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Maths 5 Marks Questions

Simple Equations · CBSE STD 7 (CBSE - English Medium). 3 questions.

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Question 15 Marks
S.No Column $I$   Column $II$
$i$ $x + 5 = 9$ $a$ $-\frac{5}{3}$
$ii$ $x - 7 = 4$ $b$ $\frac{5}{3}$
$iii$ $\frac{\text{x}}{12}=-5$ $c$ $4$
$iv$ $5x = 30$ $d$ $6$
$v$ The value of y which satisfies $3y = 5$ $e$ $11$
$vi$ If $p = 2$, then the value of $\frac{1}{3}(1-3\text{p})$ $f$ $-60$
    $g$ $3$
Answer
S.No Column $I$   Column $II$
$i$ $x + 5 = 9$ $c$ $4$
$ii$ $x - 7 = 4$ $e$ $11$
$iii$ $\frac{\text{x}}{12}=-5$ $f$ $-60$
$iv$ $5x = 30$ $d$ $6$
$v$ The value of y which satisfies $3y = 5$ $b$ $\frac{5}{3}$
$vi$ If $p = 2$, then the value of $\frac{1}{3}(1-3\text{p})$ $a$ $-\frac{5}{3}$
Solution:
$i.$ Given equation is $x + 5 = 9$
$\Rightarrow x = 9 - 5 [$transposing $5$ to $\text{RHS}]$
$\Rightarrow x = 4$
$ii.$ Given equation is $x - 7 = 4$
$\Rightarrow x = 4 + 7 [$transposing $(-7)$ to $\text{RHS}]$
$\Rightarrow x = 11$
$iii.$ Given equation is $\frac{\text{x}}{12}=-5$
$\Rightarrow12\times\frac{\text{x}}{12}=-5\times12 [$multiplying both sides by $12]$
$\Rightarrow\text{x}=-60$
$iv.$ Given equation is $5x = 30$
$\Rightarrow\frac{5\text{x}}{5}=\frac{30}{5} [$dividing both sides by $5]$
$\Rightarrow\text{x}=6$
$v.$ Given equation is $3y = 5$
$\Rightarrow\frac{3\text{y}}{3}=\frac{5}{3}$
$\Rightarrow\text{y}=\frac{5}{3}$
$vi.$ Given equation is $=\frac{1}{3}\times(1-3\text{p}),$ we get
$=\frac{1}{3}(1-3\times2)$
$=\frac{1}{2}\times(1-6)$
$=\frac{1}{3}\times(-5)$
$=-\frac{5}{3}$
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Question 25 Marks
The three scales below are perfectly balanced if $• = 3.$ What are the values of $\triangle$ and $^*?$
$​​​​​​​a.$

