Here,

Rate of discount 

=  \(\frac{M.P.-S.P.}{M.P.}\times100\%\)
= \(\frac{5x-4x}{5x}\times100\%\)
= \(\frac x{5x}\times100\%\)
= 20%

Hence,selling price of Rs. 5x on a marked price of 4x means 20% discount was applied on the bag.

Solution:

Thea area of an equilateral triangle (A) = \(\style{font-size:28px}{\frac{\sqrt3}4a^2}\) sq. unit

Here,

    8x³ - 27y³ 
=  (2x)³ - (3y)³ (using the formula of a³ -b³ )
= (2x -3y) {(2x)² + 2x.3y + (3y)²}
= (2x - 3y) (4x² +6xy +9y²) is the required answer.

Here,

Mean (\(\overline x\)) = \(\frac{\sum fx}N\)
Or, \(10=\frac{\sum fx}{40}\)
Or, ∑fx = 40 x 10
∴ ∑fx = 400 is the required answer

Solution:

If two parallelograms are standing on the same base and lie between same parallel lines, then their areas are equal.

Proof:

Consider we have two parallelograms ABCD and ABEF which are standing on the same base AB and in between same parallel lines CF and AB as shown in the diagram below. The question now is is there any relation between these two parallelogram.

Theorem: Parallelograms on the same base between the same parallel lines are equal in areas.

Proof: Take the figure given above. We can show the triangle ΔBCE and triangle ΔDAF are congruent. The proof is as follow

1. side BC = side AD (they are the opposite sides of parallelogram BCDA, and are equal in length)
2. angle ∠BCE = angle ∠ADF (corresponding angles, because BC//AD being sides of a parallelogram)
3. angle ∠BEC = angle ∠AFD ( same as above )

By the ASA criterion, the two triangles are now congruent and hence are equal in area. Now,

area of parallelogram ABCD = area of triangle BCE + area of quadrilateral BEDA --------- (1)
And, area of parallelogram BEFA = area of triangle ADF + area of quadrilateral BEDA --------- (2)

Clearly, from equation 1 and 2

Area of parallelogram ABCD = Area of parallelogram BEFA . 

Hence Proved.

Here,

Since the angle inscribed in the same segment of a circle are equal.

∠PQR = ∠PSR

Solution:

Given,

Rate of VAT = 10%
S.P. with VAT = Rs. 1320

To find,

1. Marked Price (MP) = ?
2. Amount of VAT = ?

Let, the marked price  be x

Here,

S.P. with VAT = MP + 10% of MP
Or, Rs. 1320 = x + 10% of x
Or, Rs. 1320 = 11x/100
∴ x = Rs. 1200

∴ Marked price (MP) of an article is Rs. 1200

and the amount of VAT = SP with VAT - MP = Rs. 1320 - Rs. 1200 = Rs. 120

Hence the VAT amount was Rs. 120.

Given:

Principal (P) = Rs. 1
Interest (I) = 1 Paisa = Rs. \(\frac1{100}\)
Time (T) = 1 Month = \(\frac1{12}\) year

To find:

Rate of interest (R) = ?

By formula,

\(R=\frac{I\times100}{P\times T}\)
= \(R=\frac{{\displaystyle\frac1{100}}\times100}{1\times{\displaystyle\frac1{12}}}\)
= \(R=\frac{12}1\)
= 12

∴ Rate of interest (R) = 12% is the required answer.

Solution:

Given:

In an isosceles triangle,
Length of base (b) = 10m
Area (A) = 60 sq m

To find:

Length of each equal sides (a)

By formula,

Area of an isosceles triangle (A) = \(\frac b4\sqrt{4a^2-b^2}\)
Or, 60m² = \(\frac{10}4\sqrt{4a^2-10^2}\)
Or, \(\frac{4\times60}{10}=\sqrt{4a^2-100}\)
Or, \(24=\sqrt{4a^2-100}\)

Squaring both sides we get,
\(\left(24\right)^2=\left(\sqrt{4a^2-100}\right)^2\)
Or, \(576=4a^2-100\)
Or, \(4a^2=676\)
Or, \(\frac{676}4=a^2\)
∴ a = 13 m

Hence, the measure of remaining equal sides of the triangular piece of the land is 13m.... is the required answer

Given:

Volume of a cylinder (V) = 1078 cm³ 
and radius of the base (r) = 7cm

To find:

Height of a cylinder (h) = ?

