SChand Composite Mathematics Class 7 Chapter 9 Percentage and it's application Exercise 9C

 Exercise 9C 

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Q1 | Ex-9C | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 1 

A shopkeeper bought a second hand car for Rs 1,50,000. He spent Rs, 10,000 on its painting and repair and then sold it far Rs, 200,000. Find his profit or loss. 

Sol: 

Cast price = Rs ,1,50,000

Spent on painting repair= Rs 10,000

Effective,  C.P = $1,50,000+10,000=Rs 1,60,000$

Selling price S.P = Rs 2,00,000

S.P > C.P = Profit = 200000- 1,60000

= Rs 40,000 Answer profit 

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Question 2

Vinay bought a house for Rs 4,50,000. HE Spent Rs.20,000 on repairs and white washing and then sold it for Rs 4,30,000. Find his gain\loss

Sol: 

C.P OF HOUSE =RS 4,50,000

spent on Repair $=220,000$

E.C.P = Rs 4,50,00+ Rs 20,000= Rs 4,70,000

S.p = Rs 4,30,000

C.P >S.P =loss= $4,70,000-4,30,000$

40,000 loss

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Question 3

A dealer bought 50 quintals of rice at the rate of Rs 800 per quintal. He paid Rs 400 charges and Rs 600 as transportation charges . Then he sold the whole stack at Rs 750 per quintal . find his gain or loss. 

Sol: C.P of per rice = Rs 800 

Charge spent = 400+ 600= Rs 1000

C.p of 50 quintal of rice = $50 \times 800$

=40,000

E.C.P = Rs $40,000+1000=Rs 41,000$

S.P = $750 \times 50=Rs 37500$

C.P >S.P  = loss= $41,000-37500$

= Rs 3500 loss

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Question 4

Find the gain or loss % in the following : 

(i) C.P = Rs 750

Sol: Expenses = Rs 50 

E.C.P = 750 + 50 = 800

Profit = Rs 80

Profit % = $\frac{\text { Profit}}{C.P} \times 100$

$\frac{80}{800} \times 100$ 

= 10% 

(ii) C.P = Rs 5200

S.P = Rs 5070

C.P >S.P 

Loss = 5200- 5070 

=Rs 130

Loss % = $\frac{130}{5200} \times 100$

= 2.5% 

(iii) CP=Rs 400

S.P. =₹ 460

Profit = 460 - 40= Rs 60

Profit =$=\frac{60}{400} \times 100$

=15

(iv) C.P = Rs 50 

S.P = Rs 42

Loss = 50 - 42= Rs 8

Loss =8 

Loss percentage $=\frac{8}{50} \times 100$

=16%

(v)C.P = Rs 46000; Over heads = 4000

E.C.P = Rs 50,000

S.P = Rs 60,000

Gain\loss %= ? 

Sol :

S.P > C.P

Profit=SP-CP=60,000-50,000

=10,000

Profit percentage $=\frac{10,000}{50,000} \times 100$

=20%

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Question 5

Find the S.P when 

(i) C.P = Rs 20

Sol: 

Gain = 10% 

Gain = 10% of 20 

$=\frac{10}{100} \times 20$ = Rs 2 

S.P = C.P + Gain = 20+2= Rs 22

(ii) C.P = Rs 12.50 ; Loss = $13 \frac{1}{3} \%=\frac{40}{3} \%$

Sol: 

