3. Akshay purchased an old mobile phone. He spent 1,000 to buy a screen guard and flip cover. He later sold it for 10,500 at a profit of 1,500. At what price did
Answers
Answer:
Answer:
12.5% increased
Explanation:
Question says that,
If the radius is increased by 50% and height is reduced by 50%. Find the change in volume (in %)
We know that,
→ Volume of a cylinder = \boldsymbol{ \pi r^2h}πr
2
h
Where,
‘r' denotes the radius.
‘h’ is the height.
Now,
\boldsymbol rr is increased by 50%
= r + (50\% \: { \rm of } \: r)=r+(50%ofr)
= r + \dfrac{r}{2}=r+
2
r
= \dfrac{3r}{2}=
2
3r
Also, \boldsymbol hh is decreased by 50%
= h - (50 \% \: { \rm of } \: h)=h−(50%ofh)
= \dfrac{h}{2}=
2
h
New volume of the cylinder:-
= \pi \times \bigg( { \dfrac{3r}{2} \bigg)}^{2} \times \bigg( \dfrac{h}{2} \bigg)=π×(
2
3r
)
2
×(
2
h
)
= \dfrac{9\pi {r}^{2}h }{8}=
8
9πr
2
h
Since,
New volume > Original volume
Hence, there's a increase in volume.
Volume increased:-
= \dfrac{9\pi {r}^{2} h}{8} - \dfrac{\pi {r}^{2} h}{1}=
8
9πr
2
h
−
1
πr
2
h
= \dfrac{9\pi {r}^{2} h - 8\pi {r}^{2}h }{8}=
8
9πr
2
h−8πr
2
h
= \dfrac{\pi {r}^{2} h}{8}=
8
πr
2
h
Increase in % is given by,
= \rm \dfrac{increase \: in \: volume}{original \: volume} \times 100=
originalvolume
increaseinvolume
×100
= \dfrac{\pi {r}^{2}h }{8 \times \pi {r}^{2}h } \times 100=
8×πr
2
h
πr
2
h
×100
= 12.5\%=12.5%