CDS Elementary Mathematics 4 Sep 2022 Official Paper (Previous Year Paper)

Shortcut TrickWe have, 

 $\rm\sqrt{20+\sqrt{20+\sqrt{20+ \dots \ \infty}}}$,

Find two factors of 20 such that their difference is 1.

The larger factor will be the solution for expression.

Factors of 20 = 4 × 5

⇒ x2 - 20 = 5

∴ x = 5

Alternate MethodConcept:

If y = $\sqrt {a + \sqrt{a + \dots \infty}}$

⇒ y =  $\frac{\sqrt{4a+1}+1}{2}$

Calculation:

We have

x² - 20 =$\rm\sqrt{20+\sqrt{20+\sqrt{20+ \dots \ \infty}}}$

Using the above concept, (a = 20)

⇒ x² - 20 = $\frac{\sqrt{4(20)+1}+1}{2}$ 

⇒ x2 - 20 = $\frac{\sqrt 81+1}{2} = \frac{10}{2}=5$

⇒ x² = 25

∴  x = 5

Given:

$\rm \frac{a+b}{b+c}=\frac{c+d}{d+a}$ a ≠ c

Concept used:

Componendo and Dividendo:

If a : b = c : d then

​(a + b ) : (a – b) = (c + d) : (c – d) 

**Calculation:**​

$\rm \frac{a+b}{b+c}=\frac{c+d}{d+a}$​

According to the above concept

⇒ $\rm \frac{a+b + b + c}{a + b - b-c}=\frac{c+d+ d + a}{c +d-d-a}$

⇒ $\rm \frac{a+2b + c}{a -c}=\frac{c+2d + a}{c -a}$

⇒ $\rm \frac{a+2b + c}{a -c}=\frac{c+2d + a}{-{(a-c)}}$

⇒ a + 2b + c = -c - 2d - a

⇒ 2a + 2b + 2c + 2d = 0

⇒ a + b + c + d = 0

∴ The required value is a + b + c + d = 0

Shortcut Trick

Given that (a³ + b³) ∝ (a² − b²).

Expand both terms using algebraic identities: (a + b)(a² − ab + b²) ∝ (a + b)(a − b).

Cancel the common factor (a + b) from both sides of the proportionality relation.

We get (a² − ab + b²) ∝ (a − b).

∴ The correct answer is (a − b).

Alternate Method

Given: (a³ + b³) is directly proportional to (a² − b²).

Formula Used: If x ∝ y, then x = k × y, where k is a constant of proportionality.

Calculations:

⇒ (a³ + b³) = k × (a² − b²)

⇒ (a + b)(a² − ab + b²) = k × (a + b)(a − b)

⇒ Divide both sides by the common factor (a + b):

⇒ (a² − ab + b²) = k × (a − b)

⇒ Since k is a constant, we can write this in variation form:

⇒ (a² − ab + b²) ∝ (a − b)

∴ The correct answer is (a − b).

Additional Information

Sum of Cubes Identity

The expression a³ + b³ is factored as (a + b)(a² − ab + b²).

Difference of Squares Identity

The expression a² − b² is factored as (a − b)(a + b).

Direct Proportionality

If two quantities x and y are related such that x = ky, they are said to be directly proportional, written as x ∝ y.

Difference of Cubes Identity

The expression a³ − b³ is factored as (a − b)(a² + ab + b²).

Given:

$\rm\frac{√{x+20}+√{x-1}}{√{x+20}-√{x-1}}=\frac{7}{3}$

Shortcut TrickAt x = 5,

$\rm\frac{√{5+20}+√{5-1}}{√{5+20}-√{5-1}}=\frac{5+2}{5-2}=\frac{7}{3}$

Hence, the required value

$⇒ \rm √{(x + 20)(x-1)} = √{(25\times4)}$

⇒ √100 =  10

Alternate Method

Concept used:

Componendo and Dividendo (reverse):

If (a + b ) : (a – b) = (c + d) : (c – d) then

a : b = c : d 

Calculation:

$\rm\frac{√{x+20}+√{x-1}}{√{x+20}-√{x-1}}=\frac{7}{3}$

⇒ $\rm\frac{√{x+20}}{√{x-1}}=\frac{7+3}{7-3}= \frac{10}{4}$

⇒ $\rm\frac{√{x+20}}{√{x-1}}=\frac{5}{2}$

Squaring both sides, we get

⇒ $\rm\frac{{x+20}}{{x-1}}=\frac{25}{4}$

⇒ 4x + 80 = 25x – 25

⇒ (25x – 4x) = (80 + 25)

⇒ 21x = 105

⇒ x = 5

The value of $\rm √{(x + 20)(x-1)}$

⇒ $\rm √{(5 + 20)(5-1)}$

⇒ $\rm √{(25\times 4)}$ = √100

⇒ $\rm √{(x + 20)(x-1)}$ = 10

∴ The required value is 10.