Loss = $\frac{40}{3} \% of  \quad 12.50$

$=\frac{40}{3 \times 100} \times \frac{+2.50}{100}$ =  $\frac{5}{3}=Rs 1.67$

S.P = C.P - Loss = Rs 12.50 - 1.67= Rs 10.82 

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Question 6

Find the C.P when, 

(i) S.P= Rs 12 ; Gain = 20% 

Let C.P = x

Gain = 20% of x = $\frac{20}{100} \times 21=\frac{x}{5}$

S.P= $x+\frac{x}{5}=\frac{5 x+4}{5}=\frac{6 x}{5}=12$

= x=  $12 \times \frac{5}{6}=Rs 10$

(ii) S.P = Rs 360 ; Loss 10% 

Loss = 10% of x = $\frac{10}{100} \times x=\frac{x}{10}$

 S.P $=x-\frac{x}{10}=\frac{10 x-x}{10}=\frac{9 x}{10}$

$\Rightarrow \quad \frac{9 x}{10}=360$

$\Rightarrow x=\frac{360}{9} \times 10=Rs 400$

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Question 7

By Selling a motor cycle for Rs 23000 a dealer gains 15% .Find its C.P 

Sol: Let C.P =  x   Given S.P = Rs 23,000

Gain = 15 % of x = $\frac{18}{100} \times 21$= $\frac{3x}{20}$

S.P = $2 x+\frac{3 x}{20}=\frac{20 x+3 x}{20}=\frac{23 x}{20}$

∵ $\frac{23 x}{20}=23000 \Rightarrow x$ = $23000 \times \frac{20}{23}$

C.P (x) = Rs 20,000

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Question 8

The C.P of an article is Rs 6250 . Sudhir sells it at a loss of 24% . Find its S.P.

Sol: C.P = Rs 6250

Loss = 24% of 6250

$\frac{24}{100} \times 6,250$ = Rs 1500

S.P = C.P - loss

$=6250-1500$ 

S.P = Rs 4750 Answer 

S Chand CLASS 10 Chapter 9 Arithmetic and Geometric Progression Exercise 9C

  Exercise 9C

Question 1 

Ans: (i) $27,9,3,1, \ldots$

$\begin{aligned}\therefore & a=27 \\r &=\frac{9}{27}=\frac{1}{3}, \frac{3}{9} \\&=\frac{1}{3}, .....\end{aligned}$

Hence, it is a G.P. and $r=\frac{1}{3}$

(ii) $-1,2,4,8, \ldots .$

$\therefore a=-1$,

$\quad b=\frac{2}{-1}=-2, \frac{4}{2} .$

$\quad=2, \frac{8}{4}=2 .$

So, it is not a G.P 

(iii) 

$\begin{aligned} & 2, \frac{1}{2}, \frac{1}{8}, \frac{1}{32}, \ldots \\ \therefore & a=2, \\ &=\frac{1}{2} \div 2 \\ &=\frac{1}{2} \times \frac{1}{2} \\ &=\frac{1}{4} . \\ \frac{1}{8} &=\frac{1}{2} . \end{aligned}$

$=\frac{1}{4} .$

$\frac{1}{32}=\frac{1}{8} .$

$=\frac{1}{32} \times 8 .$

$=\frac{1}{4}$

∴ IT is GP and  $r=\frac{1}{4}$

$(i v)-12,-6,0,6, \ldots . .$

$\begin{aligned}&\therefore a=-13 \\&r=\frac{-6}{-12}=\frac{1}{2} \\&=\frac{0}{-6} \\&=0\end{aligned}$

Hence,  it is not G.P.

Question 2

Ans:  (i) 2,6....

$\left(r=\frac{6}{2}=3\right) .$

2,6,18,54,162

(ii) $\frac{1}{16},-\frac{1}{8}, \ldots$

$\left(r=-\frac{1}{8} \div \frac{1}{10}=-\frac{1}{8} \times 16=-2 .\right)$

$\frac{1}{16},-\frac{1}{8}, \frac{1}{4},-\frac{1}{2}, 1$

(iii)0.3, 0.06......

$r=\frac{0.01}{0.3}$= $\frac{1}{5}=0-2$

$0.3,0.03,0.012,0.0024,0.00048$

Question 3

Ans: 6th term of the G.P 2, 10 , 50......

∴ a = 2 

r=  $\frac{10}{2}=5$

$\begin{aligned} So  T_{6} &=9 x^{h-1} \\ &=2 \times 5^{6-1} \\ &=2 \times 5^{5} \\ &=2 \times 3125 \\ &=6250 \end{aligned}$

(ii)  11th term of the G.P. $4,12,3, \ldots .$

$\begin{aligned} \therefore & a=4 \\ & r=\frac{12}{4}=3 . \end{aligned}$

$\begin{aligned} T_{11} &=ar^{h-1} \\ &=4 \times(3)^{11-1} \\ &=4 \times 3^{10} \\ &=4 \times 59,049 \\ &=236196 \end{aligned}$

Question 4

Ans:

 (i) $\begin{aligned} T_{n} &=4 \cdot 3^{h-1} \\ \therefore T_{1} &=4 \cdot 3^{1-1} \\ &=4 \cdot 3^{0} \\ &=4 \times 1=4 \\ T_{2} &=4 \cdot 3^{2-1} \\ &=4 \cdot 3^{1} \\ &=4 \times 3 \\ &=12 \\ T_{3} &=4 \cdot 3^{3-1} \\ &=4 \cdot 3^{2} \\ &=4 \times 9 \\ &=36 \end{aligned}$

$\begin{aligned} T_{4} &=4.3^{4-1} \\ &=4.3^{3} \\ &=4 \times 27 \\ &=108 \\ T_{5} &=4.3^{5-1} \\ &=4.3^{4} \\ &=4 \times 81 \\ &=324 \end{aligned}$

Hence, the terms are 4, 12, 36 , 108 , 324.

(ii) 

$\begin{aligned} T_{h} &=\frac{5^{h-1}}{2^{h+1}} \\ T_{1} &=\frac{5^{1-1}}{2^{n+1}} \\ &=\frac{5^{0}}{2^{2}} \\ &=\frac{1}{4} \\ T_{2} &=\frac{5^{2-1}}{2^{2+1}} \\ &=\frac{5^{1}}{2^{3}} \\ &=\frac{5}{8} \\ T_{3} &=\frac{5^{3-1}}{2^{3+1}} \\ &=\frac{5^{2}}{2^{4}} \\ &=\frac{25}{16} \end{aligned}$

$\begin{aligned} T_{4} &=\frac{5^{4-1}}{5^{9+1}} \\ &=\frac{5^{3}}{2^{5}} \\ &=\frac{125}{2^{5}} \\ &=\frac{125}{32} \\ T_{5} &=\frac{5^{5-1}}{2^{5+1}} \\ &=\frac{5^{4}}{2^{6}} \\ &=\frac{625}{64} . \end{aligned}$

Hence, the 5 terms are  $\frac{1}{4}, \frac{5}{8}, \frac{25}{16} ,\frac{125}{32}, \frac{625}{64}$

Question 5

Ans: (i) 12 , -36 .... sixth term

$\begin{aligned} \therefore \quad a &=12, \\ r &=\frac{-36}{12} \\ r &=-3 . \end{aligned}$

$\begin{aligned} \therefore T_{6} &=arn^{-1}  \\ &=12 \times(-3)^{6-1} \\ &=12 \times(-3)^{5} \\ &=12 \times(-243) \\ &=-2916 \end{aligned}$

(ii)  $3,-\frac{1}{3}, \ldots, 8$ th term.

$\begin{aligned} \therefore \quad q &=3, \\ r &=-\frac{1}{3} \div 3 . \\ &=-\frac{1}{3} \times \frac{1}{3} \\ &=-\frac{1}{9} \end{aligned}$

So , $\begin{aligned} T_{8} &=arn^{-1} \\ &=3 \cdot\left(-\frac{1}{9}\right)^{81} \\ &=3\left(-\frac{1}{9}\right)^{7} \end{aligned}$

(iii)

 $\begin{aligned} & b^{2} c^{3}, b^{3} c^{2}, \ldots, & 5th term \\ & \therefore a=b^{2} c^{3} \end{aligned}$

$r=\frac{b^{2} c^{2}}{b^{2} c^{8}}$

$r=\frac{b}{c}$

So $T_{5}= arn^{-1}$

$=b^{2} c^{3} \times\left(\frac{b}{c}\right)^{5-1}$

$=b^{2} c^{3} \times\left(\frac{b}{c}\right)^{4} .$

$=b^{2} c^{3} \times \frac{b^{4}}{c^{4}}$

$=b^{2} c^{2} \times b^{4} \times c^{-4}$

$=b^{2+4} \times c^{3}-4$

$=b^{6} \times c^{-1}$

$\frac{b^{6}}{c}$

Question 6

Ans: G.P. is $27,-18,12,-8, \ldots$ is $\frac{1024}{2187}$ (Given)

$\begin{aligned} \therefore a &=27, \\ r &=\frac{-18}{27} \\ r &=\frac{-2}{3} . \end{aligned}$

Let  $\frac{1024}{2187}$ be the nth term 

$\therefore a_{n}=a r n^{-1}$

$\frac{1024}{2187}=27\left(-\frac{2}{3}\right)^{n-1}$

$\frac{1024}{2187 \times 27}=\left(-\frac{2}{3}\right)^{n-1}$

$\frac{2^{10}}{3^{7} \times 3^{3}}=\left(-\frac{2}{3}\right)^{n-1}$

$\frac{2^{10}}{3^{10}}=\left(\frac{-2}{3}\right)^{n-1}$

$\left(\frac{-2}{3}\right)^{10}=\left(\frac{-2}{3}\right)^{n-1}$

On comparing, 

$10=n-1$

$10+1=n$

$11=n$

n=11

Hence, it is 11th term.

