The formula to find compound interest (CI) when the principal (P), rate of interest(R), and time(T) is given is:

Compound interest (CI) = \(P\left\{(1+\frac R{100})^T-1\right\}\)

Here,

The total surface area of a prism (TSA) = 2 area of base + lateral surface area of prism.

Let, the required positive number be x.

Now, by the question

\(\frac13x^2-12=15\)

Here,

In the formula Q₁ = \(L+(\frac{{\displaystyle\frac N4}-C.F}f)\times h\),

h represents the class interval of Q₁ class.

If a parallelogram and a triangle are standing on the same base and between the same parallel lines, then the area of the parallelogram is double the area of the triangle.

A line segment that passes through the center of the circle and joins any two points on the circumference is called the diameter of the circle.

Given,

Selling price(SP) = Rs. 1400

Rate of VAT = 13%

To find, SP with VAT

By formula,

SP with VAT = SP+VAT% of SP

= Rs.1400+13% of Rs. 1400

= \(Rs.1400+\frac{13}{100}\times1400\)

= Rs.1400 + Rs. 182

= Rs.1582

Therefore, Selling price with VAT is Rs.1582.

Given:

The present cost (C) = Rs. 225000
Rate of compound depreciation (R) = 10% p.a
Cost after depreciation (C_T) = Rs 175204.8

To find: Time (T)

By Formula,

\(C_T=c\left(1-\frac R{100}\right)^T\)
Or, \(175204.8=225000\left(1-\frac8{100}\right)^T\)
Or, \(\frac{175204.8}{225000}=\left(1-0.08\right)^T\)
Or, \(0.778688=\left(0.92\right)^T\)
Or, \(\left(0.92\right)^3=\left(0.92\right)^T\)
Or, 3 = T

Hence, the required time is 3 years.

Given,

Area of an equilateral triangle (A) = \(36\sqrt3\) cm² 

To find, Perimeter of the triangle (P) 

By formula,

Area of an equilateral triangle (A) = \(\frac{\sqrt3}4a^2\)

or, \(36\sqrt3\) cm² = \(\frac{\sqrt3}4a^2\)

or, \(\frac{4\times36\sqrt3cm^2}{\sqrt3}=a^2\)

or, a² = 144cm² 

or, a² = (12cm)² 

or, a = 12

Now,

Perimeter of an equilateral triangle (P) = 3a

= 3 x 12

= 36cm

Given, In a triangular based prism,

AB = 8cm

Ac = 6cm

BC = B'C' = 10cm

and lateral surface area of the prism = 288cm² 

To find, Height of  the prism (CC')

Here,

Perimeter of the base (P) = AB + BC + AC

=8cm + 10cm + 6cm

=24cm

Now, by the formula,

The lateral surface area of the prism = perimeter of the base x Height

or, 288cm² 24cm x CC'

or, cc' = \(\frac{\displaystyle288cm^2}{24cm}\)

or, CC' = 12cm

Thereforem, height of the prism (CC') = 12cm.

Given, Total surface area of a hemisphere (TSA) = 7392cm² 

By formula,

Total surface area of a hemisphere (TSA) = ² 

or,7392cm² = \(3\times\frac{22}7\times r^2\)

or, \(\frac{7\times7392cm^2}{66}=r^2\)

or, r² = 784cm² 

or, r²  = (28cm)² 

or, r = 28cm

Therefore the radius of the hemisphere (R) = 28cm.

Here, 

\(\frac{4^{x+1}+64\;\times4^{x-1}\;}{4^{x+1}+4^x}\)

or, \(\frac{4^x\times4^1+64\;\times4^x/{\displaystyle\frac1{4^1}}\;}{4^x\times4^1+4^x}\)

or, \(\frac{4^x(4+16)}{4^x\times(4+1)}\)

or, \(\frac{20}5\)

= 4

\(\sqrt{(a+b)^{-1}}\times\sqrt{(a-b)(a^2-b^2)}\)

= \(\sqrt{(a+b)^{-1}(a-b)(a+b)(a-b)}\)

= \(\sqrt{(a+b)^{-1+1}(a-b)^2}\)

=\(\sqrt{(a+b)^0(a-b)^2}\)

= \(\sqrt{1.(a-b)^2}\)

= (a - b)

Here,

First expression = 3a³ - 27a

= 3a(a² - 9)

= 3a (a² - 3²)

= 3a (a+3) (a-3)

Second expression = a³ + 27

= a³ + 3³ 

= (a + 3) (a² - 3a + 9)

Therefore, HCF = (a + 3).