Given:

The HCF of (x8 – y8) and (x7 – y7 + x5y2 – x2y5) 

Concept:

1. HCF of two numbers:  Product of the smallest power of each common prime factor in the numbers.

2. (aⁿ - bⁿ) = (a - b) (a^n-1 + ba^n-2 +  ....+ b^n-1)

Where n is a natural number. Hence,

For every value of n, (a - b) will be a factor of (aⁿ - bⁿ)

3. (a + b)(a  - b) = a² - b²

Calculation:

Using the  above formula

(x8 – y8) = (x⁴ - y⁴)(x4 + y4)

⇒ (x8 – y8) = (x² - y²)(x² + y²) (x⁴ + y⁴) 

⇒ (x8 – y8) = (x + y)(x – y)(x² + y²)(x⁴ + y⁴)

Now, (x7 – y7 + x5y2 – x2y5) 

⇒ x⁷ + x⁵y² – y⁷ – x²y⁵

⇒ x⁵(x² + y²) – y⁵(y² + x²)

⇒ (x² + y²) (x5 – y5)

Using the general formula of (aⁿ - bⁿ)

⇒ (x² + y²)(x - y)(x⁴ + yx³ + y²x² + y³x + y⁴)

Now, using the concept of HCF

The required HCF = (x – y)(x² + y²)

Multiplying the factor

⇒ (x3 – y3 – x2y + xy2) 

∴The required HCF is (x3 - y3 - x2y + xy2)

Given:

The sum of the squares of the roots of the equation x2 - 14x + k = 0 = 100

Formula used:

For a quadratic equation ax² + bx + c = 0

Sum of roots = $\frac{-b}{{a}}$

Product of roots = $\frac{c}{{a}}$

Calculation:

x2 - 14x + k = 0

α + β = $\frac{-b}{{a}}$ = 14

αβ = $\frac{c}{{a}}$ = k

Now, the sum of the squares of the roots 

⇒ α² + β² = (α + β)² - 2αβ

⇒ α2 + β2 = (14)² - 2k = 100      (Given)

⇒ 196 - 2k = 100

⇒ 2k = 96

⇒ k = 48

∴ The required value of k is 48.

Given: 

(x + k) is the HCF of x2 + px + q and x2 + qx + p

Concept:

1. HCF: Product of the smallest power of each common prime factor in the numbers.

2. If (x - a) is factor of p(x) then p(a) = 0

Calculation:

We have,

P(x) = x2 + px + q

Q(x) = x2 + qx + p

Since, (x + k) is HCF (common prime factor) of P(x) & Q(x)

⇒ P(k) =  k2 - kp + q = 0

⇒ k2 = kp - q          ------(1)

Again,

Q(k) = x2 + qx + p

⇒ k² = kq - p           ------(2)

From equation (1) & (2)

⇒ kp - q = kq - p 

⇒ k(p - q) = (q - p)

⇒ k = -1

∴ The required value of k is -1.

Additional Information

LCM: Product of the greatest power of each prime factor, involved in the numbers.

Formula used:

Dividend = Divisor ×  quotient + Remainder

Calculation:

Let the greater number be G & smaller number be S.

⇒ 3G = S × 6 + 6

⇒ 3G = 6S + 6           

⇒ G = 2S + 2           -----(1) 

Also,

⇒ 5S = G × 2 + 3

⇒ 5S = 2G + 3          

⇒ 5S = 2(2S + 2) + 3        [From equation (1)]

⇒ 5S = 4S + 7  

⇒ S = 7

Put this value in equation (1)

⇒ G = 2 × 7 + 2 = 16

∴ Difference = 16 - 7 = 9

Concept:

Factor: A number that divides another number.

