Solution,

Here, initial population (P) = X₀

Population growth rate (R) = R% p.a 

Population after Y years (P_T) = ?

By using population growth formula,

P_T = \(P\left(1+\frac R{100}\right)^T\)
     = \(X_0\left(1+\frac R{100}\right)^Y\)

∴ Population of that place after Y years is \(X_0\left(1+\frac R{100}\right)^Y\)

Solution,

Here, length of each equal sides (a) = m cm,

length of third side (b) = n cm

Area of isosceles triangle  (A) = ?

By using formula,

A = \(\frac b4\sqrt{4a^2-b^2}\)

   = \(\frac n4\sqrt{4m^2-n^2}cm^2\)

Solution,

We know from indices that x⁰ = 1,

So, applying above rule, we get (9m)⁰ = 1.

Proof,

9m⁰ = (9m)^x-x (because x-x = 0)

= (9m)^x x (9m)^-x

= \(\frac{9m^x}{9m^x}\)

= 1

The name of the quartile which divides the given continuous data below 25% is first quartile or lower quartile. It is represented as Q₁.

If a rectangle and triangle are standing on a same base MN and between the same parallel lines MN and AB, then the area of rectangle becomes double the area of the triangle.

The opposite angles of a cyclic quadrilateral are supplementary or their sum is 180°.  

Cyclic Quadrilateral Angles

Cyclic Quadrilateral is a quadrilateral which has all its four vertices lying on a circle. It is also know as inscribed quadrilateral. The sum of the opposite angles of a cyclic quadrilateral is supplementary. 

Let ∠A, ∠B, ∠C and ∠D are the four angles of an inscribed quadrilateral. Then,

∠A + ∠C = 180° 

∠B + ∠D = 180°

Therefore, an inscribed quadrilateral also meet the angle sum property of a quadrilateral, according to which, the sum of all the angles equals 360 degrees. Hence,

∠A + ∠B + ∠C + ∠D= 360° 

Solution,

Given, Selling price (SP) = Rs. 1800

Rate of VAT = 13%

To find, Selling Price with VAT included.

Using Formula,

S.P with VAT = SP + VAT % of SP

= Rs. 1800 + 13% x 1800

= Rs 1800 + Rs 234

= Rs 2043

∴ Selling price of a bag after levying 13% VAT on it becomes Rs. 2043

Solution:

Given:

Present price (P) = Rs. 20,00,000
Rate of compound depreciation (R) = 10% pa
Price after depreciation (P_T) = Rs. 14,58,000

To find:

Time taken to reach the new reduced price (T) = ?

Using Formula:

\(P_t=P\left(1-\frac R{100}\right)^T\)
Or, Rs, 1458000 = \(Rs\;2000000\left(1-\frac{10}{100}\right)^T\)
Or, \(\frac{Rs\;1458000}{Rs\;2000000}={(1-0.1)}^T\)
Or, \(0.729=\left(0.9\right)^T\)
Or, \(\left(0.9\right)^3=\left(0.9\right)^T\)
Or, T = 3

∴ Required time (T) = 3 years.

Solution,

Given:

In ΔMNO,
MN (a) = 24 cm
NO (b) = 25 cm
MO (c) = 7 cm

To find,

Area of triangle ΔMNO,

Here,

Semi-Perimeter (s) \(=\frac{a+b+c}2\)
\(=\frac{24cm+25cm+7cm}2\)
\(=\frac{56cm}2\)
=28cm

Now using formula,

Area of triangle (A) = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{28(28-24)(28-25)(28-7)}cm^2\)
= \(\sqrt{28x4x3x21}cm^2\)
=\(\sqrt{7056}cm^2\)
=\(84cm^2\)

∴ Area of triangle MNO is 84 cm².

Solution,

Given:

Volume of a hemisphere(V) =486π cm³.

To find:

Radius of the hemisphere (R) = ?

By formula,

Volume of hemisphere(V) = \(\frac23\pi R^3\)
Or, 486π cm³ = \(\frac23\pi R^3\)
Or, \(R^3\) = \(\frac{3\times486\mathrm\pi}{2\mathrm\pi}\)
Or, \(R^3\) = 729 cm³.
^{∴ R = 9 cm}

Hence, the radius of hemisphere is 9cm.

