5 Marks Questions

Question 15 Marks

Sasha solves a linear equation. Her work is shown below
$\begin{aligned} 2 x-3 & =\frac{x}{2}-5 \\ 2(2 x-3) & =x-5 \ldots \ldots \text { Step } 1 \\ 4 x-6 & =x-5 \ldots \ldots \text { Step } 2 \\ 4 x-x & =6-5 \ldots \ldots \text { Step } 3 \\ 3 x & =1 \ldots \ldots \text { Step } 4 \\ x & =\frac{1}{3}\end{aligned}$
Is Sasha's solution correct? If not, in which step did Sasha make an error?

Answer

No, Sasha's solution is not correct. She did an error in Step 1. She should transpose $\frac{x}{2}$ to LHS and -3 to RHS.

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Question 25 Marks

A linear equation can be written as $a x+b=0$, where $a$ and $b$ are constants. Write the equation $3(x-1)=5(2 x+1)$ in the form of $a x+b=0$.

Answer

We have, $3(x-1)=5(2 x+1)$
$\Rightarrow \quad 3 x-3=10 x+5 \quad$ [on opening the brackets]
$\Rightarrow \quad-5-3=10 x-3 x\quad$ [transposing 5 to LHS and $3 x$ to RHS]
$\Rightarrow \quad-8=7 x$
$\therefore \quad x=-\frac{8}{7} \quad$ [dividing both sides by 7 ]

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Question 35 Marks

Find the value of $2 m+\frac{1}{2} n$, if $m$ and $n$ are the solutions of the equations $\frac{m+3}{7-2 m}=\frac{1}{2}$ and $\frac{1}{4}(n+4)=2 n-3$, respectively.

Answer

Given, $\frac{m+3}{7-2 m}=\frac{1}{2} \Rightarrow 2(m+3)=1 \times(7-2 m)$
$\Rightarrow \quad 2 m+6=7-2 m \quad$ [on opening the brackets]
$\Rightarrow \quad 2 m+2 m=7-6\quad$ [transposing $-2 m$ to LHS and 6 to RHS]
$\Rightarrow \quad 4 m=1 \Rightarrow m=\frac{1}{4}\quad\ldots\text{(i)}$
Now, $\frac{1}{4}(n+4)=2 n-3$
$\Rightarrow \quad n+4=4(2 n-3)$
$\Rightarrow \quad n+4=8 n-12$
$\Rightarrow \quad 8 n-n=12+4$
$\Rightarrow \quad 7 n=16 \Rightarrow n=\frac{16}{7}\quad\ldots\text{(ii)}$
Then, $\quad 2 m+\frac{1}{2} n=2 \times \frac{1}{4}+\frac{1}{2} \times \frac{16}{7}$
$=\frac{1}{2}+\frac{8}{7}=\frac{7+16}{14}=\frac{23}{14}$
So, $\quad 2 m+\frac{1}{2} n=\frac{23}{14}$

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Question 45 Marks

Game : Who will be the Lakhpati?
Rohit and Saurabh are playing a game. The one who solves the following equations will be a winner. Find out if you were at their place would you have been be a winner. Till what money did you reach successfully?
Rules of the game
(a) You can only move to the next problem if the previous answer is correct.
(b) Winning amount slab

Question Number	Amount Won
1	₹ 1000
2	₹ 2000
3	₹ 3000
4	₹ 4000
5	₹ 10000
6	₹ 12000
7	₹ 14000
8	₹ 20000
9	₹ 40000
10	₹ 100000

(c) Problems:
(i) $\frac{x+1}{2 x+7}=\frac{3}{8}$
(ii) $\frac{1}{(x-1)}+\frac{2}{(x+1)}=2$
(iii) $\frac{6 x+1}{3}+1=\frac{x-3}{6}$
(iv) $3 m=7 m-\frac{8}{7}$
(v) $-x=\frac{-6}{5}(x-10)$
(vi) $5 x+\frac{7}{2}=\frac{3}{2} x-14$
(vii) $\frac{x}{3}+1=\frac{8}{15}$
(viii) $\frac{x}{2}+\frac{3 x}{4}-\frac{5 x}{6}=2$
(ix) $\frac{50}{x}+4=14$
(x) $x+\frac{2}{3} x+\frac{x}{7}=97-\frac{x}{2}$

Answer

(c) (i) $x=6 \frac{1}{2}$ $\quad$(ii) $x=1$
(iii) $x=-1$$\qquad$(iv) $m=\frac{2}{7}$
(v) $x=60$ $\qquad$ (vi) $x=-5$
(vii) $x=\frac{-7}{5}$ $\quad$ (viii) $x=\frac{24}{5}$
(ix) $x=5$ $\qquad$ (x) $x=42$