For example, 3 and 6 are factors of 12

Shortcut TrickWe have  a2 (b - c)3 + b2 (c - a)3 + c2 (a - b)3

Let, a = 1, b = 2, c = 3

⇒ 1² (2 - 3)³ + 2²(3 - 1)³ + 3²(1 - 2)³

⇒ (-1) + 32 - 9 = 22

Statement:1 

(ab + bc + ca) = 1 × 2 + 2 × 3 + 3 × 1

⇒ (ab + bc + ca) = 11

Which is a factor of 22

Statement:2

(a + b + c) = 1 + 2 + 3 = 6

Which is not a factor of 22

∴ Only statement 1 is correct.

Shortcut Trick

We have,

$\rm \frac{1}{x(x-y)(x-z)}+\frac{1}{y(y-z)(y-x)}+\frac{1}{z(z-x)(z-y)}$

Put x = 1, y = 2, z = 3

⇒ $\rm \frac{1}{1(1-2)(1-3)}+\frac{1}{2(2-3)(2-1)}+\frac{1}{3(3-1)(3-2)}$ 

⇒ $\frac{1}{(-1)(-2)}+\frac{1}{2(-1)}+\frac{1}{3(2)}$

⇒ $\frac{1}{2}-\frac{1}{2}+\frac{1}{6} = \frac{1}{6}$

From option (3), 

$\frac{1}{xyz} = \frac{1}{1\times 2\times 3} = \frac{1}{6}$

∴ $\rm\frac{1}{xyz}$ is the correct value.

Calculation:

Statement 1. 

x4(y – z) + y4(z - x) + z4(x - y) is positive if x > y > z.

We have to solve this by Value putting method

Here, x > y > z.

Then, x = 2, y = 1 or z = 0

Now,

⇒ 2⁴(1 − 0) + 1⁴(0 − 2) + 0⁴(2 − 1)

⇒ 16 + 1(-2) + 0

⇒ 16 − 2 = 14

So, Statement 1 is positive.

Statement 2.

x4(y - z) + y4(z - x) + z4(x - y) is negative if x < y < z.

Here, x < y < z

Then, x = 0, y = 1 or z = 2

Now,

⇒ 0⁴(1 − 2) + 1⁴(2 − 0) + 2⁴(0 − 1)

⇒ 0 + 2 − 16 = -14

So, Statement 2 is negative.

∴ Both 1 and 2.

Shortcut TrickPut x = 0 

$\rm \frac{x^6-24x^4+144x^2}{(x^2+4√3x+12)(x-2√3)^2} = 0$

From option (1)

x² = (0²) = 0

Except (1), all other option gives a value other than 0. 

∴ The required value is x2

Alternate Method

Formula used:

(a + b)(a − b) = a² − b²

Calculation:

$\rm \frac{x^6-24x^4+144x^2}{(x^2+4√3x+12)(x-2√3)^2}$

⇒ $\rm \frac{x^2(x^4-24x^2+144)}{(x^2+4√3x+12)(x^2 +12- 4√3x)}$

⇒ $\rm \frac{x^2(x^4-24x^2+144)}{(x^2+12)^2-(4√3x)^2}$  

⇒ $\rm \frac{x^2(x^4-24x^2+144)}{(x^4+144 + 24x^2 - 48x^2)}$

⇒ $\rm \frac{x^2(x^4-24x^2+144)}{(x^4+144 - 24x^2)}$

⇒ x²

∴ The required value is x²

Shortcut Trick

Total work = LCM of 45 and 40 = 360 units.

Efficiency of X = 360 ÷ 45 = 8 units/day | Efficiency of Y = 360 ÷ 40 = 9 units/day.

Work done by Y alone in the last 23 days = 23 × 9 = 207 units.

Remaining work done by X and Y together = 360 − 207 = 153 units.

Combined efficiency = 8 + 9 = 17 units/day ⇒ n = 153 ÷ 17 = 9.

∴ The correct answer is n = 9.

Alternate Method

Given:

Time taken by X to complete the work = 45 days.

Time taken by Y to complete the work = 40 days.

X and Y work together for n days, and Y completes the remaining work in 23 days.

Formula Used:

Total Work = LCM of given days.

Efficiency = Total Work ÷ Number of Days.

Calculations:

⇒ Combined efficiency of X and Y = 8 + 9 = 17 units/day.

⇒ Work completed by Y alone in 23 days = 23 × 9 = 207 units.

⇒ Remaining work done by X and Y together = Total Work − Work done by Y alone.

