MPTET Varg 2 Math Mock Test - 4 - MPTET MCQ

Formula Used:

We know that tan²θ + 1 = sec²θ

and sec²θ = 1/cos²θ

Calculation:

Using the identity, 1 + tan²θ = sec²θ

⇒ cos²θ (1 + tan²θ) = cos²θ (sec²θ) = cos²θ / cos²θ = 1

Hence, the value of cos²θ (1 + tan²θ) is equal to 1.

Number of mobiles sold on Wednesday and Saturday = 450 + 950 = 1400

Number of mobiles sold on Sunday, Tuesday and Friday = 300 + 600 + 800 = 1700

∴ Required ratio = 1400/1700 = 14 : 17

Shortcut Trick
The average of N numbers is x, If the average of A numbers is a and, the average of the last B numbers is b,

The summation of the remaining numbers = (N × x) -{(A × a) + (B × b)}

Note: N < (a="" +="" />

According to the question,

The summation of the remaining numbers

= (16 × 10) - {(7 × 9) + (7 × 72/7)}

The summation of the remaining numbers = 160 - {63 + 72}

The summation of the remaining numbers = 160 - 135 = 25

Therefore, '25' is the required answer.

Alternate Method

Given:

The average of 16 numbers = 10

The average of 7 numbers = 9

The average of the last 7 numbers = 72/7

Formula:

Summation of n numbers = Average of n numbers × n

Calculation:

The summation of 16 numbers = 16 × 10 = 160

The summation of first 7 numbers = 7 × 9 = 63

The summation of the last 7 numbers = 7 × 72/7 = 72

The summation of first 7 and last 7 numbers = 63 + 72 = 135

The summation of 8th and 9th number = 160 - 135 = 25

Therefore, '25' is the required answer.

Concept used:

Median = I +[(N/2 − C)/f] × h

Here, l = lower limit of the median class

C = C.F. of the class preceding the median class

h = higher limit - lower limit

f = frequency of median class

Calculation:

N = 50

⇒ N/2 = 50/2 = 25

As 29 is just greater than 25, therefore the median class is 145 - 150.

The lower limit of median class = 145

C = 11

h = 150 − 145 = 5

f = 18

Median = 145 + [(25 − 11)/18]5

⇒ 145 + 3.89

⇒ 148.89

∴ Median is 148.89

Total sales = 300 + 200 + 600 + 450 + 600 + 800 + 950 = 3900

Given:

Mean = 20

Median = 22

Formula Used:

Mode = (3 × Median) - (2 × Mean)

Calculation:

Mode = (3 × Median) - (2 × Mean)

⇒ Mode = (3 × 22) - (2 × 20)

⇒ Mode = 66 - 40 = 26

∴ The value of mode is 26.

The correct option is 2 i.e. 26

Sales on Tuesday = 600

Sales on Wednesday = 450

∴ Required percentage = 450/600 × 100 = 75%

Given:

Amount = Rs. 8800

Time = 4 years

Rate = 25%

Formula Used:

Amount = Principal + SI

P = principal

R = Rate of interest

T = Time

Calculation:

⇒ 8800 - P = P

⇒ 8800 = 2P

⇒ P = Rs. 4400.

Given:

The surface area of a cubic block = 7.26 m2

Formula used:

The surface area of a cube = 6a²

The volume of a cube = a³

Calculation:

The surface area of a cube = 6a2

⇒ 6a2 = 7.26 m2

⇒ a² = 1.21 m²

⇒ a = 1.1 m

Now, the Volume of the cube = 1.1 × 1.1 × 1.1 m³ = 1.331m3

∴ The correct answer is 1.331m3

Given:

A 6 cm cube is cut into 2 cm cubes.

Formula Used:

Surface are of cube = 6(side)2 cm2 .

Calculation:

Vol. of 6cm cube = 6³ cm³ = 216 cm³ .

Vol. of 2cm cube = 2³ cm³ = 8 cm³ .

Total number of 2 cm cube =

Surface are of 2cm cube = 6(side)² cm² = 6(2)² = 24 cm² .

Total surface are of all small cubes = 27 × 24 = 648 cm² .

