11+x=(5 x-7) .\end{array}$

(1) (D) 95
$4 x+4=28 \Rightarrow 4 x=24 \Rightarrow x=6$
Total number of sweets with Tanya
$
\begin{array}{l}
=x+(2 x-10)+(4 x+4)+(3 x+2)+(5 x-7)+16 \\
=(15 x+5)=15 \times 6+5=95
\end{array}
$

(2) (B) $(2 x-9)$
Required difference $=(5 x-7)-(3 x+2)=5 x-7-3 x-2=(2 x-9)$

(3) (A) $3(x-1)$
Total number of sweets with $B$ and $C=(2 x-10)+(4 x+4)=(6 x-6)=6(x-1)$
Number of sweets each of them got $=\frac{1}{2}[6(x-1)]=3(x-1)$.

(4) (B) $(3 x+1)$
Total number of sweets with Tanya $=(15 x+5)$.
$\therefore \text { required number }=\frac{(15 x+5)}{5}=(3 x+1)$

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