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4

I) The room number in which the treasure took place is 28
II) Place of the treasure is C H A I R
III) The name of the treasure G I F T V O U C H E R.

(i) y=x

(ii) y = 2x

(iii) y= 3x

(iv) y = 5x

The given three'lines are passing through the origin.

Since we have given that
Equal sides:
 and 
Length of third side = 2y+5
Since above two sides are equal.
5y-2 = 4y+9
5y-4y = 9+2
y = 11 units
Length of first side = 
Length of second side = 
Length of third side = 
Perimeter of triangle would be 

y = 11 units p = 133 units

The points of intersection of th lines is (0, 025)
The points of intersection of the lines in X-axis are (0.25,0) and (- 0.25,0)

From the graph, the point S is (5, - 4).

56 and 12
Take m = 56 and n = 12, Here m > n.
56 - 12 = 44
44 - 12 = 32
32 - 12 = 20
20 - 12 = 8
12 - B = 4
8 - 4 = 4
4 - 4 = A
The HCF of 56 and 12 is 4.

64.6km

27

Let x and y be the cost price of the two sarees
Saree I                             
Gain = 10%                     
x + 10% of x = 2200       
\( x+\frac{10}{100} x =2200 \)
\(\frac{110}{100} x =2200 \)
\(\frac{11}{10} x =2200 \)
\(x =\frac{2200 \times 10}{11} \)
\(x =Rs. 2000\)
Saree II
Loss = 12%
y-12% of y = 2299
\( y-\frac{12}{100} y =2200 \)
\(\frac{88 y}{100} =2200 \)
\(\frac{22 y}{25} =2200 \)
\(y =\frac{2200 \times 25}{22} \)
\(y =Rs.2500 \)
Total cost price of both sarees = 2000 + 2500
= 4500
But total selling price = 2x 2200
= 4400
Loss : C.P-S.P
= 4500 - 4400 = Rs.100
\(\text { Loss % }=\frac{100}{4500} \times 100\)
\(=\frac{20}{9}=2 \frac{2}{9} \%\)
The loss percentage of the saIes = \(2 \frac{2}{9} \%\)

P's income = 25%
The percen(age of Q's income less than P's
\( =\left(\frac{100 x}{100+x}\right) \%=\left(\frac{100 \times 25}{100+25}\right) \% \)
\(=\left(\frac{100 \times 25}{125}\right) \%=20 \% \)
i.e., The perlentage of Q's income less than P's = 20%

T3, 93\(\frac { 2 }{ 3 } \)%

P = Rs. 8000,
i = 2 years, Difference amount = Rs. 20
Difference between C.I and S.I for 2 years
\( =P\left(\frac{r}{100}\right)^{2} \)
\(20 =8000\left(\frac{r}{100}\right)^{2} \)
\(20 =8000 \times \frac{r^{2}}{10000} \)
\(20=\frac{4 r^{2}}{5} \)
\(r^{2} =\frac{20 \times 5}{4} \)
\(r^{2} =\frac{100}{4}=25\)
r = 5
the rate of interest = 57%

4%

\( \text { Amount } =P\left(1+\frac{r}{100}\right)^{a}\left(1+\frac{\frac{b}{c} \times r}{100}\right) \)
\(=4000\left(1+\frac{10}{100}\right)^{2}\left(1+\frac{\frac{1}{2} \times 10}{100}\right) \)
\(=4000\left(\frac{110}{100}\right)^{2}\left(1+\frac{5}{100}\right) \)
\(=4000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{21}{20}\)
= 2 x 11 x 11 x 2
A = Rs. 5082
C.I = A - P
= 5082 - 4000 = 1082
C.I = Rs. 1082

By using Pythagoras theorem,
AP² = AK² + PK² = AK² + AK² (since it is an isosceles A) = 2AK²
Hence it is proved 

Yes, the side of length 3.7 cm is the hypotenuse.

Let \(\triangle\) ABC is a isosceles triangle.
Draw AD perpendicular to BC, we get two right angled triangles.
In  \(\triangle\)ABD
13² = x² + h²        ..........(1)
In \(\triangle\) ADC,
13² = h² +(24 - x)²  .........(2)
From (1) and (2)
\( x^{2}+\mathrm{h}^{2} =(24-x)^{2}+\mathrm{h}^{2} \)
\(x^{2} =(24-x)^{2} \\ x^{2} =576-48 x+x^{2} \)
\(0 = 576-48 x \\ \therefore \ 48 x =576 \)
\(x =\frac{576}{48} \\ x =12 \)
Put x = 12 in (1), we get 13² = x² +h²
\( 13^{2} =12^{2}+h^{2} \)
\(169 =144+h^{2} \)
\(\therefore h^{2} =169-144 \)
\(h^{2} =25 \)
\(h =\sqrt{25}\)
= 5
Height of the isosceles triangle is 5 cm.

The given two successive discounts are 10% and 20%
The amount to be paid by the customer
\(=\frac{90}{100} \times \frac{80}{100} \times 7500=5400\)
The customer has to paid = Rs. 5400

Let x be the C.P of the speaker
x - 768 = 20% of x
\( x-768 =\frac{20}{100} x \)
\(x-768 =\frac{x}{5} \)
\(x-\frac{x}{5} =768 \)
\(\frac{5 x-x}{5} =768 \)
\(\frac{4 x}{5} =768 \)
\(x=\frac{768 \times 5}{4}=960\)
C.P of the speaker is Rs. 960. 
In order to get gain 20%
Then the rate of the speaker = 960 + 20 % of 960
\(=960+\frac{20}{100} \times 960\)
= 960 + 192 = 1152
To get 20% gain,he should sell the speaker for Rs. 1152.

Decrease value = Original price - Sold Price
= Rs. 3,00,00 - 2,00,00
= Rs. 1,00,000
\(\therefore \text { Percentage of decrease }=\frac{100000}{300000} \times 100\)
\(=33 \frac{1}{3} \%\)

\(\frac{Pnr}{100}\)
\(240 = \frac{3000\times2 \times r}{100}\\ = \frac{240}{60}\)
R = 4%

(i) \(\text { one half }=\frac{1}{2} \times 100 \%=50 \%\)
(ii) \(\frac{7.5}{10} \times 100 \%=75 \%\)
(iii) The word pure means 100%
(iv) \(\frac{48}{50} \times 100 \%=96 \%\)
(v) \(\frac{2}{3} \times 100 \%=66 \frac{2}{3} \%\)