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Question 55 Marks

Ranika wanted her friend Radhika's mobile number. But Radhika played a trick. She gave her the number as
$9 X Y Z P 1 Q 2 R 3$
and told her to decode it with the help of following equations:
(a) $16 X-35=7 X-8$
(b) $\frac{6 Y-7}{3 Y+9}=\frac{1}{3}$
(c) $\frac{4 Z-5}{8+6 z}=\frac{3}{20}$
(d) $P+\frac{3}{10} P=\frac{13}{10}$
(e) $4(Q+4)=5(Q+2)$
(f) $3(R+10)+200=236$

Answer

(a) $X=3$ $\quad$ (b) $Y=2$
(c) $Z=2$ $\quad$ (d) $P=1$
(e) $Q=6$ $\quad$ (f) $R=2$

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Question 65 Marks

$\text {Find the value of } x \text {, if } \frac{x}{2}+\frac{3 x}{8}+9=x$

Answer

We have, $\frac{x}{2}+\frac{3 x}{8}+9=x$
On multiplying by 8 (LCM of denominator) on both sides, we get
$8 \times \frac{x}{2}+\frac{3 x}{8} \times 8+9 \times 8=8 \times x$
$\Rightarrow \quad 4 x+3 x+72=8 x$
$\Rightarrow \quad 7 x+72=8 x$
$\Rightarrow \quad 72=8 x-7 x \quad$ [transposing $7 x$ to RHS]
$\therefore \quad 72=x$

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Question 75 Marks

Solve $13(y-4)-4(y-6)+5(y-9)=0$

Answer

We have,
$13(y-4)-4(y-6)+5(y-9)=0$
$\Rightarrow \quad 13 y-52-4 y+24+5 y-45=0\quad$ [on opening the brackets]
$\Rightarrow \quad 14 y-73=0$
$\Rightarrow \quad 14 y=73 \quad$ [transposing 73 to RHS]
$\Rightarrow \quad y=\frac{73}{14} \quad$ [dividing both sides by 14]

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Question 85 Marks

Simplify and solve the following equation.
$16(y+3)-3 y+4(y+5)=0$

Answer

We have, $16(y+3)-3 y+4(y+5)=0$
Let us open the brackets,
$ \text { LHS }=16 y+48-3 y+4 y+20 $
The equation is $16 y+48-3 y+4 y+20=0$
$\Rightarrow \quad 17 y+68=0$
$\Rightarrow \quad 17 y=-68 \quad[$transposing 68 to RHS$]$
$\Rightarrow \quad y=\frac{-68}{17} \quad$ [dividing both sides by 17]
$\therefore \quad y=-4$

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Question 95 Marks

Solve $0.35(4 x-2)=0.05(12 x+0.4)$

Answer

We have, $0.35(4 x-2)=0.05(12 x+0.4)$
Let us open the bracket,
$ \begin{array}{l} \text { LHS }=0.35 \times 4 x-0.35 \times 2=1.4 x-0.7 \\ \text { RHS }=0.05 \times 12 x+0.05 \times 0.04=0.6 x+0.02 \end{array} $
Now, the equation is
$14 x-0.7=0.6 x+0.02 \Rightarrow 1.4 x-0.6 x=0.7+0.02\quad$
[transposing $0.6 x$ to LHS and -0.7 to RHS]
$\Rightarrow \quad 0.8 x=0.72 \Rightarrow x=\frac{0.72}{0.8}$
$\therefore \quad x=0.9$

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Question 105 Marks

Solve the following equations and check your result
(i) $\frac{x}{5}+11=\frac{1}{15}\quad$ (ii) $\frac{1}{3} x-\frac{1}{2}=4$

Answer

(i) We have, $\frac{x}{5}+11=\frac{1}{15}$
On multiplying by 15 , i.e. ICM of $(5,15)$, we get
$\frac{x}{5} \times 15+11 \times 15=\frac{1}{15} \times 15$
$\Rightarrow \quad 3 x+165=1$
$\Rightarrow \quad 3 x=1-165$ [transposing 165 to RHS]
$\Rightarrow \quad 3 x=-164$
$\therefore \quad x=\frac{-164}{3}$
Check LHS $=\frac{x}{5}+11=\frac{-164}{3 \times 5}+11$
$=\frac{-164+165}{15}=\frac{1}{15}=$ RHS
(ii) We have, $\frac{1}{3} x-\frac{1}{2}=4$
On multiplying by 6 i.e. LCM of $(3,2)$ both sides, we get
$6 \times \frac{1}{3} x-6 \times \frac{1}{2}=6 \times 4$
$\Rightarrow \quad 2 x-3=24$
$\Rightarrow \quad 2 x=24+3 \quad[$transposing -3 to RHS$]$
$\Rightarrow \quad 2 x=27 \Rightarrow x=\frac{27}{2}$
Check LHS $=\frac{1}{3} x-\frac{1}{2}$
$=\frac{1}{3} \times \frac{27}{2}-\frac{1}{2}=\frac{9}{2}-\frac{1}{2}=\frac{8}{2}=4=$ RHS

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MATHS 5 Marks Questions

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