By formula,

Volume of a cylinder (V) = πr²h
Or, 1078 = (22/7) x 7² x h
Or, h = 1078 / 154
Or, h = 7 cm

Hence the required height of the cylinder is 7 cm.

Given,

In a triangular based prism,
PQ ⊥ RQ, PQ = 8 cm, QR = 6 cm and RR' = 15 cm

To find:

Lateral surface area of the prism (LSA) 

Here, In the right angled triangle PQR,

PR = \(\sqrt{PQ^2+QR^2}\) 
= \(\sqrt{8^2+6^2}\)
= \(\sqrt{64+36}\)
= \(\sqrt{100}\)
= 10 cm

∴ Perimeter of the triangular base = PQ + QR + PR = 8 cm + 6 cm + 10 cm = 24 cm

Now, by formula

Lateral surface area of the prism (LSA) = perimeter of base x height
= 24 cm x 15 cm
= 360 cm² is the required answer.

Here,

First expression

= 6 ( a³ + 8 )
= 6 ( a³ + 2³ )
= 2 x 3 ( a + 2 ) ( a² - 2a + 4 )

Second expression

= 8 ( 4a² - a - 18 )
= 2 x 2 x 2 { 4a² - (9 - 8)a - 18 }
= 2 x 2 x 2 { 4a² - 9a + 8a - 18 )
= 2 x 2 x 2 { a(4a - 9) + 2(4a - 9)}
= 2 x 2 x 2 (4a - 9) ( a+ 2 )

∴ HCF of two given expressions is the highest common factor = 2 (a + 2)

Here,

\(\frac{2^{x-1}+4\times2^x}{2^{x-1}}\)
= \(\frac{2^x\times{\displaystyle\frac12}+4\times2^x}{2^x\times{\displaystyle\frac12}}\)
= \(\frac{2^x\left({\displaystyle\frac12}+4\right)}{2^x\times{\displaystyle\frac12}}\)
= \(\frac{\displaystyle\frac{1+8}2}{\displaystyle\frac12}\)
= \(\frac92\times\frac21\)
= 9 is the required answer

Here,

\(\frac1{\sqrt[3]a-\sqrt[3]b}\)

Multiplying both numerator and denominator by \(\left(\sqrt[3]a\right)^2+\sqrt[3]a.\sqrt[3]b+\left(\sqrt[3]b\right)^2\), we get

\(\frac1{\sqrt[3]a-\sqrt[3]b}\times\frac{{\displaystyle\left(\sqrt[3]a\right)^2}{\displaystyle+}{\displaystyle\sqrt[3]a}{\displaystyle.}{\displaystyle\sqrt[3]b}{\displaystyle+}{\displaystyle{\displaystyle\left(\sqrt[3]b\right)}^2}}{\left(\sqrt[3]a\right)^2+\sqrt[3]a.\sqrt[3]b+\left(\sqrt[3]b\right)^2}\)

= \(\frac{\sqrt[3]{a^2}\times\sqrt[3]{ab}\times\sqrt[3]{b^2}}{\left(\sqrt[3]a\right)^3-\left(\sqrt[3]b\right)^3}\)

= \(\frac{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}{a-b}\)

is the required answer

Here,

\(9-\sqrt{x-4}=5\)
Or, \(9-5=\sqrt{x-4}\)
Or, \(4=\sqrt{x-4}\)

Squaring both sides, we get

\(\left(4\right)^2=\left(\sqrt{x-4}\right)^2\)
Or, 16 = x - 4
Or, 16 + 4 = x
Or, x = 20

Hence x = 20 is the required answer

Let, the smallest even number be x. 

