Mathematics and Science PYQs

Shortcut Trick

For a unique solution in a system of linear equations, the lines must be intersecting.

Logic: The ratio of the coefficients of x and y must not be equal.

Condition: a₁/a₂ ≠ b₁/b₂

1/3 ≠ −K/2

K ≠ −2/3

∴ The correct answer is K ≠ −2/3.

Alternate Method

Given:

Equation 1: x − Ky = 2

Equation 2: 3x + 2y = 5

Formula Used:

For a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the condition for a unique solution is:

a₁/a₂ ≠ b₁/b₂

Calculations:

⇒ From the given equations, we identify the coefficients:

⇒ a₁ = 1, b₁ = −K

⇒ a₂ = 3, b₂ = 2

⇒ Applying the condition: 1/3 ≠ −K/2

⇒ Cross-multiplying the values: 1 × 2 ≠ 3 × (−K)

⇒ 2 ≠ −3K

⇒ K ≠ −2/3

∴ The correct answer is K ≠ −2/3.

Additional Information

Condition for Infinite Solutions

A system has infinitely many solutions (coincident lines) if a₁/a₂ = b₁/b₂ = c₁/c₂.

Condition for No Solution

A system has no solution (parallel lines) if a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Consistency of Equations

A system is called consistent if it has at least one solution (unique or infinite) and inconsistent if it has no solution.

Given:

Time taken by A to complete the work = 6 days

Time taken by B to complete the work = 8 days

Formula used:

Time = Total work/Efficiency

Calculation:

A's 1 day work = 1/6

B's 1 day work = 1/8

C's 1 day work = 1/6 + 1/8 = 7/24

Work done by B and C in 1 day = 1/8 + 7/24 = 10/24

Thus, time taken by B and C to complete the work together = 24/10 = 12/5 days

∴ Time taken by B and C to complete the work together = \(2 \frac 2 5\) days

Alternate Method

The efficiency of work by C = (4 + 3) = 7 units in 1 day

Amounts of work is done by B and C together in 1 day = 3 + 7 = 10 units

Time is taken to do 24 units of work = 24/10 = \(2 \frac 2 5\) days

∴ Time is taken by B and C to complete the work together = \(2 \frac 2 5\) days

GIVEN:

Number of students = 30

Mean = 58.5

FORMULAE USED:

Mean = Total marks/Number of students

CALCULATION:

Total marks = 30 × 58.5 = 1755

New Total marks after correction = 1755 – 57 + 75 = 1773

New Mean = 1773/30 = 59.1

Shortcut Trick

Increase in marks after correction = 75 – 57 = 18

Mean = 18/30 = 3/5 = 0.6

New mean = 58.5 + 0.6 = 59.1

∴ The corrected mean of marks scored by the student is 59.1

GIVEN:

Points = (3, 5), (6, y) and (-3, 4)

CONCEPT:

Considering all three points as the vertices of the triangle.

The centroid or geometric center of a plane figure is the arithmetic mean position of all the vertices of the triangle

FORMULAE USED:

Centroid = [(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3]

CALCULATION:

x = x1 + x2 + x3/3 = 3 + 6 + (-3)/3 = 2

y = y1 + y2 + y3/3 = 5 + y + 4/3 

3y = 9 + y

y = 9/2

∴ x = 2, y = 9/2

Given:

The given polynomial = ax3 + bx2 + x - 6

Concept used:

Factors satisfy the polynomial 

Calculation:

Let f(x) = ax3 + bx2 + x - 6

According to question

f(-2) = 0

f(2) = a(-2)³ + b(-2)² - 2 - 6 = 0

⇒ -8a + 4b - 8 = 0   

⇒ 2a - b + 2 = 0      ----(1)

f(2) = 4

a(2)³ + b(2)² + 2 - 6 = 4

⇒ 8a + 4b - 4 = 4

⇒ 2a + b - 2 = 0      ----(2)

Solving (1) and (2) simultaneously

∴ a = 0, b = 2

Concept:

Non-terminating but recurring/repeating decimal is defined as a decimal fraction that never comes to an end, but repeats one or more numbers after the decimal point .

10/3 = 3.33333333… etc.

10/3 never terminate, but keep on repeating their patterns consisting of 6 numbers or just one respectively. Thus they are non-terminating repeating decimals. 

