MPTET Varg 2 Math Mock Test - 10 - MPTET MCQ

Given:

The ratio of cost price and selling price of an article is 20 ∶ 21.

Formula Used:

Calculation:

Let the S.P be 21x and C.P be 20x .

⇒ 5

∴The answer is 5 .

Given:

In this figure ∠COE = 2y, ∠BOF = 5y, ∠DOA = 2y

Concept:

One round angel at a point = 360°

When two lines intersect each other then Vertically opposite angles are Equal.

Here, θ₁ + θ₂ + θ₃ + θ₄ = 360^∘

θ1 = θ3 and θ2 = θ4 (Vertically opposite angles)

Solution:

⇒ ∠AOD = ∠COB ⇒ 2y = 2y, (Vertically Opposite Angles)

⇒ ∠AOE = ∠BOF ⇒ 5y = 5y, (Vertically Opposite Angles)

⇒ ∠COE = ∠DOF ⇒ 2y = 2y. (Vertically Opposite Angles)

Angle formed around a point = 360°

⇒ 2y + 5y + 2y + 2y + 5y + 2y = 360^∘

⇒ 18y = 360°

⇒ y = 20°

Given

A (k + 1, 1), B (4, -3) and C (7, -k)

Formula

The area of triangle = 1/2[(x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Solution

Area of ΔABC,

⇒ 6 = 1/2[(k + 1)(−3 + k) + 4(−k − 1) + 7(1 + 3)]

⇒ 12 = [− 3k + k² − 3 + k − 4k − 4 + 28]

⇒12 = [k² − 6k + 21]

⇒ k² - 6k + 21 − 12 = 0

⇒ k² − 6k + 9 = 0

⇒(k − 3)² = 0

⇒ k − 3 = 0

⇒ k = 3

The correct option is 1.

Given:

Selling price = Rs. 450

Loss 10%

Formula Used:

Cost Price = (Selling Price × 100)/(100 - Loss percent)

Selling Price = Cost Price - Loss

Calculation:

According to question,

⇒ Cost Price = 50 x 10

⇒ Cost Price = Rs.500

Alternate Method Let the cost price of the watch be Rs. 100.

So, the loss on the watch is Rs. 10

So, the selling price of the watch is 100 - 10 = Rs. 90

⇒ 90 → Rs. 450

⇒ 100 → (450 × 100) ÷ 90

⇒ Rs. 500

∴ A man loses 10% by selling his watch for Rs. 450. The cost price of the watch is Rs. 500

Given:

AB ⊥ BC

DC ⊥ BC

DE ⊥ AC

Concept:

Alternate Interior Angle: The angles which are formed inside the two parallel lines, when intersected by a transversal, are equal to its alternate pairs. These angles are called alternate interior angles.

AA similarity rule: Two triangles are similar if they have 2 pairs of corresponding angles that are congruent.

Calculation:

In Δ CED and Δ ABC

∠ABC = ∠DEC (Right Angle)

∠BAC = ∠ECD (Alternate Interior Angle)

(AB ∥ CD, So that ∠BAC = ∠ECD)

Δ CED ∼ Δ ABC (AA criterion)

The correct option is 1.

Mathematics is a branch of science that deals with counting, calculating, and studying numbers, shapes, and structures. It is the study of numbers, shapes, quantities, and patterns. It relies on logic and connects learning with children's day-to-day life.

Key Points

Problems in teaching and learning of mathematics:

• Fear and Failure: Most of the students, peers, teachers, parents, etc. have given priority to teaching and learning mathematics at the elementary level although most of them thought that it is a difficult subject. Lack of awareness of objectives is also another cause of fear and failure. Failure to recognize place value leads to failure in four operations in mathematics.
• Disappointing Curriculum: Unattractive and Loaded mathematics curriculum created disappointment among the students.
  • Most of the mathematics curriculum emphasizes procedure, formulas, mathematical facts, and memorization of concepts.
  • The textbook and syllabi on mathematics are rigidly prescribed. The mathematics curriculum is far away from real life.
• Inadequate Learning Materials: For the majority of children in elementary schools textbook in mathematics is the only resource material available to them.
  • Further, most of the textbooks, mathematics textbooks as well, are mostly content loaded and prescriptive. The student finds very little scope for pleasure and fun in learning mathematics from the textbooks.
  • There is hardly any other materials available to the children especially those who are in rural and remote areas.
• Crude Assessment: Most of our mathematics curriculum emphasized on memorization of formulas. Our classroom teaching process is also examination-oriented. In our school, different tests are designed to assess student’s knowledge of procedure and memory of formulas and facts. Questions are set not to expose student’s experiences but to get a fixed answer.
  • For example students free to answer — + — = 8 than 2 + 6 = —? Moreover, similar types of assessment procedures are applied in the formative assessment as well as a summative assessment.
  • This type of crude methods of assessment encourages the perception of mathematics as mechanical computation.
• Inadequate teacher preparation: Teaching and learning of mathematics in the elementary level are purely depends on the preparation of teachers, her own understanding, the preparation of teachers on pedagogic techniques, and the student’s preparedness.
  • As there is an acute dearth of mathematics teachers, it forced other teachers to teach mathematics in the classes in compulsion.
  • They are mostly depending on the textbooks. Most of the teachers also assume that they know all the mathematics required for the elementary level. So there is a lack of teacher preparation in the teaching of mathematics.
• The teaching-learning process: The teaching-learning process in mathematics at the elementary level are not attractive because
  • i) Bookish knowledge in the class creates dissatisfaction,
  • ii) School mathematics learning becomes charmless, dull, uninteresting, and stereotype,
  • iii) Emphasis on rote learning,
  • iv) Emphasis on teaching, not on learning,
  • v) Development of Understanding, Application and skill are ignored.
• Lack of interest: Most of the school children find learning of mathematics difficult and lose their confidence in mathematics.
  • The teaching-learning process in mathematics is not joyful and attractive.
  • Even the students don’t know what benefit they will get after learning mathematics. So students lack their interest and attitude towards mathematics.

Hence, the above statement is about fear and failure.

Concept:

Euclid’s Division Algorithm: If a and b are positive integers such that a = bq + r, then every common divisor of a and b is a common divisor of b and r, and vice-versa.

HCF (Highest Common Factor), is the highest possible number that divides both numbers. (Remainder = 0)

Calculation:

Using Euclid’s Division Algorithm:

Step 1: H.C.F. of 29 and 24

29 = 24 × 1 + 5

5 ≠ 0

Step 2: H.C.F. of 24 and 5

24 = 5 × 4 + 4

24 = 5 × 4 + 1 + 3

1 and 3 ≠ 0

Step 3: H.C.F. of 5 and 1

5 = 1 × 5 + 0

Hence, correct order H.C.F. of 29 and 24 by using Euclid's division algorithm will be (b), (c), (a).

Given:

Length of train is 120 m and it takes 5 seconds to pass a pole.

Concept used:

Distance = Speed × Time

Calculation:

Speed of train = 120/5 = 24 m/s

Distance to be travelled to cross platform = 120 + 180 = 300 m.

⇒ Time taken = 300/24 = 12.5 seconds

∴ If the length of the train is 120 metres, the time taken by it to cross a railway platform 180 metres long, is 12.5 seconds.

PQ is the tower and the length of shadow RQ changes to OQ when sun’s elevation changes from 60° to 30°;

∴ tan30° = PQ/OQ

⇒ OQ = PQ√3 m

∴ tan60° = PQ/QR

⇒ QR = PQ/√3 m

∴ (QO - QR) = 30

⇒ PQ√3 - PQ/√3 = 30

⇒ PQ(3 - 1) = 30√3

⇒ PQ = 15√3

∴ Height of tower = 15√3 meter

Given:

Concept used:

Calculation:

cos (90° + θ) = – sin θ

sec (- θ) = sec θ

tan (180° – θ) = – tan θ

sec (360° – θ) = sec θ

sin (180° + θ) = – sin θ

cot (90° - θ) = tan θ

So,

⇒ -1

∴ The required answer is -1

The teaching of mathematics should be done in the way, in which a student learns the best i.e., following the child-centred approach by engaging students actively in the learning process.

