Mathematics and Science PYQs

Mathematics is a branch of science which deals with counting, calculating, and studying numbers, shapes, and structures. 

Key Points

Note that:

• The main branches of mathematics are algebra, number theory, geometry and arithmetic.
• Based on these branches,  other branches have been discovered.
• Before the advent of the modern age, the study of mathematics was very limited.
• But over a period of time, mathematics has been developed as a vast and diverse topic.
• Development in Math continues making large contributions to the field of technology.
• Hence, it is better termed as the "Queen of Science".
• So, if we solve a geometrical figure we used algebra for variables, arithmetic for arithmetical operation, numbers used for calculation, trigonometry used for angles etc.

Hence, we conclude that different branches of Mathematics, viz., algebra, geometry, etc., are dependent on each other.

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.

Key Points

• Pedagogy refers to the set of principles that influence the approaches to the teaching-learning process. It consists of learning environment, teaching-learning arrangements, methods, general educational principles, etc.

Motivating the learners to develop their existing capabilities is the best pedagogical way to teach Mathematics as it will guide them to achieve a goal and will enable them to:

• develop the skill of systematic reasoning.
• develop mathematical and logical thinking.
• initiate mathematical talk in the classroom.
• guess-and-verify the solutions to problems.
• assimilate a sound understanding of mathematics.
• recognize and employ patterns of mathematical thought.

Hence, it is clear that motivating the learners to develop their existing capabilities is the best pedagogical way to teach Mathematics.

Solution:

Let the income of B be 100x.

∴ income of A = 125x

⇒ decrease in income of B = (125x) - 100x = 25x

⇒ percentage decrease in salary of B = {(25x)/(125x)} × 100 = 20%

∴ B's salary is 20% less than that of A.

Given Data:

Sum of two natural number is 8.

The sum of their reciprocals is 8/15.

Concept Used:

(a + b)² = (a – b)² + 4ab

Calculation:

Let the number be a and b.

a + b = 8  …(i)

1/a + 1/b = 8/15

⇒ (b + a)/ab = 8/15

⇒ 8/ab = 8/15

⇒ ab = 15

(a + b)² = (a – b)² + 4ab

⇒ 8² = (a – b)² + 4 × 15

⇒ 64 = (a – b)² + 60

⇒ (a – b)² = 4

⇒ a – b = 2    ….(ii)

Adding (i) and (ii), 2a = 10

⇒ a = 5 and b = 3

∴ The two natural numbers are 5 and 3.

Shortcut:

Directly we can get the value from option 4, as a + b = 8 and ab = 15.

Given Data:

The difference in expenditure between education and travel Rs. 4,000.

Calculation:

The family spends all the money what they earned. So, 100% of the money is utilised.

The difference in expenditure between education and travel Rs. 4,000.

⇒ 25% of total income – 15% of total income = 10% of total income = 4000

⇒ 1% of total income = 400

⇒ 100% of total income = 40,000

∴ Total income of the family = 40,000.

Given:

Ladies ∶ Gents = 3 ∶ 2

Concept used:

x/y = a/b

⇒ bx = ay

Calculation:

Let, the number of Ladies and Gents in the gathering be 3x and 2x respectively.

According to question,

3x/(2x + 20) = 2/3

⇒ 9x = 2 × (2x + 20)

⇒ 9x = 4x + 40

⇒ 5x = 40

⇒ x = 8

The number of ladies in gathering = 3x = 3 × 8 = 24

∴ The number of ladies was 24.