Here,

\(\sqrt{y-5}=\sqrt[3]{216}\)

or, \(\sqrt{y-5}=\sqrt[3]{6^3}\)

or, \(\sqrt{y-5}=6\)

Squaring both sides, we get

\((\sqrt{y-5})^2=(6)^2\)

or, y - 5 = 36

or, y = 36 + 5

or, y = 41

Therefore, the value of y is 41.

Let, the cost of a pencil = Rs.x

Then, the cost of a pen = Rs. (x + 15)

Now, By question

x + (x + 15) = 35

or, x + x + 15 = 35

or, 2x = 35-15

or, x = 20/2

or, x = 10

and x + 15 = 10 + 15 = 25

Therefore, the cost of the pen is rs. 25 and the cost of a pencil is Rs. 10.

Given, PQRS is a rectangle

and PQ = 3 Ps = 12cm

To find, Area of triangle PQM

Here,

(I) 3PS = 12cm

or, PS = 12/3 = 4cm

(II) Area of rectangle PQRS = PQ x PS [Area of rectangle = length x breadth]

= 12cm x 4cm

= 48cm² 

(III) Area of triangle PQM = 1/2 Area of rectangle PQRS [Both are standing on the same base and between the same parallel lines.]

or, Area of Triangle PQM = 1/2 x 48cm² 

= 24cm² 

Given,

O is the center of the circle

And BOD = 160°

To find: The value of x and y

Here,

(I) BAD = 1/2 BOD [Therefore, BAD is a circumference angle and BOD is a central angle standing on the same arc of a circle.]

or, x = (1/2) x 160°

or, x = 80°

(II) BAD + BCD = 180° [Therefore, being opposite angles of a cyclic quadrilateral]

or, 80° + y = 180°

or, y = 180° - 80°

or, y = 100°

Therefore, 80° and y = 100°

Given,

RT is a tangent, R is the point of contact,

QRT = 70° and PRT = 30° 

To find, The measure of RPQ.

Here,

(I) QRS + QRT = 180° [Therefore, sum of the adjacent angles standing on a straight line]

or, QRS + 70° = 180° 

or, QRS = 180° - 70° 

=110° 

(II) RPQ = QRS [Therefore, the angle made by a chord to the tangent of a circle at the point of contact is equal to the angle in the alternate segment.]

or, RPQ = 110° 

Given,

QR = 27cm, PR = 30cm and PRQ = 30°

To find, Area of kite PQRS

By formula,

Area of Triangle PQR = 1/2 .QR.PR. Sin 30°

= (1/2) x 27cm x 30cm x (1/2)

= 202.2cm² 

Here,

Area of triangle PQR = (1/2) area of kite PQRS [ Diagonal so formed by joining the vertices made by equal sides of kite bisects the kite.]

or, 202.5cm² = 1/2 area of kite PQRS

or, Area of kite PQRS = 405cm² 

In the given data,

No. of terms (N) = 35

Here,

Third quartile (Q₃) = The value of \(\frac{3N^{th}}4\) item

= the value of \(\frac{(3\times35)^{th}}4\) item

= the value of 26.25^th item

Here, CF just greater than 26.25 is 28 and it lies in the class ( 60- 80 )

Therefore, Third quartile class (Q₃) = (60 - 80).

Let, the event of getting a face card be F.

And the event of getting ace of spades be C.

Then, n(S) = 52, n(F) = 12 and n(C) = 1

To find, P(F∪C)

Here, F and C are mutually exclusive events.

Then, by formula

P(F∪C) = P(F) + P(C)

= \(\frac{n(F)}{n(S)}+\frac{n(C)}{n(S)}\)

= \(\frac{12}{52}+\frac1{52}\)

= \(\frac{12+1}{52}\)

= \(\frac{13}{52}\)

= \(\frac14\)

Therefore, the probability of getting a face card or an ace of spades is \(\frac14\).

Let, the event of getting a yellow marble be Y.

And the event of getting green marble be G.