 

Question 7

 

Ans: In a G.P 

$T_{4}=54$

$T_{7}=1458$

Let a be the first term and r be the common ratio

$So T_{4}=a r^{n-1}$

$54=ar^{4-1}$

$54=ar^{3}$

$a r^{3}=5 4$.........(i)

Dividing ,eqn (i) and (ii)

$r^{3}=\frac{1458}{54}$

$r^{3}=27$

$r=\sqrt[3]{27}$

$r=3$

Put the value of r = 3 in eq (i)

$a r^{3}=54$

$a \times(3)^{3}=54$

$a \times 27=54$

$a=\frac{54}{27} .$

$a=2$

$\therefore a=2, r=3 .$

Hence , G.P will be 2, 6 ,18 ,54....

 

Question 8

Ans: In a G.P. $3,3 \sqrt{3}, 9, \ldots$

Last term (l) = 2187

∴ a = 3

r =  $\frac{3 \sqrt{3}}{3}$

$T_{n}=l=arn^{-1}$

$2187=3(\sqrt{3})^{n-1}$

$\frac{2187}{3}=(\sqrt{3})^{n-1}$

$729=(\sqrt{3})^{n-1}$

$3^{6}=(\sqrt{3})^{n-1}$

$(\sqrt{3})^{6 \times 2}=(\sqrt{3})^{n-1}$

$(\sqrt{3})^{12}=(\sqrt{3})^{n-1} .$

On comparing,

$12=n-1$

$12+1=n$

$13=n$

$n=13$

Hence, it is 13th term

Question 9

Ans: Given , 

1, x, y , z 16 are in G.P 

So, first term (a) = 1, 

Common ratio(r) = $\frac{x}{1}$

$T_{5}=16$

$\begin{aligned} \therefore T_{5} &=a r(h-1) . \\ 16 &=9 r(5-1) \\ 16 &=9 r^{4} \\ 16 &=1 \times r^{4} \\(2)^{4} &=r^{4} \end{aligned}$

On comparing, 

r= 3

 So, the common ratio is 2 

Then, 

$\begin{aligned} r=\frac{x}{1} &=\frac{2}{1} \\ x &=2 \end{aligned}$

Also , the common ratio

$\frac{16}{z}=2$

$\begin{aligned}&z=\frac{16}{2} \\&z=8\end{aligned}$

Also, the common ratio 

$\begin{aligned} \frac{2}{y} &=2 \\ \frac{y}{y} &=2 \\ y &=\frac{8}{2} \\ y &=4 . \end{aligned}$

According to the question 

$\therefore x+y+z$

$=2+4+8$

=14 

Question 10

 

Ans:  In a G.P 

$T_{3}=18$,

$T_{7}=3 \frac{5}{9}=\frac{32}{9}$

Let a be the first term and r be the common ratio 

∴ $T_{n}=a r^{n-1}$

$T_{3}=a r^{3-1}$

$18=a r^{2}$

$a r^{2}=18$ ........(i)

$T_{7}=a r^{7-1}$

$\frac{32}{9}=a r^{6}$

$a r^{6}=\frac{32}{9}$...........(ii)

On dividing , 

$\frac{ar^{6}}{ar^{2}}=\frac{32}{9\times 18}$

$r^{4}=\frac{16}{81}$

$r^{4}=\left(\frac{2}{3}\right)^{4}$

On comparing, 

$\begin{aligned} & r=\frac{2}{3} \\ \therefore \quad & a r^{2}=18 \end{aligned}$

$a \times\left(\frac{2}{3}\right)^{2}=18$

$a \times \frac{4}{9}=18$

$a=\frac{81}{2}$

Then , 

$T_{10}=ar^{9}$

$=\frac{81}{2} \times\left(\frac{2}{3}\right)^{9}$

$=\frac{81}{2} \times \frac{2^{9}}{3^{9}}$

$\begin{aligned} &=\frac{3^{4} \times 2^{9}}{2 \times 3^{9}} \\=& \frac{2^{9-1}}{3^{9-4}} \\=& \frac{2^{8}}{3^{5}} \\=& \frac{256}{243} \end{aligned}$