⇒ Remaining work = 360 − 207 = 153 units.

⇒ Time taken by X and Y to complete 153 units (n) = 153 ÷ 17 = 9 days.

⇒ Hence, n = 9.

∴ The correct answer is n = 9.

Additional Information

LCM Method in Time and Work

Total work is considered to be the Lowest Common Multiple (LCM) of the distinct days taken by the individuals to complete the job. Efficiency of a person is given by: Total Work ÷ Time Taken.

Work and Wages

Wages are strictly distributed in the ratio of the total work done by each individual. If they work for the same number of days, the wages are distributed in the ratio of their efficiencies.

Pipes and Cisterns

The concept remains similar to time and work, with the only variation being that an inlet pipe does positive work (filling) while an outlet or leakage pipe does negative work (emptying).

Given:

480 persons working 10 hours per day complete one-fourth of a work in 10 days.

Formula used:

M₁ × D₁ × H₁/W₁ = M₂ × D₂ × H₂/W₂

Where, M = Number of men,

D = Number of days, H = Number of hours,

W = Work done

Calculation:

Let, y additional workers are required;

M₁ = 480, D₁ = 10, H₁ = 10, W₁ = $\frac{1}{{4}}$

M₂ = (480 + y), D₂ = 20, H₂ = 8, W₂ =  (1 − $\frac{1}{{4}}$) = $\frac{3}{{4}}$

Using the above formula

⇒ $\frac{480\times 10\times10}{\frac{1}{4}} = \frac{(480+y)\times 20\times 8 }{\frac{3}{4}}$

⇒ $\frac{480\times 10\times10}{1} = \frac{160(480+y) }{3}$

⇒ 480 + y = 900

⇒ y = (900 − 480) = 420

∴ The required number of persons is 420.

Shortcut Trick

Equate total work: 5 × (12W + 16M) = 4 × (13W + 24M)

⇒ 60W + 80M = 52W + 96M ⇒ 8W = 16M ⇒ 1W = 2M

Convert Group 1 to men only: 12(2M) + 16M = 40M, taking 5 days.

Target group in men only: 25(2M) + 50M = 100M.

Using M₁D₁ = M₂D₂ ⇒ 40 × 5 = 100 × D ⇒ D = 2 days.

∴ The correct answer is 2 days.

Alternate Method

Given:

Group 1: 12 Women (W) and 16 Men (M) can do the work in 5 days.

Group 2: 13 Women (W) and 24 Men (M) can do the work in 4 days.

Method Used:

We use the LCM method to determine the efficiency of each group and then solve for the individual efficiencies of men and women.

Calculations:

⇒ 12W + 16M = 4    (Equation 1)

⇒ 13W + 24M = 5    (Equation 2)

⇒ Multiply Equation 1 by 3 and Equation 2 by 2 to balance M:

⇒ 36W + 48M = 12

⇒ 26W + 48M = 10

⇒ Subtracting the two equations: 10W = 2 ⇒ W = 0.2 units/day

⇒ Substituting W = 0.2 into Equation 1: 12(0.2) + 16M = 4

⇒ 2.4 + 16M = 4 ⇒ 16M = 1.6 ⇒ M = 0.1 units/day

⇒ Efficiency of target group (25W + 50M) = 25(0.2) + 50(0.1) = 5 + 5 = 10 units/day

⇒ Time required = Total Work ÷ Efficiency = 20 ÷ 10 = 2 days

∴ The correct answer is 2 days.

Additional Information

Total Work Concept

Total work is the product of efficiency and the time taken to complete it. For equivalent groups completing the same task, the work relation is W = M₁ × D₁ = M₂ × D₂.

Combined Efficiency

When individuals with different capabilities work together, their one-day work (efficiency) is added directly. The combined efficiency of a mixed group is (Number of men × Men's efficiency) + (Number of women × Women's efficiency).

Pipes & Cisterns Analogy

The efficiency concept is closely mirrored in Pipes & Cisterns problems, where the work done corresponds to the volume filled, and efficiency equates to the filling rate of different pipes.

Given:

Average speed of first train = 60 km/hr

Average speed of second train = 80 km/hr

Formula used:

Distance = speed × time

Average speed = Total Distance between two trains/ Total Time

Calculation:

According to the question;

The first train starts from station X at 6 a.m.

