navbar brand

Model Set

23 Mar 2023, Thu
  • Applications
    Dictionary
    Quiz
    Converter
  • User Profile
    • image profile

      Hello Guest 👋 !

      Login Register
UI
Namaste, Guest Become a member?
  • Login
  • Register
  • Ask Question
  • Quick Access

    • Home
    • Questions
    • Profile
  • Questions

    • Ask Question
    • All Subjects
    • Accountancy
    • Astronomy
    • Biology
    • Chemistry
    • Computer
    • English
    • Environment
    • Geography
    • Geology
    • Health
    • History
    • Math
    • Nepali
    • Physics
    • Population
    • Social Studies
  • Exam Capsule

    • All Entrances
    • IOE Entrance
    • SEE Model Set
  • MCQ

    • GK Practice
    • Categories
    • Exam
    • Submit Quiz
  • Articles

    • Essay Contest
    • Latest
    • Health
    • Education
    • Sports
    • Abroad
    • Politics
    • Environment
    • Technology
  • Institutes

    • All Institutes
    • Schools
    • High Schools
    • Colleges
    • Tuition Centers
    • Library
  • Notices

    • Admission Open
    • Result
    • Loksewa
  • Top Topics

    • Electricity
    • Computer Fundamentals
    • Pressure
    • Heat
    • Waves
Quick Navigation

SEE Compulsory Math Question Paper With Solution
SEE 2077 (2021)

SEE Compulsory Mathematics Model Set – I


: 180 Mins
FM: 100
PM: 40
Group A - [ 3 x ( 1 + 1 ) = 6 ]
1(a)

If initial population of any place is X0 and annual population growth is R%, then what is the population of that place after Y years?


Solution,

Here, initial population (P) = X0

Population growth rate (R) = R% p.a 

Population after Y years (PT) = ?

By using population growth formula,

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

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

1(b)

 Write the area of an isosceles triangle having length of equal sides 'm' cm and third side 'n' cm,


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

   = \(\frac n4\sqrt{4m^2-n^2}cm^2\)

2(a)

What is the value of (9m)0 ?


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

 

2(b)

Write the name of the quartile which divides the continuous data below 25%


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

3(a)

Write down the relation between the area of rectangle and triangle standing on bases MN and between the parallel lines MN and AB.


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.

3(b)

Write the relation between opposite angles of a cyclic quadrilateral.


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° 

Group B - [ 4 x ( 2 + 2 ) + 3 x ( 2 + 2 + 2 ) = 34 ]
4(a)

What is the price of price of a bag costing Rs. 1800 after levying value added tax VAT? Find it.


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

4(b)

The present price of a car is Rs. 2000000. If its price reduces by 10% annually, after how many years it's price will be Rs. 1458000? Find it?


Solution:

Given:

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

5(a)

Find the area of triangle MNO in the given figure


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.

5(b)

If the volume of a hemisphere is 486π cubic cm, then find its radius.


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.

5(c)

In the given figure, the area of rectangular surfaces of the prism is 180 sq. cm, ∠PQR = 90°, PR = 5 cm, and P'Q' = 3cm. Find the length of PP'


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

6(a)

Prove that: \(\frac{4^{a+2}+4^a}{17\times4^a}=1\)


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.

6(b)

Rationalize the denominator of : \(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\)


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

7(a)

Find the LCM of p2 - q2, (p+q)2 


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]

7(b)

Solve: \(\sqrt{w+4}=\sqrt[3]{64}\)


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.

7(c)

If the sum of two consecutive even numbers is 32, find the numbers.


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.

8(a)

In the given figure, WXYZ is a parallelogram. If ∠XAY (angle XAY) = 90°, AY = 8 cm and the area of parallelogram WXYZ is 64 sq. cm, find the length of AX.


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.

8(b)

In the given figure, M is the center of the circle. If ∠GEF = 2x° and ∠EGF = 3x°, find the value of ∠FGH.


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°

8(c)

In the given figure, X is the center of the circle, AB and AC are two tangents to the circle. If angle ∠BAC = 105°, find the value of ∠BAC


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°.

9(a)

In triangle ΔDEF, DE = 12√3 cm, ∠DEF = 60° and the area of ΔDEF is 36 sq cm, find the measurement of EF.


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

9(b)

In a continuous series the average marks of some students is 40 and the sum of their marks is 1200. Then find the number of students.


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

10(a)

If a dice is rolled and coin is tossed at the same time, find the probability of occurring odd number on on dice and head on the coin.


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 \(\frac14\)

10(b)

Two cards drawn randomly from a well shuffled deck of 52 cards in succession without replacement. Show the probabilities of possible outcomes of getting Ace and not getting Ace in tree diagram.


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

Group C - [ 10 x 4 = 40 ]
11

In a survey of people of a community, it was found that 70% liked curd, 60% liked milk, 20% don't like both curd and milk and 550 liked both curd and milk then,

  1. Draw a venn-diagram to illustrate above information
  2. Find total number of people participated in the survey
  3. Find the total number of people who liked curd only.

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. no(C) = ?

Again let, n(C ∩ M) = x%

Representing above information in venn-diagram we get,

venn diagram

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

no(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.

12

A businessman exchanged Nepali currency Rs. 6,60,000 into US dollar at the rate of $1 = NRs. 110. After four days Nepali currency is revaluated by 10% and he exchanged the dollars into Nepali currency again. What is his gain or loss? Find it.


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

13

The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the solid object is 6 cm, length of the cylindrical part is 116 cm and the total length of the solid object is 120 cm. Find the total surface area of the solid.

cylinder and cone

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

14

Find the HCF of: p2 + 4pq + 4q2 , p4 + 8pq3 and 3p4 - 10p2q2 + p3q


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)

15

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


Solution:

Here,

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

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

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

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

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

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

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

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

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

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

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

16

In the given figure, EF//GH, EI//FG and FJ//GI, then prove that:

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

Solution:

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

To prove:

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

Proof:

SN Statements Reasons
1.