Then, the consecutive greater even numbers are (x+2) and (x+4)

Now, according to the question,

x + (x+2) + (x+4) = 36
Or, 3x + 6 = 36
Or, 3x = 30
Or, x = 30/3 = 10

Then, x + 2 = 10 + 2 = 12

And, x + 4 = 10 + 4 = 14

Hence the required numbers are 10, 12 and 14

Solution:

Given:

D is the mid-point of BC, DE⊥AC
AC = 12 cm, DE = 5 cm

To find: 

Area of triangle ΔABC

Here,

1. Area of triangle ΔADC = \(\frac12\times AC\times DE\) [∵ Area of triangle is half the product of base and height ]
   = \(\frac12\times12\times5\)
   = 30 cm² 
   
   
2. Area of triangle ΔABC = 2 x Area if triangle ΔADC [∵ median AD bisects ΔABC into two equal halves by area ]
   Or, Area of triangle ΔABC = 2 x 30 cm² = 60 cm²

Hence the required area of triangle ΔABC is 60 cm²

Given:

O is the center of the circle
and ∠DAC = 25°

To find:

The values of the angles ∠BDC and ∠BAC

Here,

1. ∠DBC = ∠DAC = 25° [ being angles inscribed in the same segment of a circle ]
2. ∠BCD = 90° [ being angle inscribed in a semi-circle]
3. ∠DBC + ∠BCD + BDC = 180° [ sum of the interior angles of a triangle BCD]
   Or, 25° + 90° + ∠BDC = 180°
   Or, ∠BDC = 180° - 115°
   Or, ∠BDC = 65°
4. ∠BAC = ∠BDC  = 65° [ being angles inscribed in the same segment of a circle ]

∴ ∠BDC  = 65° and ∠BAC = 65° is the required answer

Given:

O is the center of circle, PG is tangent to the circle,
G is the point of contact and PG=DG

To find: The value of angle ∠DPG

Here,

1. angle ∠DGP = 90° [ Diameter drawn from the point of contact to a circle is perpendicular to the tangent of the circle. ] 
   
   
2. angle ∠PDG = angle ∠DPG [ being the base of isosceles triangle ΔDGP ]
   
   
3. ∠DGP + ∠PDG + ∠DPG = 180° [ being sum of the angles of triangle ΔDGP ]
   Or, 90° + ∠DPG + ∠DPG = 180°
   Or, 2 ∠DPG = 90°
   Or, ∠DPG = 90° / 2 = 45°

∴ the value of ∠DPG is 45°, required answer.

Given,

In triangle ΔXYZ, XY = 8 cm, YZ = 12 cm and area of triangle ΔXYZ is 24cm² 

To find: The size of angle ∠XYZ

Let,

angle ∠XYZ be θ

Now using formula

Area of triangle ΔXYZ = \(\frac12XY.YZ\sin\left(\theta\right)\)
Or, 24 = \(\frac12\cdot8\cdot12\cdot\sin\left(\theta\right)\)
Or, \(24=48\cdot\sin\left(\theta\right)\)
Or, \(\frac{24}{48}=\sin\left(\theta\right)\)
Or, \(\frac12=\sin\left(\theta\right)\)
Or, \(\sin\left(30^\circ\right)=\sin\left(\theta\right)\)
Or, 30° = θ

Hence the size of angle ∠XYZ is 30°

Given:

In a continuous series,
mean (\(\overline x\)) = 14 + p,
the number of terms (N) = 30
and ∑fx = 600

To find:

The value of p,

By formula

Mean (\(\overline x\)) = \(\frac{\sum fx}N\)
Or, 14 + p = \(\frac{600}{30}\)
Or, 14 + p = 20
Or, p = 20 - 14
Or, p = 6

Hence the value of p is 6.

Solution:

Let, the event of numbers divisible by 3 be A
and the event of square number be B.

Then,

S = {1, 2, 3, 4, 5, 6}
A = {3, 6}
B = {1, 4}
∴ n(S) = 6, n(A) = 2 and n(B) = 2

To find: n(A∪B)

Here,

A and B are mutually exclusive events. Then by formula
P(A∪B) = p(A) + p(B) 
= \(\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}\)
= \(\frac26+\frac26\)
= \(\frac23\)

∴ The probability of getting a number divisible by 3 or a square number is 2/3.