Solution,

Given:

Area of the rectangular faces of a prism (L.S.A) = 180 cm²., ∠PQR = 90°, PR = 5cm

To find:

The length of PP'

Here, PQ = P'Q' = 3cm

Then, in the right angles triangle PQR,

PQ² + QR² = PR²

Or, (3cm)² + QR² = (5cm)²
Or, 9cm² + QR² = 25cm²
Or, QR² = 16cm²
Or, QR = 4cm

Again,

Perimeter of triangle PQR = PQ + QR + PR
                                          = 3 cm + 4 cm + 5 cm
                                          = 12 cm

Now, by formula

L.S.A of a prism = perimeter of the base x height

Or, 180 cm² = 12 x PP'
Or, \(\frac{180cm^2}{12cm}=PP'\)
Or, 15 cm = PP'

∴ The length of PP' = 15 cm

Solution,

Here:

L.H.S = \(\frac{4^{a+2}+4^a}{17\times4^a}\)
= \(\frac{4^a\times4^2+4^a}{17\times4^a}\)
= \(\frac{4^a(16+1)}{17\times4^a}\)
=\(\frac{4^a\times17}{17\times4^a}\)
= 1
= R.H.S, hence proved.

Solution:

Here,

\(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\)
Multiplying both numerator and denominator by \(\sqrt3+\sqrt2\) , we get
\(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\)
= \(\frac{{(\sqrt3+\sqrt2)}^2}{{(\sqrt3)}^2-{(\sqrt2)}^2}\)
= \(\frac{{(\sqrt3)}^2+2\times\sqrt3\times\sqrt2+{(\sqrt2)}^2}{3-2}\)
= \(\frac{3+2\times\sqrt6+2}{3-2}\)
= \(\frac{5+2\times\sqrt6}{1}\)
= \(5+2\sqrt6\)

Is the required answer

Solution:

Here, first expression = p² - q² 
 = (p+q)(p-q)

Second expression = (p+q)² 
= (p+q)(p+q)

∴ L.C.M = (p+q)(p+q)(p-q) = (p+q)² (p-q) [ LCM is product of common term from both expressions and remaining terms]

Solution:

Here,

\(\sqrt{w+4}=\sqrt[3]{64}\)
Or, \(\sqrt{w+4}=\sqrt[3]{4^3}\)
Or, \(\sqrt{w+4}=4\)

Squaring both sides, we get

\(\left(\sqrt{w+4}\right)^2=4^2\)
Or, w + 4 = 16
Or, w = 16 - 4
Or, w = 12

∴ The value of w is 12.

Solution:

Let, the smaller even number be x. Then the consecutive greater even number will be x + 2.

Now, according to the question,

x + (x + 2) = 34
Or, 2x + 2 = 34
Or, 2x = 34 -2 
Or, 2x = 32
Or, x = 32/2 
Or, x = 16,
Then, x + 2 = 16 + 2 = 18 

∴ The two required consecutive even numbers whose sum is 34 are 16 and 18.

Solution:

Given,

angle XAY (∠XAY ) = 90°,
AY = 8cm
and Area of parallelogram WXYZ = 64cm² 

To find,

The length of AX = ?

Here,

Area of triangle ΔAXY = 1/2 area of parallelogram WXYZ [ ∵ Both are standing on same base between the same parallel lines.]
Or, Area of ΔAXY  = 1/2 x 64 cm² 
Or, Area of ΔAXY  = 32 cm² 

Now using formula,

Area of ΔAXY = 1/2 x base x height
Or, 32 = 1/2 x AX x AY
Or, 32 = 1/2 x AX x 8
Or, AX = 32/4 = 8 cm

∴ The length of AX = 8 cm.