Question 11

 

Ans: Given, 

In a G.P 

$T_{5}=P$,

$T_{8}=Q$,

$T_{11}=S$

Let a be the first term and r be the common ratio 

$\begin{aligned} \therefore \quad T_{5} &=ar^{5-1} \\ &=ar^{4} \\ &=p \end{aligned}$

$\begin{aligned} T_{8} &=a_{r} 8-1 \\ &=a^{7} \\ &=a \end{aligned}$

$\begin{aligned} T_{11} &=a r^{11-1} \\ &=a r^{10} \end{aligned}$

=5

$\begin{aligned} Q^{2} &=\left(ar7^{2}\right)^{2} \\ &=a^{2} r^{14} . \end{aligned}$

and p $\times s$ = $a r^{4} \times a r^{16} .$

$=a^{2} r^{4}$

Hence , prove, 

$Q^{2}=P \times S$

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 9 Variation Exercise 9C

 Exercise 9C

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Question 1

6 men take 12 hours to weed a certain field. How long would 9 men take to do so, if all work at the same rate?

Sol :

M₁=6 men,  D₁=12 , M₂=9

M₁.D₁=M₂.D₂

or D₂$=\frac{6\times 12}{9}$=8 hr 

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Question 2

12 men can hoe a field in 10 days. How long will 15 men take?

Sol :

M₁=12 men,  D₁=10 , M₂=15

M₁.D₁=M₂.D₂

or D₂$=\frac{12\times 10}{15}$=8 days 

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Question 3

A and B can do a piece of work in 6 and 12 days respectively. They (both) will complete the work in how many days?

Sol :

A do work in 6 days

B do work in 6 days

∴efficiency of A$=\frac{1}{6}$

∴efficiency of B$=\frac{1}{12}$

∴efficiency of (A+B)$=\frac{1}{6}+\frac{1}{12}$ $=\frac{2+1}{12}=\frac{3}{12}=\frac{1}{4}$

∴(A+B) do the work in 4 days

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Question 4

Rekha can finish a work in 18 days and Prema can do the same work in half the time taken by Rekha. Then, working together, what part of the same work they can finish in a day?

Sol :

Rekha can finish a work in 18 days

∴efficiency of rekha $=\frac{1}{18}$

Prema can do a work in $=\frac{18}{2}$=9 days 

∴efficiency of prema $=\frac{1}{9}$

∴efficiency of (Rekha + Prema)$=\frac{1}{18}+\frac{1}{9}$ $=\frac{1+2}{18}=\frac{3}{18}=\frac{1}{6}$

∴A work can done by (rekha and prema) together $\frac{1}{6}$

It take 6 days to complete work when they work together.

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Question 5

A alone can complete a work in 12 days and B alone can complete the same work in 24 days. In how many days can A and B together complete the same work?

Sol :

A can do a work =12 days

B can do a work =24 days

∴efficiency of A$=\frac{1}{12}$

∴efficiency of B$=\frac{1}{24}$

∴efficiency of (A+B)$=\frac{1}{12}+\frac{1}{24}=\frac{2+1}{24}$ $=\frac{3}{24}=\frac{1}{8}$

(A+B) do work= 8 days

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Question 6

A and B together can do a piece of work in 12 days, while B alone can finish it in 30 days. In how many days can A alone finish the work?

Sol :

(A+B) do a work=12 days

∴efficiency of (A+B)$=\frac{1}{12}$

B can do alone a work=30 days

∴efficiency of B$=\frac{1}{30}$

∴efficiency of A$=\frac{1}{12}-\frac{1}{30}=\frac{5-2}{60}$ $=\frac{3}{60}=\frac{1}{20}$

∴A can do a work alone=20 days

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Question 7

A, B and C can complete a work in 2 h. If A does the job alone in 6 h and B in 5 h, how long will it take C to finish the job alone?