Solution:

Let,

the event of getting red ball be R
and the event of getting black ball be B

Then, n(R) = 10, n(B) = 6 and total number of balls = 10 + 6 = 16.

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

Let, the set of students who like cricket be C and the set of students who like basketball be B.

Then, n(U) = 180, n_o(C) = 50, n_o(B) = 30 and \(n\left(\overline{C\cup B}\right)\) = 50

To find: n(C):n(B) = ?

Let, n(C∩B) = x.

Now, representing above information in Venn-diagram we get,

From above Venn-diagram 

n_o(C) + n_o(B) + n(C∩B) + \(n\left(\overline{C\cup B}\right)\) = n(U)
Or, 50 + 30 + x + 50 = 180
Or, 130 + x = 180
Or, x = 180 - 130
Or, x = 50

∴ n(C∩B) = x = 50

Here, n(C) = n_o(C)  + n(C∩B) = 50 + 50 = 100

And, n(B) = n_o(B) + n(C∩B) = 30 + 50 = 80

So, n(C):n(B) = 100:80 = \(\frac54\) = 5:4

Hence, the ratio of the students who like cricket game and basketball game is 5:4

Solution:

Given:

Rate of discount = 20%
Rate of VAT ( value added tax ) = 13%
Amount of VAT = Rs. 83.20

To find: Marked price (M.P)

Let, marked price of cap be Rs. x

By formula,

Selling price (SP) = MP - Discount % of MP
= x - 20% of x
= \(x-\frac{20}{100}\times x\)
= \(x-\frac x5\)
= \(\frac{5x-x}5\)
= \(\frac{4x}5\)

Again, by formula

Amount of VAT = VAT % of SP
Or, \(83.20=13\%\times\frac{4x}5\)
Or, \(83.20=\frac{\displaystyle13}{100}\times\frac{4x}5\)
Or, \(83.20=\frac{13x}{125}\)
Or, \(x=\frac{83.20\times125}{13}\)
Or, x = Rs. 800

Hence the marked price of the cap is Rs. 800 is the required answer

Solution:

Given:

Perimeter of triangular piece of land (P) = 1800m
Ratio of the length of triangular sides (a:b:c) = 5:6:7

To find: Area of the triangular piece of land (A)

Let, the common unit for the length of sides be x.

The, a = 5x, b = 6x and c = 7x

We know that,

Perimeter of a triangle (P) = a + b + c
Or, 1800 m = 5x + 6x + 7x
Or, 18x = 1800 m
Or, x = 100 m

∴ a = 5x = 5*100 m =  500 m
b = 6x = 6*100 m = 600 m 
c = 7x = 7*100 m = 700 m

Again,

Semi-peremeter (s) = half of the perimeter = \(\frac P2\)
= \(\frac{100}2\)
= 900 m

Now, by formula,

Area of triangle (A) = \(\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\)
= \(\sqrt{900\left(900-500\right)\left(900-600\right)\left(900-700\right)}\)
= \(\sqrt{21600000000m^4}\)
= \(\sqrt{\left(146969.38m^2\right)^2}\)
= 146969.38 m² 

Hence the area of triangular piece of land is 146969.38 m² is the required answer

Solution:

Here, first expression is

a³ + 1 +2a² + 2a
= a³ + 1³ + 2a(a+1)
= (a+1)(a² - a +1) + 2a(a+1)
= (a+1) (a² - a +1 + 2a)
= (a+1) (a² + a +1)

Second expression,

= a³ - 1
= a³ - 1³ 
= (a-1)(a² + a +1)

Third expression,

a4 + a2 + 1
= (a²)² + (1)² + a² 
= (a² + 1)² - 2a² + a² 
= (a² + 1)² - a² 
= (a² + a +1) (a² - a +1)

Hence highest common factor (HCF) from all three expressions is (a² + a +1)

