Question
Evaluate: $\frac{-2}{3}+\frac{3}{4}$
✓
Answer
$\frac{-2}{3}+\frac{3}{4}$
L.C.M. of 3 and $4=3 \times 4=12$
$ \begin{aligned} & \Rightarrow \frac{-2 \times 4}{3 \times 4}+\frac{3 \times 3}{4 \times 3} \\ & (\because \text { L.C.M. of } 3 \text { and } 4=12) \end{aligned} $
$ \Rightarrow \frac{-8+9}{12}=\frac{1}{12} $
| 3 | 3,4 |
| 4 | 1,4 |
| 1,1 |
L.C.M. of 3 and $4=3 \times 4=12$
$ \begin{aligned} & \Rightarrow \frac{-2 \times 4}{3 \times 4}+\frac{3 \times 3}{4 \times 3} \\ & (\because \text { L.C.M. of } 3 \text { and } 4=12) \end{aligned} $
$ \Rightarrow \frac{-8+9}{12}=\frac{1}{12} $
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Draw a line segment AB = 8 cm. Mark a point P in AB so that AP = 5 cm. At P, construct angle APQ = 30°.
State, giving reason, whether the pair of triangle are congruent or not : $\triangle DEF$ in which $\angle E =48^{\circ}, DE =6 cm, EF =8 cm$ and $\triangle MNR$ in which $\angle R =48^{\circ}$, $MN =6 cm, MR =8 cm$.
→Evaluate: $-5+\frac{5}{-8}+\frac{-5}{-12}$
→In the given figure, angle $A D B=90^{\circ}, A C=A B=26 cm$ and $B D=$ $D C$. If the length of $A D=24 cm$; find the length of $B C$.
→Use the given figure to express:
(i) $a$ in terms of $b$ and $f$;
(ii) $e$ in terms of $f$ and $g$;
(iii) $d$ in terms of $a$ and $e$;
(iv) $c$ in terms of $e$ and $g$.
→(i) $a$ in terms of $b$ and $f$;
(ii) $e$ in terms of $f$ and $g$;
(iii) $d$ in terms of $a$ and $e$;
(iv) $c$ in terms of $e$ and $g$.
Simplify: $5 \frac{3}{4}-\frac{3}{7} \times 15 \frac{3}{4}+2 \frac{2}{35} \div 1 \frac{11}{25}$
→State, giving reason, whether the pair of triangle are congruent or not : $\triangle ABC$ in which $AB =3 cm, \angle A =90^{\circ}, BC =5 cm$ and $\triangle KLM$ in which $KM =3 cm$, $\angle K =90^{\circ} cm , LM =5 cm$.
→Arrange the given fraction in descending order by making numerator equal:
$\frac{1}{10}, \frac{6}{11}, \frac{8}{11} \text { and } \frac{3}{5}$
→$\frac{1}{10}, \frac{6}{11}, \frac{8}{11} \text { and } \frac{3}{5}$
What should be subtracted from $\frac{5}{9}$ to get $\frac{9}{5} ?$
→In the given figure, $\angle S P R=\angle Q R P$ and $\angle R S P=\angle P Q R$.
Prove that :
(i) $PQ = RS$
(ii) $PS = QR$
→Prove that :
(i) $PQ = RS$
(ii) $PS = QR$