Mathematics and Science PYQs

Concept used:

The distance of a point (x, y) from the y-axis means the distance of the point (x, y) from (0, y).

Formula used:

Distance between two points (a, b) and (c, d) = \( {\sqrt{(a - c)^2 + (b - d)^2}}\) 

Calculation:

The distance of a point (-7, -2) from the y-axis is the distance of the point (-7, -2) from (0, -2).

⇒ Required distance = \( {\sqrt{(-7 - 0)^2 + (-2 + 2)^2}}\) =

⇒ \( {\sqrt{49}}\) = 7 unit

∴ The distance of the point (-7, -2) from the y-axis is 7 units.

Given:

The total surface area of the cylinder = 462 sq. cm.

The curved surface area of the cylinder is one-third of its total surface area.

Formula used:

The curved surface area of cylinder = 2πrh

Total surface area of cylinder = 2πr²+ 2πrh = 2πr(r + h)

Where r and h are the radius and height of the cylinder respectively.

Calculation:

Let the radius of the cylinder be r and height be h.

As the total surface area of cylinder = 462 sq. cm

⇒ 2πr(r + h) = 462      ----(1)

Curved surface area = 1/3 × (total surface area)

⇒ 1/3 × 462 = 154 sq. cm

⇒ Curved surface area = 2πrh = 154      ----(2)

Diving equation (1) by (2), we get

⇒ (r + h)/h = 3

⇒ r + h = 3h

⇒ r = 2h      ----(3)

From equation (2) and (3),

⇒ 2 × (22/7) × 2h × h = 154

⇒ h² = 49/4

⇒ h = 7/2

Putting the value of h in equation (3), we get

⇒ r = 2h = 2 × 7/2 = 7

Now, volume of cylinder = πr²h 

⇒ (22/7) × 7² × 7/2 =

⇒ 11 × 49 =539 sq. cm

∴ The volume of the cylinder is 539 sq. cm

Given:

The dimension of the rectangular piece of paper is 11 cm × 4 cm

The height of the cylinder is 4 cm

Concept used:

When a rectangular piece of paper is folded to make a cylinder, then the length of the rectangular piece of paper becomes the circumference of the base of the cylinder and breadth becomes the height of the cylinder. 

Formula used:

Circumference of circle = 2πr

And, Volume of cylinder = πr²h

Where 'r' is radius and 'h' is height.

Calculation:

The length of the rectangular piece of paper becomes the circumference of the base of the cylinder and breadth becomes the height.

Let the radius of the cylinder be r and height be h.

⇒ h = 4 cm

⇒ Circumference of base of cylinder = 11 cm

⇒ 2πr = 11

⇒ 2 × 22/7 × r = 11

⇒ r = (11 × 7)/(2 × 22) = 7/4 cm

Now, volume of cylinder = πr²h 

⇒ 22/7 × (7/4)² × 4 

⇒ 77/2 = 38.5 cm²

∴ The volume of the cylinder made from rectangular piece of paper is 38.5 cm².

Given:

Equations of the lines are x + 2y - 3 = 0 and 2x + 4y + 11 = 0

Concept used:

 

Calculation:

According to given equations x + 2y - 3 = 0 and 2x + 4y + 11 = 0

Here, a₁ = 1, a₂ = 2, b₁ = 2, b₂ = 4, c₁ = -3 and c₂ = 11

Now, a₁/a₂ = 1/2,

⇒ b₁/b₂ = 2/4 = 1/2

⇒ c₁/c₂ = -3/11

⇒ \((\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}≠ \dfrac{c_1}{c_2})\)

⇒ The concept serial no. 3, shows that given lines are parallel lines.

∴ The graphs of lines x + 2y - 3 = 0 & 2x + 4y + 11 = 0 are parallel to each other.

The National Curriculum Framework 2005 is the fourth National Curriculum Framework published in 2005 by the National Council of Educational Research and Training in India.

• It served as a guideline for syllabus, textbooks, and teaching practices for the schools in India.
• The NCF 2005 has based its policies on previous government reports on education, such as Learning Without Burden and National Policy of Education 1986–1992 and focus group discussion.