Here, 

\(\frac{6x-49}{7+\sqrt{6x}}=6-\frac{\sqrt{6x}{\displaystyle-}{\displaystyle11}}3\)
Or, \(\frac{\left(\sqrt{6x}\right)^2-\left(7\right)^2}{\sqrt{6x}+7}=\frac{\displaystyle18-\left(\sqrt{6x}-11\right)}3\)
Or, \(\frac{\left(\sqrt{6x}+7\right)\left(\sqrt{6x}-7\right)}{\sqrt{6x}+7}=\frac{\displaystyle18-\left(\sqrt{6x}-11\right)}3\)
Or, \(3\left(\sqrt{6x}-7\right)=29-\left(\sqrt{6x}-11\right)\)
Or, \(3\sqrt{6x}-21=29-\sqrt{6x}\)
Or, \(3\sqrt{6x}+\sqrt{6x}=29+11\)
Or, \(4\sqrt{6x}=50\)
Or, \(\sqrt{6x}=\frac{\displaystyle50}4\)
Or, \(\sqrt{6x}=\frac{25}2\)

Squaring both sides we get,
\(\left(\sqrt{6x}\right)^2=\left(\frac{25}2\right)^2\)
Or, \(6x=\frac{625}4\)
Or, \(x=\frac{625}{24}\)
Or, \(x=26\frac1{24}\)

is the required answer.

Solution:

Given:

O is the center of the circle and ABCD is a cyclic quadrilateral.

To prove: 

i. ∠ABC + ∠ADC = 180°
ii. ∠BAD + ∠BCD = 180°

Construction: Radii OA and OC are drawn

Proof:

hence proved

Given,

In ΔABC,
AB = 5.4 cm,
BC = 7 cm,
and ∠ABC = 60°

To construct: ΔABC of given data and a parallelogram equal in area to the triangle ΔABC and having a side of length 7.2 cm

hence ΔABC of given data and parallelogram CDEF equal in area to the triangle ΔABC  and having side DE = CF = 7.2 cm is constructed.

Statement: The circumference angles of a circle standing on the same arc are equal.

Construction figures:

Two circles of different radii with center 'O' are drawn. In each circle, circumference angle ∠ABC and ∠ADC are drawn which are standing on the same arc AC

To verify :- ∠ABC = ∠ADC

Method of verification:

By using a protractor, ∠ABC and ∠ADC are measured and the result are tabulated below.

Conculsion:- hence, the circumference angles of a circle standing on the same arc are equal.

Hence verified

Solution:

Let, AB be the height of the house CD be the height of a tree, GC = BD be the distance between the house and the tree, and ∠FAC be the angle of depression of the top of the tree from the house as shown in the figure below.

Then, according to the question

GB = CD = 20 ft, GC = BD = 10√3 ft, ∠FAC = 30°, AB = ?

here, in right angled triangle ΔAGC,

\(\tan\left(30^\circ\right)=\frac{AG}{GC}\)
Or, \(\frac1{\sqrt3}=\frac{AG}{10\sqrt3ft}\)
Or, \(\sqrt3AG=10\sqrt3ft\)
Or, \(AG=\frac{10\sqrt3}{\sqrt3}ft\)
Or, AG = 10 ft.

Now, AB = AG + GB = 10 ft + 20 ft = 30 ft

hence, the height of the house (AB) = 30 feet is the required answer.

Solution:

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

Here, median (Md) = 19 lies in (18 - 24)^th class

∴ Median class = 18 - 24

Median (Md) = \(L+\frac{{\displaystyle\frac N2}-cf}f\times i\)

Where, 

L = 18, \(\frac N2=\frac{24+p}2\), cf = 14, f = p, i = 6

∴ \(19=18+\frac{\left({\displaystyle\frac{24+p}2}-14\right)}p\times6\)
Or, \(19-18=\left(\frac{24+p-28}2\times\frac1p\right)\times6\)
Or, \(1=\frac{3\left(p-4\right)}4\)
Or, p = 3p - 12
Or, 2p = 12
Or, p = 6

Hence the required value of p is 6

Solution:

Here,

In the first case,
Principal (P) = Rs. 150000
Time (T) = 5 years
Simple amount (A) = Rs. 252600
Rate of interest (R) = ?