Given:

\(\frac{{\rm{a}}}{2} = {\rm{;}}\frac{{\rm{b}}}{3} = {\rm{;}}\frac{{\rm{c}}}{4} = \frac{{2{\rm{a;}} - {\rm{;}}3{\rm{b;}} + {\rm{;}}5{\rm{c}}}}{{\rm{k}}}\)

Calculation:

Let, \(\frac{{\rm{a}}}{2} = {\rm{;}}\frac{{\rm{b}}}{3} = {\rm{;}}\frac{{\rm{c}}}{4} = \frac{{2{\rm{a;}} - {\rm{;}}3{\rm{b;}} + {\rm{;}}5{\rm{c}}}}{{\rm{k}}} = {\rm{x}}\)

a/2 = x

⇒ a = 2x       .……(1)

b/3 = x

⇒ b = 3x          .…….(2)

c/4 = x

⇒ c = 4x           ……..(3)

\(\frac{{2{\rm{a;}} - {\rm{;}}3{\rm{b;}} + {\rm{;}}5{\rm{c}}}}{{\rm{k}}} = {\rm{x}}\)

⇒ 2a – 3b + 5c = kx

Using equation (1), (2) and (3)

⇒ (2 × 2x) – (3 × 3x) + (5 × 4x) = kx

⇒ 4x – 9x + 20x = kx

⇒ 15x = kx

⇒ k = 15

∴ The value of k is 15.

Given:

A Δ B = A

Concept:

The Δ in set theory is the symmetric difference of two sets.

A Δ B = (A – B) ∪ (B – A)

Calculation:

A Δ B = A

\( \Rightarrow \left( {{\rm{A}} - {\rm{B}}} \right) \cup \left( {{\rm{B}} - {\rm{A}}} \right) = {\rm{A}}\)

\( \Rightarrow \left( {{\rm{A}} \cap {\rm{B'}}} \right) \cup \left( {{\rm{B}} \cap {\rm{A'}}} \right) = {\rm{A}}\)

\( \Rightarrow \left( {{\rm{A}} \cup {\rm{B}} \cap {\rm{A'}}} \right) \cap \left( {{\rm{B'}} \cup {\rm{B}} \cap {\rm{A'}}} \right) = {\rm{A}}\)

\( \Rightarrow \left[ {\left( {{\rm{A}} \cap {\rm{A'}}} \right) \cup {\rm{B}}} \right] \cap \left[ {\left( {{\rm{B'}} \cup {\rm{B}}} \right) \cap {\rm{A'}}} \right] = {\rm{A}}\)

\( \Rightarrow \left[ {\emptyset \cup {\rm{B}}} \right] \cap \left[ {{\rm{U}} \cap {\rm{A'}}} \right] = {\rm{A}}\)

\( \Rightarrow {\rm{B}} \cap {\rm{A'}} = {\rm{A}}\)

\( \Rightarrow {\rm{B}} - {\rm{A}} = {\rm{A}}\)

\( \Rightarrow {\rm{A}} \cap {\rm{B}} = \emptyset \)

\(\therefore {\bf{A}} \cap {\bf{B}} = \emptyset \)

Concept used:

General quadratic equation is ax² + bx + c 

Then, Sum of the roots = -b/a

Product of the roots = c/a

Formula used:

(a + b)² = a² + b² + 2ab

Calculation:

Let the roots of the equation be α and α + 1,

because as mentioned in the question we have two consecutive integers as the roots.

⇒ α + (α + 1) = b = sum of roots

⇒ 2α + 1 = b = sum of roots

⇒ Product of two roots = α × (α + 1) = c

⇒ Now b² - 4c = (2α + 1)² - 4 (α² + α)

⇒ 4α² + 1 + 4α - 4α² - 4α = 1

Given:

HCF and LCM of two numbers are 13 and 1989 respectively

One of the numbers is 117

Concept used:

First number × Second number = HCF × LCM

Calculation:

Let the second number be m

⇒ 117 × m = 13 × 1989

⇒ m = \(\frac{{13 \times 1989}}{{117}}\)

⇒ m = 221

Given Data:

a, a + 2, a + 4 are prime numbers.

Concept Used:

3, 5, 7 are the only set of three prime numbers which are in arithmetic progression with common difference = 2

Calculation:

a, a + 2, a + 4, these three terms are in Arithmetic progression with common difference = 2.