Solution:

Given,

M is the center of the circle.
∠GEF = 2x° and ∠EGF = 3x°

To find:

The value of ∠FHG

Here,

1. ∠EFG = 90°, being angle inscribed in a semi-circle
   
   
2. ∠GEF + ∠EFG + ∠EGF = 180, being sum of the angles in a triangle.
   Or, 2x° + 90° + 3x° = 180°
   Or, 5x° = 180° - 90°
   Or, 5x° = 90°
   Or, x° = 90/5 = 18°
   
   
3. ∠FHG = ∠GEF , being circumference angles of a circle standing on a same arc.
   = 2x°
   = 2 x 18°
   = 36°

∴ The value of ∠FGH = 36°

Solution:

Given: 

X is the center of the circle,
AB and AC are two tangents to the circle.
∠BXC = 105°

To find:

The value of ∠BAC

Here,

1. ∠XBA = 90° and ∠XCA = 90° [because, a tangent to a circle is perpendicular to the radius of the circle drawn at the point of contact]
   
   
   
2. ∠XBA + ∠BXC + ∠XCA + ∠BAC = 90° [because, sum of the angles of quadrilateral ABXC]
   Or, 90° + 105° + 90° + ∠BAC = 360°
   Or, 285° + ∠BAC = 360°
   Or, ∠BAC = 75°

∴ The value of ∠BAC is 75°.

Solution:

Given, In ΔDEF, DE = 12√3 cm, ∠DEF = 60° and the area of ΔDEF = 36cm2

To find:

The measurement of EF

By formula,

Area of ΔDEF = \(\frac12DE\times EF\times\sin\left(60\right)\)
Or, 36 cm² = \(\frac1212\sqrt3\times EF\times\frac{\sqrt3}2\)
Or, 36 cm² = 9cm x EF
Or, EF = 4cm

∴ The measurement of EF is 4cm is the required answer

Solution:

Given: In a continuous series,

Average marks (\(\overline x\)) = 40
and sum of their marks (∑fm) = 1200.

To find: No of students (N)

By formula,

\(\overline x=\frac{\sum fm}N\)
Or, \(40=\frac{1200}N\)
Or, \(40\times N=1200\)
Or, N = 30

∴ The number of students (N) = 30 is the required answer

Solution,

Let, the event of occurring odd number on dice be A and the event of occurring head on a coin be H.

Then, when a dice is rolled,

S = {1, 2, 3, 4, 5, 6} and A = {1, 3, 5}

∴ n(S) = 6 and n(A) = 3

By formula P(A) = \(\frac{n(A)}{n(S)}\)
= \(\frac36\)
= \(\frac12\)

And,

When a coin is tossed,
P(H) = \(\frac12\)

To find: P(A ∩ H)

here, A and H are independent events

∴ P(A ∩ H) = P(A) x P(H)

= \(\frac12\times\frac12\)
= \(\frac14\)

∴ The probability of occurring odd number on dice and head on a coin is 0.25 or \(\frac14\)

Solution:

Let, the event of getting an ace be A and the event of not getting an ace be \(\overline A\)

Then, n(A) = 4, n(\(\overline A\)) = 48

and total number of cards = 52

Now drawing probability tree diagram when two cards are drawn randomly in succession without replacement, we get

Solution:

Let, the set of people who like curd be C and the set of people who like milk be M. Then,

n(U) = 100%
n(C) = 70%
n(M) = 60%
n(\(\overline{C\cup M}\)) = 20%
and n(C ∩ M) = 550

To find,

1. n(U) = ?
2. n_o(C) = ?

Again let, n(C ∩ M) = x%

Representing above information in venn-diagram we get,

From above venn-diagram,

no(C) + no(M) + n(C ∩ M) + n(\(\overline{C\cup M}\)) = n(U)
Or, (70% -x%) + (60% - x%) + x% + 20% = 100%
Or, 150% - x% = 100%
Or, x% = 50%
∴ n(C ∩ M) = 50%

Here, 

50% of n(U) = n(C ∩ M)
Or, 0.5 x n(U) = 550
∴ n(U) = 550/0.5 = 1100

Again, from the venn diagram above, we get

n_o(C) = (70% - x%) of n(U)
= (70% - 50%) x 1100
= 20% x 1100
= 0.2 x 1100
= 220

∴ Total number of people participated in survey = 1100 and the number of people who like curd only are 220.