We know, 

Simple Interest (I) = Simple Amount (A) - Principle (P) 

  = Rs. 252600 -Rs. 150000
  = Rs. 112500

Now, using formula for simple rate (R)

R = \(\frac{I\times100}{P\times T}\)
= \(\frac{112500\times100}{150000\times5}\)
= \(\frac{1125}{75}\)
= 15

∴ Rate of simple interest (R) = 15% p.a. (per annum)

Again, according to the second condition,

Rate of interest (R) = 15% p.a,
Time (T) = 2 years,
Compound amount (CA) = Rs. 198375,
Principal (P) = ?

By formula,

\(CA=p\left(1+\frac R{100}\right)^T\)
Or, \(198375=p\left(1+\frac{15}{100}\right)^2\)
Or, \(198375=p\left(1+0.15\right)^2\)
Or, \(198375=p\left(1.15\right)^2\)
Or, \(198375=1.3225P\)
Or, P = 150000

Hence principal (P) is Rs 150000 is the required answer.

Given:

Total height of the tank = 14 m
Area of the circular base = 38.5 m² 

To find: Per liter cost of water required to fill the tank (T)

By formula, 

Area of the circular base = πr² 
Or, 38.5 = \(\frac{22}7\times r^2\)
Or, \(r^2=\frac{269.5}{22}\)
Or, \(r^2=\left(3.5m\right)^2\)
Or, r = 3.5 m

Hence the radius of cylindrical base is 3.5 m

Here, height of the cylindrical part = total height - radius of the hemispherical part = 14 m - 3.5 m = 10.5 m

Now, by formula,

Volume of the cylindrical part (V₁) = Area of the base x height
= 38.5 x 10.5
= 404.25 m³  

and volume of the hemispherical part

(V₂) = \(\frac23\mathrm{πr}^3\)
= \(\frac23\times\frac{22}7\times\left(3.5m\right)^3\)
= \(\frac{44\times42.875m^3}{21}\)
= 89.8333 m³ 

So, total volume of the tank (V) = V₁ + V₂ = 404.25 m³ + 89.8333 m³ = 494.0833 m³

We know that, 1m³ = 1000 liters

So, 494.0833 m³ = 494083.3 liters

∴ Total cost of water required to fill the tank (T) = per liter cost x quantity of water = 24 paisa x 494083.3 liters = 24/100 x 494083.3 liters = Rs. 118580 ans. 

Here,

\(\sqrt{\frac x{1-x}}+\sqrt{\frac{1-x}x}=\frac{13}6\)
Or, \(\frac{\sqrt x}{\sqrt{1-x}}+\sqrt{\frac{1-x}x}=\frac{13}6\)
Or, \(\frac{\left(\sqrt x^2\right)+\left(\sqrt{1-x}\right)^2}{\sqrt{1-x\;}\sqrt x}=\frac{13}6\)
Or,\(\frac{\displaystyle x+\left(1-x\right)}{\sqrt{x-x^2}}=\frac{13}6\)
Or,\(\frac{\displaystyle x+1-x}{\sqrt{x-x^2}}=\frac{13}6\)
Or,\(\frac{\displaystyle1}{\sqrt{x-x^2}}=\frac{13}6\)
Squaring both sides, we get
\(\left(\frac{\displaystyle1}{\sqrt{x-x^2}}\right)^2=\left(\frac{13}6\right)^2\)
Or,\(\frac1{x-x^2}=\frac{169}{36}\)
Or,36=169(x-x²)
Or,36=169x-169x²
Or,169² - 169x+36=0
Or,169² - (117 - 52)x + 36 = 0
Or,169² - 117x - 52x + 36 = 0
Or,13x(13x-9)-4(13x-9) = 0
Or,(13x-9) (13x-4) = 0

Either, 13x-9=0
Or,13x=9
Or,\(x=\frac9{13}\)


Or, 13x - 4 = 0
Or, 13x=4
Or, \(x=\frac4{13}\)

Solution:

Given:

M is the mid point of AE and ABCD is a parallelogram

To prove: Area of triangle ΔABE = area of parallelogram ABCD

Construction: BM is joined.

Proof:

Hence proved