Key Points

• The aim of school education is to provide general education – basically five curricular areas the languages, social science, science, mathematics and Health and Physical education.
• The syllabus committees will have to redefine the objectives of teaching each of the subjects as per the guidelines contained in the NCF 2005.
• As attempt has been made in the next chapter to provide guidelines for the selection of content by the respective syllabus committees covering general education up to secondary level.
• The co-scholastic activities have to be provided more space to help the learners acquire the necessary life skills and creative abilities.

Thus, it is concluded that according to the National Curriculum Framework 2005, the sequence of Mathematics is after language.

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.

Important Points

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.
• It starts from examples and reach towards generalizations.
• Example: Square of an odd number is odd and the square of an even number is even.
• Inductive approach proceeds from-
• Particular to general
• Known to unknown
• Simple to complex
• Example to formula

Hence, it could be concluded that in the Inductive method of teaching Mathematics we proceed from Known to unknown.​

Additional Information 

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.

Analysis Method:

• It is a generic process combining the power of the Scientific Method with the use of a formal process to solve any type of problem.
• Analysis means breaking up into components.

All scientifically based problem-solving approaches use the analytical method.

Synthesis Method: 

• The word “synthetic’ is derived from the word ‘synthesis which means to combine.
• In this method, we combine several facts, perform cer­tain mathematical operations, and arrive at the solution.

Concept used:

Adjacent angles are the angles with a common arm(side) and a common vertex. An angle is formed by two rays meeting at a common endpoint.

 In the given figure, ∠1 and ∠2 are adjacent angles.

Calculation:

Adjacent angles have a common vertex and a common arm, but uncommon arms are always opposite in linear pairs.

So, they always do not need to be opposite.

**∴ Statement

(iii) is false**.

Given:

The area of an equilateral triangle is a and height is b.

Formula used:

Area of an equilateral triangle = \({ √{3} \over 4} × (side)^2\)

⇒ H² = 3S², where H and 2S are the height and side of the equilateral triangle respectively.

Area of triangle = \({ 1\over 2} × Base × Height\)

Calculation:

Let the side of an equilateral triangle be '2s' 

⇒ H2 = 3S2

⇒ b² = 3s²

⇒ s = \({b \over √{3} }\)

⇒ 2s = side of the equilateral triangle = \({2b \over √{3} }\)

⇒ Area of equilateral triangle = \({ 1\over 2} × Base × Height\)

⇒ a = \({ 1\over 2} × {2b \over √{3} } × b\)

⇒ a = \({b^2\over √{3}} \)

⇒ \(\frac{b^2}{a}\) = √3

∴ The value of \(\frac{b^2}{a}\) is √3.

 Important Points

Properties of equilateral triangle:

• Number of Sides = 3
• Number of angles = 3
• Each interior angle = 60°
• Each exterior angle = 120°
• Perimeter = 3 times of side-length
• Area = √3/ 4 x (side)²

Given:

The total number of persons at the party is 10.

Formula used:

The formula for the number of handshakes possible with various people say “n” can be solved by a formula \({n × (n - 1) \over 2}\).

This is because that each of the "n" numbers of people can shake hands with "(n - 1)" people and they will not shake their own hand.

The handshake between two people cannot be counted twice.

Calculation:

As the total number of persons in the party is 10.

⇒ The total number of handshakes when each one of the persons shakes had with other exactly once =

⇒ \({10 × (10 - 1) \over 2}\)

⇒ \({10 × 9 \over 2}\)

⇒ 45

∴ The total number of handshakes is 45.

.n×(n−

Given:

Length of the rope = 30 m

Angle traced by the horse with rope = 105°

Formula used:

Circumference of circle = 2πr,  where r is the radius of the circle.

Length of arc =  \({θ\ \over 360°}× 2πr\)

Calculation:

Here, the length of rope is the radius of the circle

⇒ The radius of circle = r = 30 m

⇒ Circumference of circle = 2 × 22/7 × 30

⇒ Distance covered by rope when the horse has traced the angle of 105° = 

⇒ \({105°\over 360°} × 2 × {22\over7} × 30\) =

⇒ 55 m

∴ The horse has gone 55 m far from the initial point if the rope has traced an angle of 105° at the center.

Given:

The probability of husband's selection = \(\frac{1}{7}\)

The probability of wife's selection = \(\frac{1}{5}\)

Formula Used:

P(A) = \( {n(A) \ \over n(S)}\) 

P(A) is the probability of an event 'A'

n(A) is the number of favorable outcomes

n(s) is the number of events in the sample space.