∴ A non-terminating but recurring decimal is always an rational number

Given:

PT = 5 cm

PA = 3 cm

Formula used

PT² = PA × PB

Here, PT is tangent

Calculation:

Let AB be "x"

PT² = PA × PB

5² = 3(3 + x)

25 = 9 + 3x

3x = 16

∴ x = 16/3

Given:

ABCD is parallelogram

AC, BD diagonals intersecting at “O”

∠DAC = 32°, ∠AOB = 70°

Concept used:

The alternate angle is equal

Sum of two interior opposite angles = Exterior angle

Calculation:

AB is parallel to DC, AC is transversal.

∠DAC = ∠BCA = 32° (alternate angle theorem)

In ΔBCO, CO is extended to A.

Sum of interior opposite angles = exterior angle at O

⇒ ∠BCA + ∠OBC = ∠AOB

⇒ ∠BCA + ∠DBC = ∠AOB

⇒ 32° + ∠DBC = 70°

⇒ ∠DBC = 70° – 32°

⇒ ∠DBC = 38°                

∴ The value of ∠DBC is 38°

Given:

\(a^{m^{n}}=(a^m)^n\)

Concept:

(a^b)^c = a^bc

If ab = a^c then b = c

log ab = b log a

Calculation:

\(a^{m^{n}}=(a^m)^n\)

∵ Bases are same

⇒ mⁿ = mn

⇒ n = (mn)/m

⇒ n = m^{(n – 1)}

Taking log on both sides,

log n = (n - 1)log m

⇒ log m = {1/(n-1)} log n

⇒ log m = log n^{{1/(n - 1)}}

Comparing both sides,

\(m = n^{\frac {1}{(n-1)}}\)

∴ The value of m is n{1/(n - 1)}.

Given:

 \(\left(1-\frac 1 5\right)\left(1-\frac 1 6\right)\left(1-\frac 1 7\right)...\left(1-\frac 1 {100}\right)\)

Concept used:

This type of questions should be arranged in such a way that calculation becomes easy 

Calculation:

\(\left(1-\frac 1 5\right)\left(1-\frac 1 6\right)\left(1-\frac 1 7\right)...\left(1-\frac 1 {100}\right)\)

⇒ \(\left({5 - 1\over 5}\right)\left({6 - 1\over 6}\right)\left({7 - 1\over 7}\right)...\left({100 - 1\over 100}\right)\)

⇒ \(\left({4\over 5}\right)\left({5\over 6}\right)\left({6\over 7}\right)...\left({99\over 100}\right)\)

All the numbers of the given series will be canceled by themselves except 4 and 100 so,

⇒ {4/100}

⇒ 1/25

∴ The required value is 1/25

Concept:

If a number lies between 0 to 1

On increase the power its value decrease.

Calculation:

(0.9)² = 0.81

(0.9)³ = 0.729

And when the number greater than one with increase in power its value also increase.

(1.1)² = 1.21

(1.1)³ = 1.331

Given:

Rate = 8%

Simple interest = (2/5) of principal

Formula used:

Simple interest = (Principal × Time × Rate)/100

Calculation:

Let the Principal be 5

Rate% = 8%

According to question

Simple interest = 2/5 of principal

⇒ S.I. = 2/5 × 5 = 2

∴ S.I. = 2

⇒ 2 = (5 × 8 × time)/100

⇒ Time = (2 × 100)/ (5× 8)

⇒ Time = 5 years

∴ The required time is 5 years

Given:

The given number = 33275

Concept used:

Prime factorization is the process of finding the prime numbers, which are multiplied together to get the original number. For example, the prime factors of 16 are 2 × 2 × 2 × 2. This can also be written as 2⁴

Calculation:

Prime factorization of 33275 = 5 × 5 × 11 × 11 × 11

⇒ 5² × 11³

From the above prime factorization, we can say if we multiply the number by 5, it will become a perfect cube

∴ The smallest number multiplied to make it perfect cube is 5

Calculation:

As (p - a) is HCF of the given expression i.e. (p - a) must be a factor of both the expression, putting p = a in the given first expression we get

⇒ a² - 8a + 12 = 0

⇒ a² - 6a - 2a + 12 = 0

⇒ a(a - 6) -2(a - 6) = 0

⇒ (a - 6)(a - 2) = 0

⇒ a = 6, 2

putting p = a in the given second expression we get

⇒ a² + 4a - 12 = 0

⇒ a² + 6a - 2a - 12 = 0

⇒ a(a + 6) -2(a + 6) = 0

⇒ (a + 6)(a - 2) = 0

⇒ a = -6, 2

since value of a = 2 is common in both the expression

∴ required value of a = 2.