Formula used:

The volume of cuboid = L × B × H

Total surface area of cuboid = 2(LB + BH + HL)

Calculation:

L × B × H = 1536

Let the length breadth and height of a cuboid are in the ratio 4x: 3x: 2x

∴ 24x³ = 1536

1536/24 = 64 = x³ or x = 4

Total surface area of cuboid = 2 × (12x² + 6x² + 8x²) = 52x²

∴ Total surface area of cuboid = 52 × 16 = 832 cm²

Given:

Radius of hemisphere = 6.3 cm

Formula used :

Volume of hemisphere = 

Calculation:

According to the formula

⇒ Volume of hemisphere =

⇒ Volume of hemisphere = 523.90 cm³

⇒ Hence, The Volume of the hemisphere is 523.90 cm³

Given:

Two vessel contain milk and water in the ratio 7 ∶ 8 and 13 ∶ 5.

Calculation:

Let 90x liter of each solution is taken [As LCM of 15, 18 is 90]

So, quantity of milk from 1st mixture = 90x × (7/15)

⇒ 42x

Quantity of milk from 2nd mixture = 90x × (13/18)

⇒ 65x

Quantity of water from 1st mixture = 90x × (8/15)

⇒ 48x

Quantity of milk from 2nd mixture = 90x × (5/18)

⇒ 25x

Total quantity of milk = 42x + 65x

⇒ 107x

Total quantity of water = 48x + 25x

⇒ 73x

Ratio = 107x : 73

⇒ 107 : 73

∴ Thre required ratio is 107 : 73.

Given:

Radius

r₁ : r₂ = 2 : 3

height

h₁ : h₂ = 5 : 4

Formula Used:

The curved surface area of the cylinder = 2πrh

The volume of a cylinder = πr²h

Calculation:

Let a be multiple of radius

and b be multiple of heights

r₁ : r₂ = 2a : 3a and h₁ : h₂ = 5b : 4b

The ratio of curved surface area = 2πr₁h₁: 2πr₂h₂

⇒ The ratio of curved surface area = (2a)(5b) : (3a)(4b) = 5 : 6

The ratio of volume = ​πr₁²h1 : ​πr2²h₂

⇒ The ratio of volume = (2a)²(5b) :(3a)2(4b) = 5 : 9

∴ The ratio of their curved surface areas and volumes are 5 : 6, and 5 : 9.

Calculation:

⇒ 42 - 25 = 17

⇒ 72 - 55 = 17

⇒ 84 - 67 = 17

Now, LCM (42, 72, 84) = 504

Hence, required number = 504 - 17 = 487

∴ The correct answer is 487.

Concept used:

Divisibility rule of 9:

Sum of all digits in a number is divisible by 9, then the number is also divisible by 9

Divisibility rule of 11:

Difference between the sum of even place and sum of odd places digit is divisible by 11 or 0, then the number is divisible by 11

Calculation:

If the number is divisible by 99, then the number laso divisible by 11 and 9 both

Option 1:

⇒ 114345,

Sum of all digits = 1 + 1 + 4 + 3 + 4 + 5 = 18

Which is divisible by 9

Sum of odd place - Sum of even places digits = (1 + 4 + 4) - (1 + 3 + 5) = 9 = 9 = 0

∴ The number 114345 is divisible by 99.

Option 2:

⇒ 135792,

Sum of all digits = 1 + 3 + 5 + 7 + 9 + 2 = 27

∴ The number is divisible by 9.

Sum of odd place - Sum of even places digits = (1 + 5 + 9) - (3 + 7 + 2) = 15 - 12 = 3

∴ The number is not divisible by 99.

Option 3:

⇒ 3572404,

Sum of all digits = 3 + 5 + 7 + 2 + 4 + 0 + 4 = 25

Which is not divisible by 9,

∴ The number is not divisible by 99.

Option 4:

⇒ 913464,

Sum of all digits = 9 + 1 + 3 + 4 + 6 + 4 = 27

The umber is divisible by 9.

Sum of odd place - Sum of even places digits = (9 + 3 + 6) - (1 + 4 + 4) = 18 - 9 = 9

which is not divisible by 11,

∴ The number is not divisible by 99.

∴ The number 114345 is divisible by 99.

The annual payment that will discharge a debt of Rs. ‘d’ in time ‘t’ years at r% p. a. is given by

Given, debt of Rs. 8480 in 8 yrs at 8% per annum.

