5 Marks Questions

Question 15 Marks

Below $u, v, w$ and $x$ represent different integers, where $u = –4$ and $\text{x}\neq1$ By using following equations, find each of the values: $u \times v = u x \times w = w u + x = w$
$a. v$
$b. w$
$c. x$

Answer

We have, three equations $u \times v = u x \times w = w u + x = w$ and $u = -4$
$a.$ By puting the value of $u$ in $Eq.(i),$ we get $(-4) \times v = (-4)$
$\Rightarrow\text{v}=\frac{(-4)}{(-4)}=1$
$b.$ From $Eq.(ii),$ $\text{x}\times\text{w}$
$\Rightarrow\text{x}=\frac{\text{w}}{\text{w}}$
$\Rightarrow\text{x}=1$
But,
$\text{x}\neq1$
Hence, $x \times w = w, (ii)$ is possible, when $\text{w}=0(\text{x}\neq1)$
$c.$ From $Eq.(iii), u + x = w$
Put $u =$ and $w = 0$, we get
$\Rightarrow -4 + x = 0$
$\Rightarrow x = 4$
$\therefore v = 1, x = 4$ and $w = 0$

View full question & answer→

Question 25 Marks

The given table shows the freezing points in $0F$ of different gases at sea level.
Convert each of these into $0C$ to the nearest integral value using the relation and complete the table, $\text{C}=\frac{5}{9}(\text{F}-32)$

Gas	Freezing Point at (°F)	Freezing Point at Sea LevelSea Level $(^\circ C)$
Hydrogen	$-435$
Krypton	$-251$
Oxygen	$-369$
Helium	$-458$
Argon	$-309$

Answer

We have, $\text{C}=\frac{5}{9}(\text{F}-32)$ For Hydrogen, put $F = (-435)$
$\Rightarrow\text{c}=\frac{5}{9}(-435-32)$
$=\frac{5}{9}\times(-467)=-259.44$
$\Rightarrow\text{C}=-259.44$ For Oxygen, put $F = (-251)$
$\Rightarrow \text{C}=\frac{5}{9}(-251-32)$
$=\frac{5}{9}\times(-283)=-157.22$
For Oxygen, put $F = (-369)$ $\text{C}=\frac{5}{9}(-369-32)$
$=\frac{5}{9}\times(-401)$
$\Rightarrow\text{C}=-222.7$
$\Rightarrow\text{C}=-223$ For Helium put $F = (-458)$
$\text{C}=\frac{5}{9}(-458-32)$
$=\frac{5}{9}\times(-490)=-272.22$' $\text{C}=272.22$
For Argon, put F = (-309) $\text{C}=\frac{5}{9}(-309-32)$
$=\frac{5}{9}(-341)=-189.44$
$\Rightarrow\text{C}=-189.44$

View full question & answer→

Question 35 Marks

Match the following

Column $I$		Column $II$
$(a)$	$a \times 1$	$(i)$	Additive inverse of a
$(b)$	$1$	$(ii)$	Additive identity
$(c)$	$(-a) ÷ (-b)$	$(iii)$	Multiplicative identit
$(d)$	$a \times (-1)$	$(iv)$	$a ÷ (-b)$
$(e)$	$a \times 0$	$(v)$	$a ÷ b$
$(f)$	$(-a) ÷ b$	$(vi)$	$a$
$(g)$	$0$	$(vii)$	$-a$
$(h)$	$a ÷ (-a)$	$(viii)$	$0$
$(i)$	$-a$	$(ix)$	$-1$

Answer

Column $I$		Column $II$
$(a)$	$a \times 1$	$(vi)$	$a$
$(b)$	$1$	$(iii)$	Multiplicative identit.
$(c)$	$(-a) ÷ (- b)$	$(v)$	$a ÷ b$
$(d)$	$a \times (-1)$	$(vii)$	$-a$
$(e)$	$a \times 0$	$(viii)$	$0$
$(f)$	$(-a) ÷ b$	$(iv)$	$a ÷ (-b)$
$(g)$	$0$	$(ii)$	Additive identity.
$(h)$	$a ÷ (-a)$	$(ix)$	$-1$
$(i)$	$-a$	$(i)$	Additive inverse of a.

View full question & answer→

Question 45 Marks

An atom changes to a charged particle called ion if it loses or gains electrons. The charge on an ion is the charge on electrons plus chargeon protons. Now, write the missing information in the table given below:

Name of Ion	Proton Charge	Electron Charge	Ion Charge
Hydroxide ion	$+9$	$—$	$-1$
Sodium ion	$+11$	$—$	$+1$
Aluminium ion	$+13$	$-10$	$—$
Oxide ion	$+8$	$-10$	$—$

Answer

$a.$ For Hydroxide ion:
Proton charge $+$ Electron charge $=$ Ion charge
Electron charge $=$ Ion charge $-$ Proton charge
Electron charge
$= -1 - 9 $
$= -10$
Hence, the electron charge in a Hydroxide ion is $-10.$
$b.$ For Sodium ion:
Electron charge $=$ Ion charge $-$ Proton charge
$= +1 - 11 $
$= -10$
Hence, the electron charge in a Sodium ion is $-10.$
$c.$ For Aluminium ion:
Ion charge $=$ Proton charge $+$ Electron charge
$= 13 - 10 $
$= 3$
Hence, the ion charge in an Aluminium ion is $3$.
$d.$ For Oxide ion:
Ion charge $=$ Proton charge $+$ Electron charge
Ion charge
$= 8 - 10 $
$= -2$
Hence, the ion charge in an Oxide ion is $-2.$

View full question & answer→

Maths 5 Marks Questions

Take a timed test