Additional Information

HCF: The greatest common divisor of two or more positive integers happens to be the largest positive integer that divides the numbers without leaving a remainder.

LCM: Least common multiple or lowest common multiple, or smallest common multiple of two integers a and b, usually denoted by lcm(a, b), is the smallest positive integer that is divisible by both a and b.

Given:

Polynomial, f(x) = x3 - 6x2 + 11x - 6

Concept :

Factor theorem:

If f(x) is a polynomial of degree n ≥ 1 and 'a' is any real number, then, (x - a) is a factor of f(x), if f(a) = 0.

the factors of a polynomial using factor theorem:

1). If f(-a) = 0, then (x + c) is a factor of the polynomial f(x).

2). If f(d/c) = 0, then (cx - d) is a factor of the polynomial f(x).

3)  If f(-d/c) = 0, then (cx + d) is a factor of the polynomial f(x).

4)  If f(a) = 0 and f(b) = 0, then (x - a) and (x - b) are factors of the polynomial p(x).

Calculations:

Here f(x) is a polynomial of degree 3.

Using hit and trial method

Put x = 1 in f(x).

⇒ f(1) =  1³ - 6(1)² + 11(1) - 6 

⇒ f(1) =  1 - 6 + 11 - 6 

⇒ f(1) =  0

So, (x - 1) is a factor of f(x)

Similarly, For x = 2 and x = 3.

∴ (x - 1)(x - 2)(x - 3) is a factor of f(x)

Alternate method: 

Using hit and trial method and division method

we get one root as 1, so (x - 1) is one of the factor of the given polynomial

and polynomial reduces to:

⇒ (x - 1)(x² - 5x + 6)

⇒ (x - 1)(x² - 3x - 2x + 6)

⇒ (x - 1)(x(x - 3) - 2(x - 3))

⇒ (x - 1)(x - 2)(x - 3)

Formula Used:

Quadratic equation = x² – (a + b)x + (a × b) = 0; here ‘a’ and ‘b’ are the roots

Calculation:

Using the above formulae, we get

⇒ x² – ((p - q)/(p + q) - p + q/(p - q))x + ((p - q)/(p + q) × - p + q/(p - q)) = 0

⇒ x² – ((p - q)² - (p + q)²/(p² - q²))x - 1 = 0

⇒ x² – ( -4pq)/(p² - q²)x - 1 = 0

⇒ (p² - q²)x² + 4pqx - p² + q² = 0

Given: 

The marks of 16 students in math in monthly test are 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8

Concept:

The mode is the value that occurs the most often in a data set.

Calculation: 

Using the above observations, we have 5 as maximum times i.e. (4 times)

∴ The required mode is 5.

Calculation:

Given Expression is:

32x + 1 - 3x = 3x + 3 - 32

putting 3^x = y, then our equation will become

(3^x)² × 3 – 3x  = 3x  × 3³ – 32

⇒ 3y² – y – 27y + 9 = 0      ----(1)

⇒ y(3y – 1) – 9(3y – 1) = 0

⇒ (y – 9)(3y – 1) = 0

⇒ y = 9, 1/3

Now, 3x = y

3x = 9

⇒ 3x = 3² 

By comparing the power of 3 we get, x = 2 

And, 3x = 1/3

⇒ 3x = 3^-1

By comparing the power of 3 we get, x = -1 

∴ The required value of x = 2, -1

Formula used:

P(A) = Number of favourable outcomes/Total number of  favourable outcomes

Concept:

In a leap year, there are 52 weeks and 2 odd days i.e. 366 days

∵ 364 is a multiple of 7, there will be 52 Sundays, 52 Mondays, ..., 53 Saturdays.

Calculation: 

These 2 odd days can be (Sat, Sun), (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat) i.e. out of 7 possible ways we have Sunday in 2 ways

∴ Required probability = 2/7

Given:

Given a circle with a chord of length = 10 cm = AB

Distance between centre to chord = 12 cm = OP

Concept:

The line perpendicular from the center to the chord bisects the chord.

Calculations:

From ΔAPO right-angled at P, Using Pythagoras theorem

AO² = AP² + OP²

AO² =  5² + 12²

⇒ AO = 13 cm = radius of the circle

The diameter of the circle = AO + OZ = 13 + 13 = 26 cm