⇒ A = Rs. 828.125

Given:

Numbers 24, 30 and 84.

Calculation:

To find the smallest number we have to find LCM of the given three numbers.

The prime factors of the numbers are

24 = 2 × 2 × 2 × 3

30 = 2 × 3 × 5

84 = 2 × 2 × 3 × 7

∴ The LCM of (24, 30, 84) = 2 × 2 × 2 × 3 × 5 × 7 = 840.

∴ The smallest number divisible by all the three numbers 24, 30, and 84 is 840.

The correct answer is Gallium.

Key Points

• The chemical element gallium has the atomic number 31 and the symbol Ga.
• Group 13 of the periodic table contains gallium.
• Gallium is an element that is soft and silvery at ordinary pressure and temperature.
• It turns silvery white when it's liquid.
• Gallium may fracture conchoidally if excessive force is used.
• Gallium has been commonly utilized to create alloys with low melting points.
• As a dopant on semiconductor substrates, it is also utilized in semiconductors.
• The temperature reference point is gallium's melting point.
• Most people are familiar with gallium and cesium as metals that melt when held in the palm.
• Because gallium has a relatively low melting point and is quickly melted by the heat from our palms, it essentially dissolves when held in the palm.

Additional Information

• Germanium:
  • The chemical element germanium has the atomic number 32 and the symbol Ge.
  • It resembles silicon in appearance and is hard, brittle, and glossy.
• Indium:
  • With the atomic number 49 and the symbol In, indium is a chemical element.
  • The softest metal that is not an alkali metal is indium.
• Copper:
  • The chemical element copper has the atomic number 29 and the symbol Cu.
  • It is an extremely high thermal and electrical conductivity metal that is soft, malleable, and ductile.

The correct answer is All of the above.

Key Points

• All of the noble gases are included in the seventeen elements that are categorized as non-metals.
• Iodine, carbon, phosphorous, sulfur, and selenium are the other four non-metals that are solid at ambient temperature.
• A liquid is one non-metal.
• At ambient temperature, the remaining 11 non-metals exist as gases.

Additional Information

• The physical and chemical properties of two different classes of substances—metals and nonmetals—can be used to distinguish between them.
• Metals:
  • Metals are substances that develop naturally beneath the Earth's surface.
  • Because there is a strong metallic bond between the atoms, metals are typically hard elements.
  • Elements with one, two, or three electrons in their valence shell are referred to as metals.
• Nonmetals:
  • Non-metals are substances that do not possess the properties of metals.
  • On the other hand, non-metals frequently have soft components.
  • Elements with 4, 5, 6, or 7 electrons in their outermost shell are referred to be non-metals.

The correct answer is material, length, and thickness.

Key Points

• Whenever a current is flowing in the resistance then heat is produced in it.
• The phenomenon of producing heat by the electric current is called the heating effect of electric current.
• The amount of heat produced (H) in joules = I2 R t
  • Where I is current, R is resistance and t is the time taken
• The heating effect of an electric current depends on three factors:

1. The resistance R of the conductor i.e., a higher resistance produces more heat.
2. The time t for which current flows i.e., the longer the time the larger the amount of heat produced.
3. The amount of current I i.e., the higher the current the larger the amount of heat generated.

Resistance:

• ​It is the ability of the material to oppose the flow of current through the conductor.
• It depends on the following factors:
  • Length.
  • Area.
  • Temperature.
  • Type of material.
• R = p L / A.
  • Where p = resistivity of the material.
  • L = length of the conductor.
  • A = cross-sectional area of the conductor.
• SI Unit = Ohm.
• In the case of a conductor, resistance increases on heating.

Hence, heat produced in the conductor depends on material, length and thickness.

Mathematics is the study of numbers, shape, quantity, and patterns. The nature of mathematics is logical and it relies on logic and connects learning with learners' day-to-day life.

• Teaching methods of mathematics include problem-solving, lecture, inductive, deductive, analytic, synthetic, heuristic and discovery method. Teacher adopts any method according to the needs and interests of students.

Key Points

Anisha is using deductive logic.

Deductive Method: The deductive method is based on deduction. At first, the rules are given and then students are asked to apply these rules to solve more problems.

• It begins with general premises and through logical argument, comes to a specific conclusion.
• For example, while teaching mathematics, the teacher introduces a theory and explains the rules of the theory and the formula, and the students are asked to solve problems using the given formula.