6 a.m. to 8 p.m. = 8 hours

Distance travelled by X in 8 hour

⇒ (60 × 8) = 480 km  

Remaining distance = (1320 - 480) = 840 km

Now, X & Y travel toward each other

⇒ Relative Speed = 60 + 80 = 140 km/hr

The time is the same at station X and station Y i.e. 2 p.m.

Time when both trains will meet

⇒ t = $(\frac{840}{{140}})$ = 6 hours

∴ They meet at (2 p.m. + 6 p.m.) = 8 p.m.

Alternate Method Given:

Distance between X and Y = 1320 km

Train from X starts at 6 a.m., Speed = 60 km/hr

Train from Y starts at 2 p.m., Speed = 80 km/hr

Formula Used:

Distance = Speed × Time

Calculation:

The train from X travels for 8 hours before the train from Y starts (from 6 a.m. to 2 p.m.):

⇒ Distance covered by train from X = 60 km/hr × 8 hours = 480 km

Remaining distance between the two trains when the train from Y starts:

⇒ Remaining distance = 1320 km - 480 km = 840 km

Relative speed of both trains (since they are moving towards each other):

⇒ Relative speed = 60 km/hr + 80 km/hr = 140 km/hr

Time taken to meet (using the remaining distance and relative speed):

⇒ Time to meet = 840 km / 140 km/hr = 6 hours

The train from Y starts at 2 p.m.:

⇒ Meeting time = 2 p.m. + 6 hours = 8 p.m.

∴ The trains meet at 8 p.m.

Given:

The ratio of alcohol and water in the first glass = 3 : 2

The ratio of alcohol and water in the second glass = 4 : 1

Shortcut Trick

∴ The required ratio is 7 : 3

<br>

Alternate Method

According to the question

Alcohol contained in first glass = 3/5

Water contained in first glass = 2/5

Alcohol contained in second glass = 4/5

Water contained in second glass = 1/5

Now, the ratio of alcohol and water in the third glass

⇒ (3/5 + 4/5) : (2/5 + 1/5)

⇒ (7/5 : 3/5)

⇒ 7 : 3

∴ The required ratio is 7 : 3

Shortcut Trick

For fractions, HCF = HCF of numerators ÷ LCM of denominators.

For fractions, LCM = LCM of numerators ÷ HCF of denominators.

Numerators: 3, 6, 9, 27 | Denominators: 5, 25, 20, 50.

x = HCF = 3 ÷ 100 | y = LCM = 54 ÷ 5.

Ratio y/x = (54/5) ÷ (3/100) = (54/5) × (100/3) = 18 × 20 = 360.

∴ The correct answer is y = 360x.

Alternate Method

Given: Fractions are 3/5, 6/25, 9/20, and 27/50.

Formula Used:

⇒ HCF of Fractions = HCF(Numerators) / LCM(Denominators)

⇒ LCM of Fractions = LCM(Numerators) / HCF(Denominators)

Calculations:

⇒ x = HCF(3, 6, 9, 27) / LCM(5, 25, 20, 50) = 3 / 100

⇒ y = LCM(3, 6, 9, 27) / HCF(5, 25, 20, 50) = 54 / 5

⇒ To find the relation, divide y by x:

⇒ y/x = (54/5) / (3/100)

⇒ y/x = (54/5) × (100/3)

⇒ y/x = 18 × 20 = 360

⇒ y = 360x

∴ The correct answer is y = 360x.

Additional Information

HCF and LCM of Fractions

Always ensure fractions are in their simplest form before applying the HCF/LCM formulas to avoid errors.

Product Rule

For any two numbers a and b, HCF(a, b) × LCM(a, b) = a × b. Note: This rule does not directly apply to a set of more than two fractions in the same way.

LCM of Denominators

To find the LCM of 5, 25, 20, 50: 50 is the largest, its multiples are 50, 100... 100 is divisible by 5, 25, 20, and 50.

Shortcut Trick By Remainder Theorem, we can directly substitute the remainders in place of the variables. Remainder of x = 2 and Remainder of y = 3 Remainder of (x − y) ÷ 6 = (2 − 3) = −1 Since the remainder cannot be negative, add the divisor (6) to the negative remainder. Actual Remainder = −1 + 6 = 5 ∴ The correct answer is 5.