Calculation:

Let A = Event that husband is selected

And, B = Event that wife is selected

Then, P(A) = \(\frac{1}{7}\) and P(B) = \(\frac{1}{5}\)

⇒ The probability of husband is not selected = \(1 - \frac{1}{7} = \frac{7 - 1}{7} = \frac{6}{7}\)

⇒ The probability of wife is not selected = \(1 - \frac{1}{5} = \frac{5 - 1}{5} = \frac{4}{5}\)

Now, Required probability = P[(A and not B) or (B and not A)]

⇒ \(\frac{1}{7}\) × \(\frac{4}{5}\) + \(\frac{1}{5}\) × \(\frac{6}{7}\)

⇒ \(\frac{4}{35} + \frac{6}{35}\) =

⇒ \(\frac{10}{35}\) = \(\frac{2}{7}\)

⇒ The probability of the selection of only one husband or wife is \(\frac{2}{7}\).∩E¯¯¯¯2)+ P(E2∩E¯¯¯¯1)

(E1∩E¯¯¯¯2)+ P(E2∩E¯¯¯¯(E1∩E¯2)+ P(E2∩E¯

1). PE¯¯¯¯2)+P(E2).P(E

P(E1). P(E¯¯¯¯2)+P(E2)P(E1). P(E¯2)+P(E2)

Calculation:

⇒ \(\frac{5^{n+3}-6\times5^{n+1}}{9\times5^n-2^2\times5^n}\) 

⇒ \(\frac{5^{n+1}(5^2-6)}{5^n(9-2^2)}\)

⇒ \(\frac{5^{n+1}(25-6)}{5^n(9-4)}\)

⇒ \(\frac{5^{n+1}(19)}{5^n(5)}\) 

⇒ \(\frac{5^{n+1} × (19)}{5^{n+1}}\)

⇒ 19

∴ \(\frac{5^{n+3}-6\times5^{n+1}}{9\times5^n-2^2\times5^n}\) is 19.

Formula used:

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

Calculation:

⇒ \(\frac{\sqrt{5}}{\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}}\)

⇒ \(\frac{\sqrt{5}}{\sqrt{(\sqrt{5})^2+ (2)^2 +2 × \sqrt{5} × 2} + \sqrt{(\sqrt5)^2 + (2)^2 - 2 × \sqrt5 × 2}}\)

⇒ \(\frac{\sqrt{5}}{\sqrt{(\sqrt{5} + 2)^2} + \sqrt{(\sqrt5 - 2)^2 }}\)

⇒ \(\frac{\sqrt{5}}{{(\sqrt{5} + 2)} + {(\sqrt5 - 2)}}\)

⇒ \(\frac{\sqrt{5}}{{\sqrt{5} + 2} + {\sqrt5 - 2}}\)

⇒ \(\frac{\sqrt{5}}{2{\sqrt5}}\)

⇒ \(\frac{1}{2}\)

∴ The value of \(\frac{\sqrt{5}}{\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}}\) is \(\frac{1}{2}\).

Confusion Points 

Use (a + b)2 identity to solve this Q

here 

Using (a + b)2 = a2 + b2 + 2ab

9 + 4√5 can be written as

(√5)2 + 22 + 2 × √5 × 2 where a = √5  and b = 2

making it in the form of the above identity, we have solved the Q

Given:

The cost price of two books = Rs. 410

Profit on one book is 15% and loss on another book is 10%.

The selling price of both books is equal.

Concept used:

Cost Price: The amount paid to purchase an article or the price at which an article is made is known as its cost price.

Selling Price: The price at which an article is sold is known as its selling price.

Formula Used:

Profit = Selling Price - Cost Price

Discount = Marked Price - Selling Price

Profit % = \({Profit \ \over Cost Price} × 100\)

Loss % = \({Loss \ \over Cost Price} × 100\)

Calculation:

Let the cost price of two books be 100x and 100y respectively.

The total cost price of two books = Rs. 410

⇒ 100x + 100y = 410      ----(1)

As there is the profit of 15% on 1^st book

⇒ The selling price of 1^st book = 100x + 15% of 100x

⇒ The selling price of 1^st book = 100x + 15x = 115x ----(2)

Also, there is a loss of 10% on 2^nd book,

⇒ The selling price of 2^nd book = 100y - 10% of 100y

⇒ The selling price of 2^nd book = 100y - 10y = 90y ----(3)

As A/Q Selling price of both the books is equal.