Comparing a, a + 2, a + 4 with 3, 5, 7, we will get the value of a as 3

⇒ Number of possible values of a is 1 i.e. 3

∴ The number of values of a is 1.

Mistake Points There is a difference between the value of 'a' and the number of values of 'a'

Here the value of a = 3 but the number of values that 'a' can have is 1 because only 1 value satisfies for a i.e 3

Concept Used:

2-digit number × 2-digit number = 4-digit number.

To find the biggest 4-digit number which is a perfect square, we have to take the square of biggest 2-digit number.

Calculation:

2-digit biggest number = 99

99² = 9801

∴ The biggest 4-digit number which is a perfect square = 9801.

Concept Used:

To find the remainder of f(x), we have to put the value of x from x – a = 0

Calculation:

x – 1 = 0 ⇒ x = 1

f(x) = x¹³ + 1

putting x = 1,

f(1) = 1¹³ + 1 = 2

Remainder = 2

∴ When x¹³ + 1 is divided by x - 1, the remainder is 2.

Calculation:

According to the question,

A is a non-empty subset = A ⊆ C

B is a non-empty subset = B ⊆ C

By Distributive property,

 A ∪ (A ∩ B)  = (A ⋃ A) ∩ (A ⋃ B) = A

Given Data:

Rate = r = 5% per annum

Present population = 4410

Calculation:

Let population of village 2 years ago be x

After the first year, the population = x × (100 + 5)/100 = 21x/20

After second year, the population (Present population) = (21x/20) × (100 + 5)/100 = (21x/20) × (21/20) = 441x/400 = 4410(Given)

⇒ x = 4410 × 400/441 = 4000

∴ The population before 2 years = 4000.

 

5% = 1/20

For the first year,              20 → 21

For the Second Year,         20 → 21

                                    ----------------------

                                         400 → 441

441 → 4410    (Given)

⇒ 400 → 4000

∴ The population before 2 years = 4000.

Formula Used:

 average = (sum of all observation)/ (number of observation)

Solution:

⇒ sum of all observation = 10 × 4.5 + 30 × 3.5 = 150

⇒ number of all observation = 40

⇒ average = 150/40

∴ average = 3.75

so, average = 3.75

Given:

side of cube = 14 cm.

∵ cylinder is cut from cube 

∴ r = 14/2 = 7 cm, h = 14 cm

Formula used:

 If side of a cube is a

∴ total surface area = 6 × a²

If the radius of a cylinder is r and height is h -

∴ total surface area = 2 × π × r × (r + h)

Solution:

total surface area of cylinder = 2 × (22/7) × 7 × (7 + 14)

⇒ total surface area of cylinder = 924 cm²

Formula used:

average = (sum of all observation)/(number of observation)

Concept used:

If we add or subtract anything in the given data, it will happen in the average also.

Solution:

there are 50 number which is subtracted from 53 and their mean (average) is - 3.5

∴ mean of given numbers = 53 - (-3.5) = 56.5

Formula Used:

If the total number of observation given is odd, then the formula to calculate the median is:

Median = {(n+1)/2}thterm

If the total number of observation is even, then the median formula is:

Median  = [(n/2)th term + {(n/2)+1}th]/2

Solution:

5, 9, 14, 15, x + 1, 2x - 13, 28, 30, 32, 34

∵ number of observation is even

∴ median = {(x + 1) + (2x - 13)}/2

⇒ 24 = (3x - 12)/2

⇒ 3x = 60

⇒ x = 20

∴ The value of x is 20

Formula Used:

  c1 × d1 = c2 × d2 

where c = number of cow and d = number of days

Given:

c1 = 44, d1 = 9 days, d2 = 12 days, c2 = x

∴ put these values in given formula:

⇒ 44 × 9 = 12 × x

⇒ x = 33

⇒ number of cows eating in 12 days is 33.

⇒ 11 cows less than the previous one