Solution:
Given,

US $1 = NRs. 110
Exchanged currency = NRs. 6,60,000
Revaluated rate of nepali currency = 10%

To find,

Profit or loss when US dollar is re-exchanged into Nepali currency

According to the initial rate of exchange,

NRs 110 = US $1
Or, NRs 1 = US $1/110
Or, NRs 6,60,000 = $6,60,000/110
= $6000

Here, when Nepali currency is revaluated by 10%, then the new rate of exchange becomes

$1 = NRs 110 - 10% of NRs 110
= NRs 110 - NRs 11
= NRs 99

Now, according to the new rate of exchange

US $1 = NRs. 99
∴ $6000 = NRs. 5,94,000

Here, Nepali currency received by re-exchange is less than the initial currency. So, there is a loss

∴ loss = Initial currency - currency received after re-exchange

= NRs 6,60,000 - NRs 5,94,000
= NRs. 66,000

∴ The loss for businessman is NRs. 66,000

Solution:

Given,

In a combined solid made of a cylinder and a cone, diameter of the base (d) = 6 cm,
length of cylindrical part = 116 cm,
Total length of the solid = 120 cm

To find:

Total surface area of the combined solid.

Here,

Radius of the base (r) = \(\frac d2\)
= \(\frac{6cm}2\)
= 3 cm

Height of the cone (h) = 120 cm - 116 cm = 4 cm

And, 

slant height of cone (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{{(3cm)}^2+{(4cm)}^2}\)
= \(\sqrt{9cm^2+16cm^2}\)
= \(\sqrt{25cm^2}\)
= 5 cm

Now, by formula

Area of circular base = πr² 
= \(\frac{22}7{(3cm)}^2\)
= \(\frac{22}7\times9cm^2\)
= 28.29 cm² 

CSA of cylinder = 2πrh

= \(2\times\frac{22}7\times3cm\times116cm\)
= 2187.43 cm² 

And,

CSA of cone= πrl
= \(\frac{22}7\times3cm\times5cm\)
= 47.14 cm² 

∴ TSA of the combined solid = 28.29cm² + 2187.43cm² + 47.14cm² = 2262.86 cm² is the required answer

Solution:

Here, first expression

= \(p^2+4pq+4q^2\)
= \(p^2+2.p.2q+{(2q)}^2\)
= \({(p+2q)}^2\)
= (p+2q)(p+2q)

Second expression

= \(p^4+8pq^3\)
= \(p(p^3+8q^3)\)
= \(p(p^3+{(2q)}^3)\)
= \(p(p+2q)(p^2-2pq+4q^2)\)

Third expression

= \(3p^4+10p^2q^2+p^3q\)
= \(p^2(3p^2-10q^2+pq)\)
= \(p^2(3p^2+pq-10q^2)\)
= \(p^2(3p^2+(6-5)pq-10q^2)\)
= \(p^2(3p^2+6pq-5pq-10q^2)\)
= \(p^2(3p(p+2q)-5p(p+2q)\)
= \(p^2(p+2q)(3p-5q)\)

∴ HCF is (2+2q)

Solution:

Here,

\(\frac1{1+p+p^2}-\frac1{1+p+p^2}-\frac{2p}{1-p^2+p^4}\)

= \(\frac{1\left(1+p+p^2\right)-1\left(1-p+p^2\right)}{\left(1-p+p^2\right)\left(1+p+p^2\right)}-\frac{2p}{1-p^2+p^4}\)

= \(\frac{1+p+p^2-1+p-p^2}{\left\{\left(1+p^2\right)-p\right\}\left\{\left(1+p^2\right)+p\right\}}-\frac{2p}{1-p^2+p^4}\)

= \(\frac{2p}{\left(1+p^2\right)^2-p^2}-\frac{2p}{1-p^2+p^4}\)

= \(\frac{2p}{1+2p^2+p^4-p^2}-\frac{2p}{1-p^2+p^4}\)