Alternate Method Given: When x is divided by 6, the remainder is 2. When y is divided by 6, the remainder is 3. x > y Formula Used: Dividend = (Divisor × Quotient) + Remainder Calculation: ⇒ Let the quotient when x is divided by 6 be a. Then, x = 6a + 2 ⇒ Let the quotient when y is divided by 6 be b. Then, y = 6b + 3 ⇒ (x − y) = (6a + 2) − (6b + 3) ⇒ (x − y) = 6(a − b) + 2 − 3 ⇒ (x − y) = 6(a − b) − 1 ⇒ To find the positive remainder when divided by 6, we rewrite the expression by adding and subtracting 6: ⇒ (x − y) = 6(a − b) − 6 + 6 − 1 ⇒ (x − y) = 6(a − b − 1) + 5 ⇒ Since x > y and x leaves remainder 2 while y leaves remainder 3, we know a > b, making (a − b − 1) a non-negative integer. ⇒ Here, 6(a − b − 1) is completely divisible by 6, so the remainder left is 5. ∴ The correct answer is 5.

Additional Information Negative Remainders Concept When a calculation yields a negative remainder −R with a divisor D, the true positive mathematical remainder is given by (D − R). Remainder Addition & Subtraction Rule The remainder of an expression (A ± B) ÷ D is equivalent to [Remainder(A ÷ D) ± Remainder(B ÷ D)] evaluated again with modulo D. Remainder Multiplication Rule The remainder of (A × B) ÷ D is identical to the remainder obtained from [Remainder(A ÷ D) × Remainder(B ÷ D)] ÷ D.

Given:

The value of a 2-digit number is 5 times the sum of the digits. 

Shortcut Trick

∴ Product of digits = 5 × 4 = 20

Alternate Method

Let the two digits be x and y respectively.

According to the question

⇒ 10x + y = 5(x + y)

⇒ 10x + y = 5x + 5y

⇒ 5x = 4y

⇒ $\frac{x}{{y}}$ = $\frac{4}{{5}}$

Now, the product of the digits

xy = (4 × 5) = 20

∴ The required product of the digits is 20.

Concept:

Total number of factors of x^py^q is (p + 1)(q + 1)

Number of odd factors of x^py^qz^r if x is even and y and z are odd then the factors are

1× (q + 1)(r + 1)

Calculation:

We have, 

⇒ N = 12⁶ × 3⁸ × 5³

⇒ N = 2⁶ × 2⁶ × 3⁶ × 3⁸ × 5³

⇒ N = 2¹² × 3¹⁴ × 5³

Total number of factors:- (12 + 1) × (14 + 1) × (3 + 1)

Total number of factors:- 13 × 15 × 4

Total number of factors = 780

Statement I: The number of odd factors of N is 60.

Number of odd factors = 1 × (14 + 1) × (3 +1)

Number of odd factors = 1 × 15 × 4

A number of odd factors = 60.

Hence, Statement I is correct.

Statement II: The number of even factors of N is 720.

Number of even factors = Total number of factors - Odd factors

Number of even factors = 780 - 60

Number of even factors = 720

Hence, Statement II is correct.

∴ Both statements are correct.

Given:

 $\rm \frac{\log_{10}a}{b-c}=\frac{\log_{10}b}{c-a}=\frac{\log_{10}c}{a-b}$ and (a ≠ b ≠ c) 

Formula Used:

log(m × n) = log m + log n

Calculation:

We have,

⇒  $\rm \frac{\log_{10}a}{b-c}=\frac{\log_{10}b}{c-a}=\frac{\log_{10}c}{a-b}$ = k (say)

⇒ log₁₀a = k(b - c)      ------ (1)

⇒ log₁₀b = k(c - a)      ----- (2)

⇒ log₁₀c = k(a - b)     ------ (3)

Adding eq. (1), (2) and (3)

⇒ log₁₀a + log₁₀b + log₁₀c = k(b - c + c - a + a - b)

⇒ log₁₀a + log₁₀b + log₁₀c = k × 0

⇒ log₁₀a + log₁₀b + log₁₀c = 0

⇒ log₁₀(abc) = 0

⇒ abc = 10⁰

⇒ abc = 1

∴ If $\rm \frac{\log_{10}a}{b-c}=\frac{\log_{10}b}{c-a}=\frac{\log_{10}c}{a-b}$, (a ≠ b ≠ c), then the value of abc is 1.