• Rote-learning and drill are the conventional way of teaching mathematics. At present, the teacher focuses more on providing practical exercises and abstract knowledge to students to foster their individualized discovery-oriented learning.

​Important Points

• NCF-2005 is the national curriculum framework published in 2005, by the National Council of Education, Research, and Training (NCERT).
• It is the official document which states that the curriculum must be student-centric and beyond the textbooks. It also emphasized making classroom activities more flexible and related to daily life.

Key Points

According to NCF-2005​

• The main goal of Mathematics education in school is the mathematization of the child’s thought process as mathematics relies on logic, reasoning, problem-solving, creativity, and mathematical way of thinking.
• The narrow aim of school mathematics is to develop 'useful' capabilities, particularly those relating to numeracy–numbers, number operations, measurements, decimals, and percentages.
• The higher aim is to develop the child's resources to think and reason mathematically, to pursue assumptions to their logical conclusion, and to handle abstraction.
• The teacher should provide students more opportunities to experience typical processes of mathematical activity like looking for patterns, making quizzes, puzzles, and proving arguments, etc. to develop logical and quantitative thinking in learners.

Hence, it could be concluded that teaching mathematics should only focus on building qualitative reasoning in learners is not an aim of teaching mathematics.

Given:

Diagonals of rhombus are 10 cm and 8cm respectively.

Formula Used:

Calculation:

We have,

Area of rhombus = 4 × 10

Area of rhombus = 40 cm²

∴ The area of Rhombus with diagonals 10 am and 8 cm are 40 cm².

Fraction understanding is developmental in nature. Fraction experiences should begin as early as first grade. In the Common Core State Standards students partition shapes and refer to the fractional amounts in grades 1 and 2 as “equal shares.” In grade 3, fractions are a major emphasis, with attention to using fraction symbols, exploring unit fractions (fractions with numerator 1), and comparing fractions. Grade 4 focuses on fraction equivalence and begins work on fraction operations. As the students have difficulty in understanding the fractions, the teacher needs to help the students by clarifying how fractions work and their significance.

Key Points

• The students have wrongly pointed out that 1/7 is greater than 1/3.
• To clear their confusion, the teacher needs to draw two equal circles. Ask students to divide one circle into five equal parts and the other into eight equal parts. Ask them part of which circle is bigger.
• This activity will help them visualize the bigger and smaller parts and will clear out their confusion.

Hint

• ​ Asking a student to pick one block from seven given blocks and another student to pick one block from three given blocks. Then asking the other students to comment on which is a bigger fraction is not going to clear any confusion as the size of the bricks is the same.
• Draw two equal circles, each with four equal parts. Shade two parts in one and three parts in the other. Ask the students which is a bigger fraction to test whether they can compare fractions with the equal number of total parts. This activity will not help as the number of parts in which both the circles are divided is the same. it will not clarify the problem students have faced during fractions. They need to understand that when one object is divided into more equal parts. The parts formed are smaller as compared to when it is divided into lesser parts.
• Explain to the students that while comparing unit fractions, the fraction with the bigger number in the denominator is always smaller. This explanation is true but the students may not be able to understand the logic behind the statement as they are unable to visualize it. They may rote memorize the concept but may not be able to understand it.

Thus, from the above-mentioned points, it is clear that Draw two equal circles. Ask students to divide one circle into five equal parts and the other into eight equal parts. Ask them part of which circle is bigger exercise should Ms. Parul conduct to address the misconception.

Concept:

Temperature:

• It is defined as the degree of hotness or coldness of the body.
• Temperature is a measure of the average kinetic energy of the particles in an object.
• When the temperature increases, the motion of these particles also increases.
• It is measured with a thermometer or a calorimeter. In other words, temperature determines the internal energy within a given system.

Heat and Heat flow:

• It is a form of energy that produces a sensation of warmth and flows when there is a temperature difference between two bodies.
• Heat energy can be converted to other forms of energy like kinetic energy, mechanical energy, etc.​
• Heat will flow from the high-temperature region to the low-temperature region until the temperature distribution is the same throughout the body.
• Or, there may be a temperature gradient across an object. Heat will flow so as to equalize the temperature throughout the object.