Then, n(Y) = 10, n(G) = 12

And the total number of marbles = 10 + 12 = 22

A probability tree diagram when two marbles are drawn randomly in succession without replacement is given below.

Possible outcomes

Let, the set of students who like maths be M.

and the set of students who like science be S.

Then, n(U) = 50, n_o(S) = 15 and  \(n(\overline{M\cup S})=2n(M\cap S)\)

To find, \(n(\overline{M\cap S)}\)

Let, \(n(M\cap S)\) = x

Then, \(n(\overline{M\cup S)}\) = 2x

Now, representing the above information in a Venn diagram, we get

From the above Venn diagram,

n_o (M) + n_o (S) + \(n(M\cap S)\) + \(n(\overline{M\cup S})\) = n(U)

or, 20 + 15 + x + 2x = 50

or, 35 + 3x = 50

or, 3x = 50-35

or, x = 15/3

or, x = 5

Therefore, \(n(\overline{M\cap S)}\) = n(U) - \(n(M\cap S)\) = 50 - 5 = 45

Therefore, The number of students who like at most one subject is 45.

Given,

Marked price = Rs. 75000

Discount percent = 15%

VAT percent = 15%

To find, SP with VAT and amount of VAT

By formula,

Selling price (SP) = MP - discount % of MP

= Rs.75000-15% of Rs.75000

= Rs. 75000 - (15/100) x Rs. 75000

= Rs. 75000 - Rs. 11250

= Rs. 63750

Again,

Amount of VAT = VAT % of SP

= 15% of Rs.63750

= (15/100) x Rs. 63750

= Rs. 9562.50

Given,

In an umbrella, 

No. of triangular pieces of cloth = 12

Length of each of two equal sides of triangular pieces (a) = 42cm = 42m/100 = 0.42m

and the length of third side (b) = 21cm = 21/100 = 0.21

and the cost of per sq. meter of cloth (C) = Rs. 1250

To find: (I) Total area of cloth

   (ii) Total cost of cloth

By formula,

Area of each isoceles triangle piece of cloth = \(\frac b4\sqrt{4a^2-b^2}\)

= \(\frac{0.21m}4\times\sqrt{4\times(0.42m)^2-(0.21m)^2}\)

= 0.0525m x \(\sqrt{0.7056m^2-0.0441m^2}\)

= 0.0525m x \(\sqrt{0.6615m^2}\)

= 0.0525m x 0.8133m

= 0.0427m² 

Total area of cloth (A) = Area of each piece x No. of pieces cloth

= 0.0427m² x 12

= 0.5124m² 

Now,

The total cost of cloth (T) = Cost of per sq. meter of cloth x area

= Rs. 1250 x 0.5124

= Rs. 640.50

Total area of cloth (A) = 0.5124m² 

and total cost of cloth (T) = Rs.640.50

Here,

First expression

= 8m⁴ + 27mn³ 
= m(8m³ + 27n³)
= m{(2m)³ + (3n)³}
= m(2m + 3n) (4m² - 6mn + 9n²)

Second expression

= 8m³n + 2m² n² - 15mn³ 
= mn(8m² + 12mn - 10mn - 15n²)
= mn{4m(2m+3n)- 5n(2m + 3n)}
= mn(2m+3n) (4m-5n)

Third expression

= 4m³n - 9mn³ 
= mn(4m² - 9n²)
= mn{(2m)² - (3n)²}
= mn(2m + 3n) (2m -3n)

∴ HCF of the given expressions is = m(2m+3n)

Here,

9^a - 10 x 3^a + 9 = 0
or, (3^a)² - 10 x 3^a + 9 = 0
Let, 3^a = x, we get
X² - 10x + 9 = 0
x² - 9x -x + 9 = 0
or, x(x - 9) - 1(x - 9) = 0
or, (x - 9) (x - 1) = 0
Now, substituting the value of x, we get
(3^a - 9) (3^a - 1) = 0

Therefore a = 0 or 2.

Given: O is the center of the circle, AB is diameter and ACB is an angle inscribed in a semi-circle.

To prove: ACB = 90°

Given,

In quadrilateral ABCD
AB = 4.5cm
BC = 5.5cm
CD = 5.7cm
DA = 4.9cm
and diagonal BD = 5.9cm

To Construct: Quadrilateral ABCD of given data and triangle DAE whose area is equal to the area of the given quadrilateral ABCD.