= \(\frac{2p}{1+p^2+p^4}-\frac{2p}{1-p^2+p^4}\)

= \(\frac{2p\left(1-p^2+p^4\right)-2p\left(1+p^2+p^4\right)}{\left(1+p^2+p^4\right)\left(1-p^2+p^4\right)}\)

= \(\frac{2p-2p^3+2p^5-2p-2p^3-2p^5}{\left\{\left(1+p^4\right)+p^2\right\}\left\{\left(1+p^4\right)-p^2\right\}}\)

= \(\frac{-4p^3}{\left(1+p^4\right)^2-\left(p^2\right)^2}\)

= \(\frac{-4p^3}{1+2p^4+p^8-p^4}\)

= \(\frac{-4p^3}{1+p^4+p^8}\)

Solution:

Given: EF//GH, EI//FG and FJ//GI

To prove:

1. ΔEFJ ≅ ΔHGI
2. Area of parallelogram EFGH = Area of parallelogram FGIJ

Proof:

Hence proved.

Solution:

Given:

In quadrilateral MNOP,
NO = MN = 4.2cm,
OP = PM = 5.2 cm,
and ∠NOP = 75°.

To construct:

Quadrilateral MNOP pf given data and ΔPQO equal in area to the quadrilateral MNOP.

Hence, quadrilateral MNOP of given data and ΔPQO equal in area to the quadrilateral MNOP is constructed.

Solution:

Statement:

The circumference angle BED is half of the central angle BCD standing on the same arc BD of a circle.

Construction of figures:

Two circles of different radii with center C are drawn. In each circle, circumference angle BED and central angle BCD are drawn which are standing on the same arc BD.

To verify:-

By using a protractor, ∠BED and ∠BCD are measured and the results are tabulated below.

Conclusion:

Hence, the circumference angle of a circle is half of the central angle standing on the same arc.

Hence verified.

Solution:

Let, AB be the height of the house, CD be the height of the temple, AF = BD is the distance between house and temple, ∠ECA be the roof of the house from the roof of the temple. Then, AB = FD = 13 m, CD = 25 m, ∠ECA = ∠CAF = 45°, AF = BD = ?

Here, CF = CD - FD
= 25 m - 13 m
= 12 m

Now, in the right angled triangle ΔCFA,

tan(45°) = CF/AF
Or, 1 = 12m/BD
Or, BD = 12 m

∴ The distance between the house and the temple is BD = 12 m

Solution:

Here,

Preparing a cumulative frequency table to calculate median from the given data, we get

Using formula,

Median (M_d) = The value of (N/2)^th item
= (30/2)^th item
= 15^th item

here,

The corresponding class of cf 15 is 50-60

∴ Median class = 50-60

Again, by formula

Actual median (M_d) = \(L+\frac{\left(\frac{N}{2}-cf\right)}{f}\cdot i\)

Where,

L = 50,
N/2 = 15
cf = 7
f = 8
i = 10

So, Actual median (M_d) 

= \(50+\frac{\left(15-7\right)}{8}\cdot10\)
= 60

∴ Median of the given data is 60.

Solution:

Here,

For the first year:
Principal (P1) = Rs. 55,000
Rate of interest (R1) = 10% p.a.
Time (T1) = 1 year
Annual compound interest (CI1) = ?

By formula,

CI1 = \(p1\left\{\left(1+\frac{R1}{100}\right)^{T1}-1\right\}\)
= \(55000\left\{\left(1+\frac{10}{100}\right)^1-1\right\}\)
= \(55000\left\{\left(1.1\right)-1\right\}\)
= Rs. 5500

∴ Compound interest of the first year (CI1) = Rs. 5500

And, compound Amount after 1 year (CA1) = P1 + CI1 = Rs. 60,500

Again,

For the second year:
Principal (P2) = CA1 = Rs. 60,500
Rate of interest (R2) = 10% p.a.
Time (T2) = 1 year
Semi-annual interest (CI2) = ?