$​​​​​​​b.$

$​​​​​​​c.$​​​​​​​
Answer
Let the value of $\triangle$ and $^*$ be $x$ and $y$, respectively and it is given that $•= 3.$
$a.$ From $y + y + y + y + y = x + 3x + 3$
$\Rightarrow 5y - 2x + 6$
$\Rightarrow 2x - 5y = -6 ...(i)$
$b.$ From $x + x = y + y + 3 + 3$
$\Rightarrow 2x = 2y + 6$
$\Rightarrow 2x - 2y + 6$
$\Rightarrow 2x - 2y = 6$
$\Rightarrow x - y = 3 [$dividing both sides by $2] ....(ii)$
$c.$ From $y + y + y^3 + 3 + 3 = x + x + x$
$\Rightarrow 3y + 9 = 3x$
$\Rightarrow 3x - 3y = 9$
$\Rightarrow x - y = 3 [$dividing both sides by $3] ....(iii)$
From $Eq. (iii) x - y = 3$
$\Rightarrow x = y + 3$ One putting $x = y + 3$ in $Eq. (i),$
we get $2(y + 3) - 5y = -6 $
$\Rightarrow 2y + 6 - 5y = -6 -3y + 6 = -6 $
$\Rightarrow -3y = -6 - 6 = -12$
$\Rightarrow\text{y}=\frac{12}{3}=4$
On putting $= 4$ in $Eq. (ii),$ we get $x - y = 3 $
$\Rightarrow x - 4 = 3 $
$\Rightarrow x = 3 + 4 = 7 $
$\Rightarrow x = 7$
$\therefore$ Value of $\triangle=\text{x}=7$ and
value of $^*  = y = 4.$
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Question 35 Marks
What does a duck do when it flies upside down? The answer to this riddle is hidden in the equation given below: If $i + 69 = 70$, then $i = ?$ If $8u = 6u + 8$, then $u = ?$ If $4a = –5a + 45$, then $a = ?$ if $4q + 5 = 17$, then $q = ?$ If $- 5t - 60 = -70$, then $t = ?$ If $\frac{1}{4}\text{s} + 98 = 100$, then $s = ?$ If $\frac{5}{3}\text{p} + 9 = 24$, then = ______? If $3c = c + 12$, then $c =$ ______? If $3(k + 1) = 24$, then k = ______? For riddle answer: substitute the number for the letter it equals.
Answer
We have, $i + 69 = 70$ [transposing $69$ to $RHS]$
$\Rightarrow i = 1$ and $8u = 6u + 8$
$\Rightarrow8\text{u}=6\text{u}+8$ [transposing $6u$ to $RHS]$
$\Rightarrow2\text{u}=8$
$\Rightarrow\frac{2\text{u}}{2}=\frac{8}{2}$ [dividing both sides by $2]$
$\Rightarrow\text{u}=4$
We have, $4a = -5a + 45$
$\Rightarrow4\text{a}+5\text{a}=45$ [transposing $(-5a)$ to $LHS]$
$\Rightarrow9\text{a}=45$
$\Rightarrow\frac{9\text{a}}{9}=\frac{45}{9}$ [dividing both sides by $9]$
 $\Rightarrow\text{a}=5$ and $4q + 5 = 17$
 $\Rightarrow4\text{q}=17-5$ [transposing $5$ to $RHS] $
$\Rightarrow4\text{q}=12$
$\Rightarrow\frac{4\text{q}}{4}=\frac{12}{4}$ [dividing both sides by $4]$
$\Rightarrow\text{q}=3$
We have,$ -5t - 60 = -70$
$\Rightarrow-5\text{t}=-70+60$ [transposing $(-60)$ to $RHS$]
$\Rightarrow-5\text{t}=-10$
$\Rightarrow\frac{-5\text{t}}{-5\text{}}=\frac{-10}{-5}$ [dividing both sides by $(-5)]$
$\Rightarrow\text{t}=2$ and $\frac{1}{4}\text{s}+98=100$
$\Rightarrow\frac{1}{4}\text{s}=100-98$ [transposing $98$ to $RHS]$
$\Rightarrow\frac{1}{4}\text{x}=2$
$\Rightarrow\frac{4}{4}\text{s}=4\times2$ [multiplying both sides by $4]$
$\Rightarrow\text{s}=8$
We have, $\frac{5}{3}\text{p}+9=24$
$\Rightarrow\frac{5}{3}\text{p}=24-9$ [transposing $9$ to $RHS]$
$\Rightarrow\frac{5}{3}\text{p}=15$
$\Rightarrow\frac{3}{5}\times\frac{5}{3}\text{p}=\frac{3}{5}\times15$
$\big[$ multiplying both sides by $\frac{3}{5}\big]$
$\Rightarrow\text{p}=9$
We have, $3c - c = 12$
$\Rightarrow3\text{c}=\text{c}+12$ [transposing c to $LHS]$
 $\Rightarrow2\text{c}=12$
$\Rightarrow\frac{2\text{c}}{2}=\frac{12}{2}$ [dividing both sides by $2]$
We have, $3(k + 1) = 24$
$\Rightarrow\frac{3(\text{k}+1)=24}{3}=\frac{24}{3}$ [dividing both sides by $3]$
$\Rightarrow\text{k}+1=8$
$\Rightarrow\text{k}=8-1$ [transposing $1$ to $RHS]$
​​​​​​​ $\Rightarrow\text{k}=7$ By substituting the number for the letter it equals, we get
​​​​​​​
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