Hence,

Quadrilateral ABCD of given data and triangle DAE equal in area to the quadrilateral ABCD is constructed.

Statement: The opposite angles of a cyclic quadrilateral are supplementary.

Construction of figures:

Two circles of different radii with center 'O' are drawn. In each circle, a cyclic quadrilateral ABCD is drawn.

To verify:

i) ABC + ADC = 180°
ii)BAD + BCD = 180°

Method of verification:

By using a protractor, ABC, ADC, BAD, and BCD are measured and the results are tabulated below.

Hence, The opposite angles of a cyclic quadrilateral are supplementary.

Verified

Let, AB be the height of a tower, BC be the length of the shadow of the tower and ACB be the angle made by the sun's rays with the ground.

Then, by the question

AB = 50m, ACB = 30°, BC = ?

Here, in right-angled triangle ABC

Tan 30° = AB/BC

or, \(\frac1{\sqrt3}\) = \(\frac{50m}{BC}\)

or, BC = \(50\sqrt3m\)

or, BC = 5- x 1.732m

or, BC = 86.6m

Therefore, the length of the shadow of the tower is 86.6m.

Here, preparing a table to calculate the mean, we get.

Now, by formula

mean (X) = Σfm/N

= 620/22

= 28.18

Therefore, the mean is 28.18.

Given,

Principal (P) = Rs.8000
Rate of interest (R) = 10% p.a
Time (T) = 1(1/2) years = 3/2 years

To find C. I compounded half-yearly- Simple interest.

By the formula,

Simple interest (SI) = \(\frac{P\times T\times R}{100}\)

\(\frac{8000\times{\displaystyle\frac32}\times10}{100}\)
= Rs. 120000/100
= Rs. 1200

Again by formula,

CI compounded half yearly = \(p\left\{(1+\frac R{200})^{2T}-1\right\}\)

= Rs. 8000 \(\left\{(1+\frac{10}{200})^{2\frac32}-1\right\}\)
= Rs. 8000 {(1.05)³ - 1}
= Rs. 8000 (1.15-1)
= Rs. 8000 x 0.15
= Rs. 1261

Therefore, the difference between C.I. compounded half-yearly and simple interest = Rs.1261 - Rs.1200

Rs.61. Ans

Given,

In a combined solid,
Area of the circular base = 100cm²
Height of the cylinder (h₁) = 3cm
Volume of the combined solid (V) = 600cm²

To find: Height of the combined solid (h) 

By formula,

The volume of the cylinder (V₁) = area of the base x height

= 100cm² x 3cm

= 300cm³ 

Therefore volume of the come (V₂) = combined volume - volume of cylinder

= 600cm³ - 300cm³

= 300cm³  

Again, by formula

Volume of the cone (V₂) = 1/3 area of base x height

or, 300cm³ = (1/3) x 100cm² x h₂ 

or, 900cm³ / 100cm² = h₂ 

or, h₂ = 9cm

Therefore, the volume of cone (h₂) = 9cm

Now, h = h₁ + h₂

= 3cm + 9cm

= 12cm

∴ The height of the combined solid (h) = 12cm. 

Let, length of the rectangular room (l) = x m

and breadth of the room (b) = y m

Here,

According to the first condition,

xy = 45 ..........(i)

Again, according to the second condition,

x - 3 = y + 1
x = y + 1 + 3
x = y + 4 ........(ii)

Putting the value of x from equation(ii) and equation (i), we get

(y + 4)y = 45
or, y² + 4y - 45 = 0
or, y² + (9-5)y - 45 = 0
or, y² + 9y - 5y - 45 = 0
or, y(y + 9) - 5(y + 9) = 0
or, (y + 9) (y - 5) = 0 

Either y + 9 = 0 ..........(iii)

Or, y - 5 = 0 ..........(iv)

From equation (iii)

y + 9 = 0
or, y = -9 which is impossible because length and breadth cannot be negative.

or, y = 5

Now, putting the value of y in equation (ii), we got,

x = 5 + 4

= 9

∴ Length of the room (l) = 9m and breadth of the room (b) = 5cm.

Given,

AD//BC and area of
Triangle ABE = area of triangle ACF

To prove: EF//AC

Construction: EC is joined.

Proof:

Verified.