From equations (2) and (3), we get

⇒ 115x = 90y

⇒ 115x - 90y = 0 ----(4)

Multiplying equation (1) by 9 and (4) by 10 and adding both, we get

⇒ 900x + 1150x = 3690

⇒ 2050x = 3690

⇒ x = 3690/2050 = 1.8

⇒ Cost price of 1^st book = 100x = 100 × 1.8 = Rs. 180

Putting the value of 100x in equation (1), we get

⇒ 180 + 100y = 410

⇒ 100y = 410 - 180 = 230

⇒ Cost price of 2^nd book = 100y = Rs.230

∴ The cost price of both the books are Rs.180 and Rs.230.

Alternate MethodLet the selling price of both the books be 2070x each. (LCM of 115 and 90)

Now, As, in the first book, there is a profit of 15%,

⇒ Cost price of first book = (2070x/115) × 100 = 1800x ----(1)

Now, As, in the second book, there is a loss of 10%,

⇒ Cost price of second book = (2070x/90) × 100 = 2300x ----(2)

As, the total sum of cost price of two books = Rs. 410

From equations (1) and (2),

⇒ 1800x + 2300x = Rs. 410

⇒ 4100x = Rs. 410

⇒ x = 410/4100 = 1/10

From equation (1) and (2),

Cost price of first book = 1800 × 1/10 = Rs. 180

Cost price of second book = 2300 × 1/10 = Rs. 230

∴ The cost price of both the books are Rs.180 and Rs.230.

Calculation:

Here, we, have to compare between \(\left(\frac{1}{2}\right)^{1/2}\) and \(\left(\frac{1}{3}\right)^{1/3}\)

⇒ LCM of 2, and 3 is 6.

⇒ \(\left(\frac{1}{2}\right)^{1/2} = \left(\frac{1}{2}\right)^{3/6} = \left((\frac{1}{2})^{3}\right)^{\frac{1}{6}} = \left(\frac{1}{8}\right)^{\frac{1}{6}} \)

⇒ \(\left(\frac{1}{3}\right)^{1/3} = \left(\frac{1}{3}\right)^{2/6} = \left((\frac{1}{3})^{2}\right)^{\frac{1}{6}} = \left(\frac{1}{9}\right)^{\frac{1}{6}} \)

From (1) and (2), we conclude that, \(\left(\frac{1}{8}\right)^{1/6}>\left(\frac{1}{9}\right)^{1/6}\)

∴ \(\left(\frac{1}{2}\right)^{1/2}>\left(\frac{1}{3}\right)^{1/3}\)

Given:

The total number of observations is 12.

The sixth and seventh observation is 14 and 15 respectively.

Concept used:

The median is the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average.

Formula used:

For an even set of numbers

⇒  Median = [(n/2)th observation + (n/2 +1)th observation] / 2.

Calculation:

Here, A/Q

Total number of observation is 12, which is even.

⇒ n = 12

⇒ (n/2)^th term is 12/2 = 6^th term and

⇒ (n/2 + 1)^th term is (12/2 + 1)^th = 7^th term

⇒ Median = (6^th term + 7^th term)/2

⇒ Median = (14 + 15)/2 = 29/2 = 14.5

∴ Median of 12 observation is 14.5.

Given:

The length of the floor of the room is 15 m 17 cm

And, Breadth of 9 m 2 cm

Concept used:

Number of tiles used = Area of the floor/Area of the tile

1 meter = 100 centimeter

Formula used:

Area of rectangle = Length × Breadth

Calculation:

⇒ Length of the room = 15m 17 cm

⇒ 15m 17 cm = 15 × 100 + 17 cm = 1517 cm

⇒ Breadth of the room = 9 m 2 cm

⇒ 9 m 2 cm = 9 × 100 + 2 cm = 902 cm

Now, the HCF of 1517 and 902 is 41.

⇒ Side of tile = 41 cm

⇒ Area of tile = 41 × 41 m² = 1681 cm²

⇒ Area of floor = 1517 × 902 cm²

⇒ Number of tiles = 1517 × 902/ 1681 = 1368334/1681 = 814

∴ Number of tiles required to pave the floor is 814.

Given:

The speed of the motorboat is 15 km/hr

Upstream distance covered by motor boat is 36 km

Time taken by the motor boat to go and come back to the starting point is 5 hours.