• Deductive approach proceeds from-

  • General to specific
  • Unknown to know
  • Complex to simple
  • Formula to example

In the above example Anisha proves it by taking the case n i.e, she uses the method from unknown to known.

Thus it is clear that Anisha is using deductive logic.

Additional Information

Inductive Method: Inductive approach is based on the process of induction. It is a method of constructing a formula with the help of a sufficient number of concrete examples.

• Induction means to provide a universal truth by showing, that if it is true for a particular case.
• Inductive approach proceeds from-
  • Particular to general
  • Known to unknown
  • Simple to complex
  • Example to formula

Given:

ΔABC and ΔDEF are two triangles such that ΔABC ≅ ΔFDE.

AB = 5 cm, ∠B = 40° and ∠A = 80°

Concept used:

Congruency of triangle = if a triangle is congruent to another triangle then angle or side must be equal to the same triangle

Sum of angles of triangle = 180°

Calculation:

According to the question,

ΔABC ≅ ΔFDE then,

⇒ AB = FD

⇒ BC = DE

⇒ CA = EF

⇒ ∠A = ∠F = 80°

⇒ ∠B = ∠D = 40°

⇒ ∠C = ∠E

Through the Sum of angles,

180° = ∠A + ∠B + ∠C

⇒ 180° = 80° + 40° + ∠C

⇒ 180° = 120° + ∠C

⇒ 180° - 120° = ∠C

⇒ ∠C = 60°

Since ∠C = ∠E then ∠E = 60°

∴ The correct option is DF = 5 cm, ∠E = 60°.

Given:

A triangle ABC in which DE // BC

Concept Used:

Sum of the angles in a triangle is equal to 180 degrees

If there are two parallel lines then corresponding angles are equal

Corresponding angles : The angles that are formed when two parallel lines are intersected by the transversal

Calculation:

Here, we have a triangle ABC in which DE // BC

So, the ∠ACB = ∠AED = 40° (corresponding angles as DE//BC and AC is a transversal line which intersects them)

Now, consider the triangle ADE as we know that

Sum of the angles of a triangle = 180

⇒ ∠ADE +∠ DAE + ∠AED = 180

⇒ ∠ADE + ∠DAE + 40 = 180

⇒ ∠ADE + ∠DAE = 140^o

Hence, the required angle is 140 degree .

Concept:

Power:

The rate of work done is called power (P).

The SI unit of power is the watt (W) or J/s.

Calculation:

Given:

Case 1: Power of old man

Wok done by old man W_O = 1000 J, Time is taken by old man t_O = 50 s,

Power of old man

Case 2: Power of young boy

Wok done by old man WY = 2000 J, Time is taken by old man tY = 80 s,

Power of old man

The ratio of the power of old man P_O to young boy P_Y is

⇒ P_O : P_Y = 20 : 25 = 4 : 5

Mathematics is a science that involves dealing with numbers, different kinds of calculations, measurement of shapes and structures, organization and interpretation of data, establishing relationships among variables, etc.

Key Points

• Rote memorization and rote learning in mathematics can help learners in getting good marks on exams but they won't be able to perform the computational process of mathematics well and also the learners will not be able to use mathematical knowledge in practical life.
• Solving the problem, and motivating the students towards learning.
• Presenting the problem and solving the problem with different methods with the help of teachers, and friends is also a joyful activity for the learners.

​So in this way, we can say that Rote Memorisation is not key to a joyful mathematics experience.

Mathematics education must have to be affordable to every child and at the same time, it needs to be enjoyable so that children can enjoy learning mathematics instead of fearing away from it.

• The mathematical learning makes a learner capable of handling abstractions by enhancing their logical thinking so that they can use problem-solving approaches to reach the solutions.

Key Points

Appropriate strategy for enhancing mathematical learning:

• Mathematical education aims to provide the ability to mathematize the thought process of children i.e., use of mathematical knowledge in everyday life.
• It makes students capable of using and understanding mathematical terminology in their real-life such as the use of words "profit", "discount", and "interest", etc.
• The students begin to understand and use these mathematical terminologies in communicating the thought processes i.e., reasoning, thinking, perception, memory, etc. which develops their mathematical thinking that further enhances their learning.
• Mathematical games such as puzzles and sudoku that involve numbers and shapes help the learners to internalize the fundamentals of mathematical concepts which facilitates and enhances their learning.
• In mathematics, the children learn various number operations such as the concept of average, calculating temperature, and recording the weather that includes several processes like exploring, sharing, narrating, collecting, and recording information pertaining to numbers and number operations.
• All these processes will make the acquisition of knowledge an interesting experience that will also enhance their learning.