By formula,

Semi-annual compound interest (CI2)
= \(P2\left\{\left(1+\frac{R2}{200}\right)^{2\cdot T2}-1\right\}\)
= \(60500\left\{\left(1+\frac{10}{200}\right)^2-1\right\}\)
= \(60500\left\{1.1025-1\right\}\)
= \(\left\{60500\cdot0.1025\right\}\)
= Rs. 6201.25

Here, CI2 - CI1 = Rs. 6201.25 - Rs. 5500
= Rs. 701.25

∴ Interest of the second year is more than the interest of the first year by Rs. 701.25

Now, difference in interest in percentage = \( \frac{diffininterest}{CI1}\cdot100\%\)

= \(\frac{Rs701.25}{Rs5500}\cdot100\%\)
= 12.75%

∴ Interest of the second year is 12.7% more that interest of the second year.

Solution:

Given:-

In pillars formed by prism and pyramid,
Length of the square base (l) = 6 feet
Height of prism = 8 feet
Height of pyramid = 4 feet
No of pillars = 2
Rate of painting = Rs. 75/sq.ft

To find: Total cost of painting on two pillars.

Here,

Slant height of the square based pyramid (h1)
= \(\sqrt{h^2+\left(\frac{l}{2}\right)^2}\)
= \(\sqrt{4^2+\left(\frac{6}{2}\right)^2}\)
= 5 feet

And, Perimeter of the square base = 4l = 4 x 6 feet = 24 feet

By formula, L.S.A of a square based prism 

= Perimeter of base x height
= 24 feet x 8 feet
= 192 sq. ft

Again, by formula

LSA of a square based pyramid = 2lh₁ 
= 2 x 6 ft x 5 ft
= 60 sq ft

Here, Surface area of each pillar 

= LSA of prism + LSA of pyramid
= 192 sq ft + 60 sq ft
= 252 sq ft

∴ Surface area of two pillars = 2 x 252 sq ft = 504 sq ft

Now, total cost of painting = Cost per sq ft x total area = Rs 75 x 504 = Rs. 37,800

∴ The total cost of painting two pillars = Rs. 37,800

Solution:

Given:

Area of a rectangular land = 3000 sq m
and perimeter of the land = 220 m

To find:

Percentage of length or breadth to be decreased to make the square land.

Let, length (l) = x m and breadth (b) = y m

Then, by formula, area of rectangle = l x b i.e., xy = 3000 ------- ( equation 1)

and , perimeter of a rectangle = 2(l + b) = 2 ( x + y) = 220 i.e., x + y = 110 -------- (equation 2)

Substituting the value of y = 110 - x from equation 2 in equation 1 we get,

x (110 - x) = 3000
Or, 110x - x² = 3000
Or, x² - 110x + 3000 = 0
Or, x² - (60x + 50x) + 3000 = 0
Or, x² - 60x -50x + 3000 = 0
Or, x(x - 60) - 50(x-60) = 0
Or, (x - 50)(x - 60) = 0

Either, x -60 = 0 ----- (equation 3)
Or, x - 50 =0 --------- (equation 4)

From, eqn (3), x - 60 = 0 => x = 60

Now, putting the value of 'x' in equation 2 we get

y = 110 - 60 = 50

∴ If x = 60, then y = 50

Again, from equation 4

x - 50 =0 => x = 50. Putting x in eqn (2) we get, y = 110 - x = 110 - 50 = 60

∴ If x = 50, then y = 60

But assuming that length is longer than breadth, we get length(l) = 60 m and breadth (b) = 50 m

∴ Difference between length and breadth = 60 m - 50 m = 10 m

So, to make the land square the length has to be decreased by 10 m.

Here, Percentage decreased in length = \(\frac{difference}{length}\cdot100\%\)

= \(\frac{10m}{60m}\cdot100\%\)
= \(\frac{50}{3}\%\)
= 16.66%

∴ To change the given rectangular land into a square, length has to be decreased by 16.66%

Solution:

Given:

Points K, L, M and N are concyclic such that arc \(\overset\frown{KL}\) and arc \(\overset\frown{NM}\) are intersected at a point P.

To Prove:

1. Area of triangle KPL = Area of triangle NPM
2. chord KM = chord LN

Proof:

Hence proved