Explanation:

• When two bodies are in thermal contact, heat flow will occur between them if they differ in temperature.
• Heat will flow from the higher temperature to the lower temperature.

Additional Information
Mass:

• ​It is one of the fundamental quantities in Physics and the most basic property of matter.
• It can define mass as the measure of the amount of matter in a body.
• The SI unit of mass is Kilogram (kg).
• The mass of a body does not change at any time.
• Only for certain extreme cases when a huge amount of energy is given or taken from a body. For example: in a nuclear reaction, tiny amount of matter is converted into a huge amount of energy, this reduces the mass of the substance.

Density:

• It is defined as the mass per unit volume.
• It is inversely proportional to the volume.
• Formula, density = mass/ volume
• The SI unit of density is kg/m³.

Given:

Ramu can do a work in 36 days.

Kanu can do the same work in 32 days.

Concept used:

Efficiency × Number of days = Total work

Calculation:

Now, they work for 12 days

Units of work completed = 17 × 12 = 204 unit

Remaining units of work = (288 - 204) = 84 unit

Fraction of work left to be completed = (84/288) = 7/24

∴ The correct answer is 7/24.

Given :

Time taken by A to complete the work = 12 days

Time taken by both A and B to complete the work = 8 days

Formula used :

Work = Efficiency x Time

Calculation :

Let the total work be 24 units

Efficiency of A = 24/12 = 2 units

Total efficiency of A and B = 24/8 = 3 units

Efficiency of B = 3 - 2 = 1 units

Time taken by B to complete a piece of work = 24/1 = 24 days

∴ The answer is 24 days.

Alternate Method

Time taken by B to complete a piece of work =  = 24 days

Given:

side of cube = 4 cm

Need to build a cube of side 12 cm using cube of 4cm.

Calculation:

Volume occupied by one cube of side 4 cm = 4×4×4 = 64 cm³

Volume of cube side 12cm = = 12×12×12 = 1728 cm³

∴ No. of cubes of side 4cm which would fit in a cube of 12 cm= 1728/64 = 27 cubes.

Given:

point (5, -3) in the line y = 3.

Concept:

point(x, y)

On y-axis equation x = 0

Explanation:

When mirror on the y = 3

In this figure we can see the Reflection of point(5, - 3) in the line y = 3 is (5, 9) as the distance between the point and the line is 6 unit so the mirror image will also be 6 unit above the line. Here the coordinates of x will be the same.

∴ The reflection of point(5, -3) is (5, 9).

Alternate Method

If a point (x, y) is reflected across that line y = a, then, the relation between coordinates of actual point and image point will be

(x, y) → [x, (2a – y)]

So the reflection of the point (5, -3) along y = 3 will be

(x, y) → [5, (2(3) – (-3))]

(x, y) → (5, 9)

Concept used:

Positive integer: Positive integers are the numbers we use to count: 1, 2, 3, 4 and so on.

Euclid division lemma: If we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r < />

Calculation:

According to the above concept,

Positive integer = 143 × k + 28

⇒ 13 × 11 × k + 13 × 2 + 2

⇒ 13(11k + 2) + 2

So, required remainder = 2

∴ The remainder obtained on dividing the same number by 13 is 2.

Given :

To find the Distance between the points A(4,-3) and B(-4,3).

Formula used :

Distance between two points (x,y) and (X,Y) =

Calculation :

Distance

∴ The answer is 10 .

Calculation:

A regular hexagonal prism 8 planes

12 vertices and 18 edges

∴ The correct answer is 8 planes

Given:

Sum = Rs. 8925

Rate = 8%

Concept used:

Installments in case of compound interest is

Here,

P = Principal or sum

x = amount of each installment

r = rate

n = time

For half yearly:

Rate = r/2

Time = 2n

Calculation:

Rate = 8/2 = 4%

Time = 2 × 1 = 2

⇒ x = 4732

So, each installment = Rs. 4732

∴ Amount each instalment is Rs. 4732.