Hint

• In the beginning, the teacher must introduce the mathematical concepts in informal ways that must be related to the real world to stimulate the interest of learners.
• If the teacher will introduce the list of formulae required for solving the given set of problems in the beginning then the children will not able to develop the clarity of concepts and can get confused. So, the formulas should be always introduce at a later level of mathematical learning.

Hence, it could be concluded that required formulae for solving the given set of problems should be provided at the beginning is the least appropriate strategy for enhancing mathematical learning.

According to National Curriculum Framework-2005, the main goal of Mathematics education in school is the mathematization of the child’s thought process.

• Mathematics relies on logic, reasoning, problem-solving, creativity, and mathematical way of thinking.
• The learner should construct knowledge by active participation rather than acquiring knowledge by observing lectures delivered by teachers.

Objectives at grade 3:

• Creates shapes through paper folding, paper cutting.
• Identifies 2-D shapes
• Describes the various 2-D shapes by counting their sides, corners, and diagonals.
• Makes shapes on the dot-grid using straight lines and curves.
• Creates shapes using tangram pieces.
• Matches the properties of two 2-D shapes by observing their sides and corners (vertices).
• Tiles a given region using a tile of a given shape.
• Distinguishes between shapes that tile and that do not tile.
• The intuitive idea of a map. Reads simple maps (not necessarily scaled)
• Draws some 3D objects.

Objectives at grade 4:

• Draws a circle freehand and with a compass.
• Identifies center, radius, and diameter of a circle.
• Uses Tangrams to create different shapes.
• Tiles geometrical shapes: using one or two shapes.
• Chooses a tile among a given number of tiles that can tile a given region both intuitively and experimentally.
• Explores intuitively the area and perimeter of simple shapes.
• Makes 4-faced, 5-faced, and 6- faced cubes from given nets specially designed for the same.
• Explores intuitively the reflections through inkblots, paper cutting, and paper folding.
• Reads and draws 3-D objects, making use of the familiarity with the conventions used in this.
• Draws intuitively the plan, elevation and side view of simple objects.

​Objectives at grade 5:

• Gets the feel of perspective while drawing a 3-D object in 2-D.
• Gets the feel of an angle through observation and paper folding.
• Identifies right angles in the environment.
• Classifies angles into right, acute, and obtuse angles.
• Represents right angle, acute angle, and obtuse angle by drawing and tracing.
• Explores intuitively rotations and reflections of familiar 2-D shapes.
• Explores intuitively symmetry in familiar 3-D shapes.
• Makes the shapes of cubes, cylinders and cones using nets specially designed for this purpose.

​Hence, it can be concluded that "Create shapes using tangram pieces" is the most suitable objective for teaching grade 3 children the concept of 3-D and 2-D Shapes.

Formula used:

Time allotted = D × 60

where "D" is the time difference in hours.

Calculation:

D = 08:15 PM - 07:00 PM = 1 hr 15 min or  hours

Time allotted =  × 60 = 5/4 × 60 = 75 min

∴ The time allotted is 75 minutes.

Given:

The number of days taken by A to complete the job = 26

The efficiency of B is 30% more than A.

Concept used,

If someone completes work in x days

Work done in 1 day = 1/x part of the work

Calculation

B is 30% more efficient than A

The ratio of number of days taken by A to B = 130 ∶ 100 = 13∶10

Let B takes X number of days.

⇒ 26 ∶ X ∶ ∶ 13 ∶ 10

⇒ 13X = 260

⇒ X = 260/13

⇒ X = 20

∴ The number of days taken by B alone = 20 days

Important Points

Working efficiency is the work done by an individual in one day and this efficiency is inversely proportional to the number of days to complete a work.

It means that a person who takes fewer days to complete a work is said to be more efficient than a person who takes more days to complete the same work.

Hence we can reject options 1 and 4 as they are greater than 26 days.