Formula used:

Angle between the hands = 11 × $\frac{m}{{2}}$ − 30H

Where, m denotes the minute hand and H denotes the hours hand

Calculation:

According to the question

Angle between the hands = 11 × $\frac{m}{{2}}$ − 30H

⇒ (11 × $\frac{40}{{2}}$ − 30 × 4)

⇒ (11 × 20 − 120)

⇒ (220 − 120)

⇒ 100° 

∴ The required angle is 100°.

Given:

log10$\rm \left[995+\sqrt{x^2-12x+60}\right]=3$

Concept:

Quadratic Equation:- A quadratic equation is a second order equation written as ax² + bx + c = 0 where a, b and c are coefficients of real numbers and a ≠ 0.

If α and β are the roots of the quadratic equation

Sum of roots;- α + β = $\frac{-b}{a}$

Formula Used:

log_ax^m = mlog_ax

Calculation:

We have,

⇒ log10$\rm \left[995+\sqrt{x^2-12x+60}\right]=3$

⇒ $log_{10} \left[ 995 \ + \ \sqrt{x^2 \ - \ 12x \ + \ 60} \right] = log_{10}10^3$

⇒ $995 \ + \ \sqrt{x^2 \ - \ 12x \ + 60} = 10^3$

⇒ $995 + \sqrt{x^2 \ - 12x \ + \ 60} = 1000$

⇒ $\sqrt{x^2 \ - \ 12x \ + 60} = 5$

On squaring both the sides, we get

⇒ x² - 12x + 60 = 25

⇒ x² - 12x + 35 = 0

⇒ Sum of roots = $\frac{-(-12)}{1}$= 12

∴ If log10$\rm \left[995+\sqrt{x^2-12x+60}\right]=3$, then the sum of the roots of the equation is 12.

Formula used:

1. For quadratic equation ax² + bx + c = 0

$x = {-b \pm \sqrt{(b^2-4ac)} \over 2a}$ 

2. (a + b)(a - b) = a2 - b2

3. (a + b)² = a² + b² + 2ab

Calculation:

qx2 – 2px + q = 0 

By above formula,

$x = {-b \pm \sqrt{(b^2-4ac)} \over 2a}$

Here, a = q, b = -2p and c = q

$x = {2p \pm \sqrt{(4p^2-4q^2)} \over 2q}$

⇒ $x = {p \pm \sqrt{p^2-q^2} \over q}$     ----(1)

Statement:1 

x = $\rm \frac {\sqrt{p + q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}}$, where p > q

Rationalizing the denominator

⇒ $\rm \frac {\sqrt{p + q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}}$ × $\rm \frac {\sqrt{p + q}+\sqrt{p-q}}{\sqrt{p+q}+\sqrt{p-q}}$

Using the identities (2) & (3)

⇒ $x = \frac{p+q+p-q+2\sqrt{(p+q)(p-q)}}{p+q+p-q}$

⇒ $x = \frac{p+q+p-q+2\sqrt{(p^2-q^2)}}{p+q+p-q}$

⇒ $x = {p \pm \sqrt{p^2-q^2} \over q}$

Hence, statement 1 is correct

Statement: 2 

From equation (2), we can say that,

$x = {p + \sqrt{p^2-q^2} \over q}$

Hence, statement 2 is also correct.

Statement 3: 

We have, $x = {p \pm \sqrt{p^2-q^2} \over q}$

For any value of p & q, 

x = $\rm \frac {\sqrt{p + q}}{\sqrt{p-q}}$ is not possible

Hence, statement 3 is wrong

∴ Only two values of x are possible.

Shortcut Trick

Use the standard polynomial identity: LCM × GCD = p(x) × q(x).

Given p(x) = x⁴ + x² + 1 and GCD = x² − x + 1.

Note that x⁴ + x² + 1 = (x² + x + 1)(x² − x + 1).

LCM = [p(x) × q(x)] ÷ GCD = [(x² + x + 1)(x² − x + 1) × q(x)] ÷ (x² − x + 1).

LCM = (x² + x + 1) × q(x). Since q(x) = (x² − x + 1)², LCM = (x² + x + 1)(x² − x + 1)².

∴ The correct answer is (x² + x + 1)(x² − x + 1)².

