A solution of salt and water contained 15% salt . After avaporating 8kg of water from it , the remaining solution has 20% salt .find the weight of original solution .
Answers
Answer:
Step-by-step explanation:
\underline\blue{\bold{Given- }}Given−
\blue{\texttt{volume \: of \: a \: right \: circular \: cone \: is }}volume of a right circular cone is 9856 {cm}^{2}9856cm2
\blue{\texttt{diameter \: of \: the \: base \: is }}diameter of the base is 28cm .
so \: \: r = \large\frac{28}{2}sor=228
= 14 cm
\underline\blue{\bold{To \: Find }}ToFind :-
a) Height of the cone.
b) Slant height.
c)Curved surface area
d)Total surface area.
\red{\textbf{(i)height}}(i)height = \underline\blue{\bold{h}}h
\red{\textbf{volume}}volume =
9856 = \large \frac{1}{3} \pi {r}^{2} h9856=31πr2h
9856 = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h9856=31×722×14×14×h
h = 48 \: cmh=48cm
\red{\textbf{(ii) \: slant \: height}}(ii) slant height =
{l}^{2} = \sqrt{ {r}^{2} + {h}^{2} }l2=r2+h2
{l}^{2} = \sqrt{ {14}^{2} + {48}^{2} } = \sqrt{196 + 2304}l2=142+482=196+2304
{l }^{2} = 2500l2=2500
l = \sqrt{2500}l=2500
l = \sqrt{(5 {0)}^{2} }l=(50)2
l \: = 50 \: cml=50cm
\red{\textbf{(iii) \: curved \: surface \: Area}}(iii) curved surface Area =>
we know that r = 14 cm and l = 50cm
(\pi \: rl \: = \large \frac{22}{7} \times 14 \times 50) {cm}^{2}(πrl=722×14×50)cm2
= (22 \: \times 2 \times 50)c {m}^{2}=(22×2×50)cm2
= 2200 \: c {m}^{2}=2200cm2
\red{\textbf{(iv) \: total \: surface \: area}}(iv) total surface area = πr(l + r)
= 22/7 × 14(50 + 14)
= 22 × 2(64)
= 22 × 2 × 64
= 8192 \: c {m}^{2}=8192cm2
•\blue{\texttt{height \: of \: cone}}height of cone 14cm
•\blue{\texttt{slant \: height \:}}slant height 50 {cm}^{2}50cm2
•\blue{\texttt{curved \: surface \: area}}curved surface area 2200 {cm}^{2}2200cm2
•\blue{\texttt{total \: surface \: area}}total surface area 8192 \: c {m}^{2}8192cm2
Step-by-step explanation:
hope it helpsful