Shortcut Trick

Interest rate = 8% p. a

Interest rate for 6 month = 8/2 = 4%

4% = 1/25

Here, 1275 units = 8925

⇒ 1 unit = 7

⇒ 676 units = 4732

So, each installment = Rs. 4732.

Mathematics is a subject that finds application in every walk of our lives. Knowingly or unknowingly, people use concepts of Mathematics in their daily lives.

Key Points

The objectives of teaching Mathematics at the secondary level are as follows:

• To develop desirable attitudes: The curriculum should be planned to implement educational aims. The curriculum should define what is to be taught, and the knowledge, skills, and attitudes which are to be delivered.
• Intellectual level: The study of mathematics enables us to develop all our intellectual power such as the power of imagination, memorization, observation, thinking, reason, concentration, creativity, etc.
• Social Value: Mathematics is a subject of great importance. Napoleon once remarked, "The progress and the improvement of mathematics are linked with the prosperity of the state."
• Moral Development: Mathematics helps in developing qualities like truthfulness, honesty, punctuality, patience, self-control, self-respect, confidence, justice, etc.
• Artistic Value: For the lover of mathematics there is beauty, art, music, and fitness in the subject.
• Disciplinary Value: A mind trained by mathematics is more disciplined to a mind that is not being trained by mathematics.
• International Value: Mathematics has international value in the sense that it helps create international understanding and brotherhood.

Thus from the above-mentioned points, it is clear that all of the above is true

Given:

A can do a work in 6 days. B takes 12 days. C takes as long as A and B would take working together.

Concept used:

Work done in 1 day = 1 / (Total number of days required to do work)

Calculation:

Work done by A in 1 day = 1/6

Work done by B in 1 day = 1/12

Work done by C in 1 day = 1/6 + 1/12  = 1/4

⇒ Work done by B and C in 1 day = 1/12 + 1/4 = 1/3

∴ B and C together will take 3 days to complete the work.

Given:

P = ₹4800

R = 5%

T = 2.5 years

∴ The amount of simple interest on ₹4800 at the rate 5% per annum in years will be ₹600.

The correct answer is option 3.

Given: The graph shows the import of a commodity X in crore rupees between 1993 and 1996.

Calculation:

⇒ The average import over the years =

Only the year 1997-98 and 1998-99 import more than the average import of all years.

So, Only 2 years, there was more import of X than the average import of X.

Given:

Selling Price = 2/3 of the original price

Loss = 10%

Formula used:

Calculation:

Let the original selling price be Rs 100.

Therefore, the gain percentage at the original price is 35%.

Concept used:

To compare two numbers, we need to look at their digits from left to right and compare them in order. If we find any digit in which the two numbers differ, we can immediately determine which number is greater.

Calculation:

a) 9999 < 100 is false because 9999 is a four-digit number and 100 is a three-digit number.

Therefore, 9999 is greater than 100.

b) 1052 < 1023 is false because The 3rd leftmost digit in 1052 is 5, while the 3rd leftmost digit in 1023 is 2.

5 > 2

Therefore, 1052 is greater than 1023.

c) 958765 > 957865 is true the leftmost digit in both numbers is the same, which is 9.

The second digit in both numbers is also the same, which is 5.

Then, we can compare the third digit, which is 8 in 958765 and 7 in 957865.

Therefore, 958765 is greater than 957865, and the statement c) is true.

∴ The false statements are a and b.

Given :

Height of the cone = 7 cm

The slant height of the cone = 10 cm

Formula used :

The volume of the cone, V = 1/3πr²h

Slant height(l) =

Calculation:

Slant height, l² = r² + h²

⇒ l²= 100 = r² + 7²

⇒ r² = 100 - 49 = 51

The volume of the cone =

⇒ 22 × 17 = 374 cm³

∴ The answer is 374 cm³ .

Given:

The circumference of a circular field = 88 km

Concept used:

Circumference of circle = 2πr

Calculation:

According to the above concept,

⇒ 2πr = 88

⇒ 2 × 22/7 × r = 88

⇒ 44r/7 = 88

⇒ r/2 = 2

⇒ r = 14 km

∴ The required radius is 14 km.