Sol :

(A+B+C) can complete a work=2 h

∴A can complete a work=6 h

∴B can complete a work=5 h

∴efficiency of (A+B+C)$=\frac{1}{2}$

∴efficiency of A$=\frac{1}{6}$

∴efficiency of B$=\frac{1}{5}$

∴efficiency of C$=\frac{1}{2}-\left(\frac{1}{6}+\frac{1}{5}\right)$ $=\frac{1}{2}-\left(\frac{5+6}{30}\right)=\frac{2}{15}$

∴C can do a work $=\frac{15}{2}=7\frac{1}{2}$ hours

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Question 8

A and B can do a piece of work in 72 days. B and C can do it in 120 days. A and C can do it in 90 days. In what time can A alone do it?

Sol :

(A+B) can do a piece of work=72 days

∴(B+C) can do a piece of work=120 days

∴(A+C) can do a piece of work=90 days

∴efficiency of (A+B)+(B+C)+(A+C)$=\frac{1}{72}+\frac{1}{120}+\frac{1}{90}$

∴efficiency of 2(A+B+C)$=\frac{5+3+4}{360}=\frac{12}{360}=\frac{1}{30}$

∴efficiency of (A+B+C)$=\frac{1}{30\times 2}=\frac{1}{60}$

∴efficiency of A=efficiency of (A+B+C)-efficiency of (B+C)

$=\frac{1}{60}-\frac{1}{120}=\frac{1}{120}$

∴A can do a work 120 days

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Question 9

A can do a piece of work in 10 days and B in 20 days. They begin together but A leaves 2 days before the completion of the work. In how many days will the whole work be completed?

Sol :

A can do a work in 10 days

B can do a work in 20 days

∴efficiency of A$=\frac{1}{10}$

∴efficiency of B$=\frac{1}{20}$

Let time take be x days

∴$\frac{x-2}{10}+\frac{x}{20}=1$

$\frac{2(x-2)+x}{20}=1$

or 2x-4+x=20

3x=20+4

x=8 days

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Question 10

Piyush and Ajit together can complete a work in 3 days. They started together but after 2 days Ajit left the work. If the work is completed after 2 more days, Piyush alone can complete it in how many days?

Sol :

Efficiency of (Ajit+Piyush)$=\frac{1}{3}$ part

Work together for 2 days$=\frac{1}{3}\times 2=\frac{2}{3}$ part

Remaining $=\left(1-\frac{2}{3}\right)=\frac{3-2}{3}=\frac{1}{3}$ part

Remaining part completed by piyush in 2 days

∴Piyush's 1 day work$=\frac{1}{3}-\frac{1}{2}=\frac{1}{6}$

∴Ajit's 1 day work$=\frac{1}{3}-\frac{1}{6}=\frac{2-1}{6}=\frac{1}{6}$

∴Complete by Piyush in 6 days

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Question 11

A can complete a work in 10 days, B in 12 days and C in 15 days. all of them began the work together, but A had to leave the work after 2 days of the start and B, 3 days before the completion of the work. How long did the work last?

Sol :

Efficiency of A$=\frac{1}{10}$

Efficiency of B$=\frac{1}{12}$

Efficiency of A$=\frac{1}{15}$

∴Total unit =60

Total unit divided in

A=6 unit, B=5 units, C=4 units

∴A's 2 days work=6×2=12 unit

or $2\times \frac{1}{10}+\frac{x-3}{12}+\frac{x}{15}=1$

or $\frac{12+5x-15+4x}{60}=1$

or $x=\frac{60+3}{7}=\frac{63}{9}=7$ days

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Q12 | Ex-9C | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

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Question 12

Sanjay and Ranbir can do a piece of work in 45 and 40 days respectively. They began the work together but Sanjay leaves after some days and Ranbir finished the remaining work in 23 days. After how many days did Sanjay leave?

Sol :

Efficiency of (Sanjay+Ranbir)$=\frac{1}{45}+\frac{1}{40}$

$=\frac{8+9}{360}=\frac{17}{360}$

∴Total work =360 {8 and 9}

Ranbir work 23 days=23×9=207 unit

∴Total work done by (Sanjay+Ranbir) together=360-207=153 units

∴Sanjay leave$=\frac{153}{8+9}=\frac{153}{17}$

=9 days