In ΔEFJ and ΔHGI

  1. ∠FEJ = ∠GHI
  2. ∠EJF = ∠HIG
  3. JF = IG

 

  1. Corresponding angles, being EF//GH
  2. corresponding angles, being FJ//GI
  3. being opposite sides of parallelogram FGIJ
2. ∴ ΔEFJ ≅ ΔHGI  By A.A.S statement.
3.  Area of ΔEFJ = area of  ΔHGI  Areas of congruent triangles, from statement 2
4. Area of Trapezium EFGI - ΔHGI = Area of Trapezium EFGI - area of ΔEFJ Subtracting equal triangles from Trapezium EFGI.
5. ∴ Area of parallelogram EFGH = Area of parallelogram FGIJ From statement (4), by whole part axiom.

Hence proved.

17

Construct a quadrilateral MNOP in which NO = MN = 4.2 cm, OP = PM = 5.2 cm, and ∠NOP = 75°. Also construct the ΔPQO which is equal in area to the quadrilateral MNOP.


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.

Construction of quadrilateral

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

18

Verify experimentally that the circumference angle BED is half of the central angle BCD standing on the same arc BD of a circle. (Two circles having radii at least 3 cm are necessary).


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.

Figure ∠BED ∠BCD ∠BCD Results
1. 40° 80° 40° ∠BED = 1/2 ∠BCD
2. 63° 126° 63° ∠BED = 1/2 ∠BCD

Conclusion:

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

Hence verified.

19

The heights of house and temple are 13 m and 25 m respectively. If a man observes the roof of a temple he finds the angle of depression 45°, find the distance between the house and the temple.


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 = ?

Height and distance

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

 

20

Find the median from the data given below:

Class Interval 40-50 50-60 60-70 70-80 80-90
Frequency 7 8 6 5 4

Solution:

Here,

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

Class (X) Frequency (f) Cumulative frequency (cf)
40-50 7 7
50-60 8 7+8=15
60-70 6 15+8=21
70-80 5 21+5=26
80-90 4 26+4=30
  N=30  

Using formula,

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

here,

The corresponding class of cf 15 is 50-60

∴ Median class = 50-60

Again, by formula

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

Where,

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

So, Actual median (Md) 

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

∴ Median of the given data is 60.

Group D - [ 4 x 5 = 20 ]
21

A person deposited Rs. 55,000 in bank 'p' for 2 years at 10% compound interest compounded annually. But after one year bank has changed the policy and decided to pay semi-annually, compound interest at the same rate. What is the percentage difference between the compound interest of first and second year? Give reason with calculation.


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.

22

Two pillars of height 8 feet each with four faces shown of the gate of a stadium have one-one pyramid of height 4 feet each having same base on their tops. The base of each pillar is 6 ft x 6 ft. If the pillars with pyramid are painted at the rate of Rs. 75 per square feet, what will be the total cost?


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

23

The area of a rectangular land is 3000 sq meters and perimeter is 220 meters. Out of length or breadth which one is to be decreased by what percent to make it square? Find it.


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%

24

Points K,L,M and N are concyclic such that arc KL = arc NM. If the chords KM and LN are intersected at a point P, then prove that.

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

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:

  Statements Reasons
1.

In triangle ΔKPL and triangle ΔNPM,

  1. ∠LKP = ∠PNM
  2. KL = NM
  3. ∠KLP = ∠PMN

 

  1. Circumference angles of a circle standing on the same arc
  2. Chords subtended by equal arcs of a circle. (by given)
  3. Same as reason number 1.(i)
2. ∴ triangle ΔKPL ≅ ΔNPM By A.S.A statement
3. ∴ Area of ΔKPL = Area of ΔNPM Area of congruent triangles are equal.
4.  arc \(\overset\frown{KL}\) = arc \(\overset\frown{NM}\) Given
5. \(\overset\frown{KL}\) + \(\overset\frown{LM}\) = \(\overset\frown{LM}\) + \(\overset\frown{MN}\) Adding arc MN on both sides of statement (4)
6. (\overset\frown{KLM}\) = (\overset\frown{LMN}\) From statement (5), by whole part axiom.
7. ∴ Chord KM = Chord LN From statement (6), chords subtended by equal arcs in the circle.

 

Hence proved

Popular Tags: 7 Days

  • development
  • political change
  • mediterranean climate region
  • winter rainfall
  • compound interest
  • Advantages of saving
  • egg sinks in pure water but floats in the salt solution
  • Cooperative
  • principles
  • functions of Legislature
  • secular
  • secular country
  • secularism
  • secularism in nepal

Upcoming MCQs

MCQ Online Exam

Computer Fundamental Multiple Choice Questions Exam Free

  • 2022-03-20 12:45
  • 60 Mins
  • 12 Enrolled
  • 25 Full Marks
  • 10 Pass Makrs
  • 25 Questions

Search

Quick Links
  • About Us
  • Terms of Use
  • Privacy Policy
  • Cookie Policy
  • Advertise With Us
  • FAQs
Our Products
  • Educational Institutes
  • Question & Answers Community
  • Online Course/Notes
  • Educational Utilities
  • Social Networking
  • Educational Store
Contact Us
Address:
Pryag Pokhari, Lagankhel, Lalitpur Nepal
Email:
[email protected]
Phones:
9745619281 or +977-9840704915

PathshalaNepal.com is a Registered Company of E. Pathshala Pvt Ltd Nepal. Registration number : 289280

© 2020. All right Reversed.E. Pathshala Pvt Ltd