Concept used:

Stream – The moving water in a river is called a stream.

Upstream speed: If the boat is flowing in the opposite direction to the stream, it is called upstream.

In this case, the net speed of the boat is called the upstream speed.

Downstream speed: If the boat is flowing along the direction to the stream, it is called downstream.

In this case, the net speed of the boat is called the downstream speed.

Formula used:

If the speed of a boat in still water is u km/hr. and the speed of the stream is v km/hr.

Then, Speed upstream = (u - v) km/hr. and downstream speed = (u + v) km/hr.

Calculation:

Shortcut Trick

Total time = Total Distance / Total time taken

\(5 =\frac{36}{u+v}+\frac{36}{u-v}\)

u = Speed of boat = 15 km/hr

To avoid solving quadratic equations we can use Hit and trial using option method

Numerator = 36, Let v = 3 (Will use the option which can easily divide 36)

\(5 =\frac{36}{15+3}+\frac{36}{15-3}\)

5 = 5, Hence LHS = RHS

∴ ∴ Speed of stream is 3 km/hr

Detailed Solution:

Let the speed of the stream be v km/hr.

As the speed of motor boat = 15 km/hr.

And, distance covered by the boat is 36 km upstream and 36 km downstream.

⇒ Upstream speed = (15 - v) km/hr.

⇒ Down steam speed = (15 + v) km/hr.

⇒ Speed = Distance/Time

⇒ 36/(15 - v) + 36/(15 + v) = 5

⇒ 36(15 + v) + 36(15 - v) = 5 × (15 + v) × (15 - v)

⇒ 540 + 36v + 540 - 36v = 5 × (225 - v²)

⇒ 1080 = 1125 - 5v²

⇒ 5v² = 1125 - 1080 = 45

⇒ v² = 45/5 = 9

⇒ v = √9 = 3

∴ Speed of stream is 3 km/hr

Given:

The sum of two natural numbers is 8 and the sum of their reciprocal is \(\frac{8}{15}\).

Concept used:

Natural numbers: Natural numbers are the numbers that are used for counting and are a part of real numbers.

The set of natural numbers include only the positive integers.

Reciprocal of a number: The reciprocal of a number is the fraction flipped upside down

Calculation: 

To find x and y if \(\frac{1}{x} + \frac{1}{y} = \frac{8}{15}\)

⇒ \(\frac{x + y}{xy} = \frac{8}{15} \)

From option 1: put x = 3 and y = 5

⇒ \(\frac{3 + 5}{3.5} = \frac{8}{15} \)

​So, LHS = RHS

Hence, x = 3 and y = 5 are the numbers.

Alternate Method Let the two natural numbers be x and y respectively.

Now, A/Q

⇒ x + y = 8      ----(1)

⇒ \(\frac{1}{x} + \frac{1}{y} = \frac{8}{15}\)

⇒ \(\frac{x + y}{xy} = \frac{8}{15} \)      ----(2)

Substituting the value of 'x + y' from equation (1) into (2), we get

⇒ \(\frac{8}{xy} = \frac{8}{15} \)

⇒ \(xy = 15\)      ----(3)

Now, using formula

(x - y)² = (x + y)² - 4xy

⇒ (x - y)² = 8² - 4 × 15

⇒ (x - y)² = 64 - 60 = 4

⇒ x - y = √4 = 2      ----(4)

⇒ Adding equation (1) and (4), we get

⇒ 2x = 10

⇒ x = 10/2 = 5

Substituting the value of x = 5 in equation (1), we get

⇒ 5 + y = 8

⇒ y = 8 - 5 = 3

Numbers are x = 5 and y = 3

∴ Numbers are 5 and 3.

Given:

The ratio of the two numbers is 5 ∶ 6 and their HCF is 12.

Concept used:

The least common multiple of two numbers is the “smallest non-zero common number” which is a multiple of both the numbers.

HCF of two or more numbers is the highest common factor of the given numbers. It is found by multiplying the common prime factors of the given numbers

Calculation:

Let the two numbers be 5x and 6x respectively.

Now, LCM of 5x and 6x = 30x      ----(1)

By using formula

⇒ 5x × 6x = 30x × 12

⇒ 30x² = 360x

⇒ x = 12

⇒ From (1),

⇒ LCM = 30x = 30 × 12 = 360

∴ LCM is 360.