Alternate Method

Given:

p(x) = x⁴ + x² + 1

q(x) = x⁴ − 2x³ + 3x² − 2x + 1

GCD = x² − x + 1

Formula Used:

LCM(p, q) = [p(x) × q(x)] ÷ GCD(p, q)

Calculations:

⇒ Factorizing p(x): x⁴ + x² + 1 = (x⁴ + 2x² + 1) − x² = (x² + 1)² − x²

⇒ p(x) = (x² + 1 + x)(x² + 1 − x) = (x² + x + 1)(x² − x + 1)

⇒ Factorizing q(x): x⁴ − 2x³ + 3x² − 2x + 1 = (x² − x + 1)²

⇒ Verification: (x² − x + 1)(x² − x + 1) = x²(x² − x + 1) − x(x² − x + 1) + 1(x² − x + 1)

⇒ x⁴ − x³ + x² − x³ + x² − x + x² − x + 1 = x⁴ − 2x³ + 3x² − 2x + 1 (Correct)

⇒ LCM = [(x² + x + 1)(x² − x + 1) × (x² − x + 1)²] ÷ (x² − x + 1)

⇒ LCM = (x² + x + 1) × (x² − x + 1)²

∴ The correct answer is (x² + x + 1)(x² − x + 1)².

Additional Information

Polynomial Product Rule

For any two polynomials p(x) and q(x), the product of the polynomials is equal to the product of their LCM and GCD.

Sophie Germain Identity

The factorization x⁴ + x² + 1 = (x² + x + 1)(x² − x + 1) is a specific case of the Sophie Germain Identity used frequently in algebra.

GCD Definition

The Greatest Common Divisor (GCD) is the highest degree polynomial that divides both p(x) and q(x) without a remainder.

Calculation:

According to the question

First 2 digit number = 10x + y

By reversing the 2-digit number = 10y + x

Adding the two numbers we get,

⇒ (10x + y) + (10y + x)

⇒ 11(x + y)

Statement:1 The sum is divisible by 9

Put x = 9, y = 9

⇒ 11 (9 + 9) = 198

Which is divisible by 9

Statement: 2 The sum is divisible by 2

Put x = 2, y = 2

⇒ 11 (2 + 2) = 44

Which is divisible by 2

∴ Both Statement-1 and Statement-2 are sufficient to answer

the question

Given:

LCM of two numbers x and y is 481 where x > y.

Concept used:

HCF: Product of the smallest power of each common prime factor in the numbers.

LCM: Product of the greatest power of each prime factor, involved in the numbers

Calculation:

The factors of 481 = (481 × 1) and (37 × 13)

According to the question x > y, then 

(x, y) = (481, 1) or (37, 13)

Statement 1:  y > 1

Hence, only possible value are

x = 37, y = 13

We can easily find the value of (3x – 2y).

Statement 2. HCF of x and y is 1

In both (481, 1) or (37, 13), HCF is 1.

Two different values of (3x – 2y)

∴ Statement-1 alone is sufficient to answer the question

Concept used:

Natural number: All the numbers start from 1 till infinity.

The factor of  xⁿ + yⁿ: 

When n is odd ⇒ (x + y)

When n is even ⇒ No factor

Explanation:

Statement-1: n is a natural number

When n is odd, a factor of (xn + yn) is (x + y) but,

No factor in the case when n is even

Therefore, with this statement, we are not sure about the factor of (x + y). For the odd value of n, (x + y) will be a factor & for the even value of n, (x + y) will not be a factor.

Statement-2: n is an even natural number

We have discussed already when n is even, there is no factor.

In this case (xn + yn) will not be divisible by (x + y)

Hence, if we consider only statement 2 (i.e. n = even), we can surely say (x + y) is not divisible by (xⁿ + yⁿ). 

∴ Statement-2 alone is sufficient to answer the question.

Important PointsThe question says, which one of the following is correct in respect of the question and the statements?

It doesn't mean that we have to determine whether statements 1 or 2 is/or correct or not. 

We have to find the correct option using two statements. Clearly, with statement 2, we are sure about the answer i.e. "if n = even, (x + y) will not be a factor of (xⁿ + yⁿ)'. So it is sufficient to answer.

Given:

Let x and y be two real numbers. 

Concept:  

1. For positive real number:

• Both even & odd power gives positive result.
• Ex: (2)² = 4, (-2)² = 4 2. For negative real number:
• Even power gives positive result
• Odd power gives negative result.
• Ex: (2)³ = 8, (-2)³ = -8

 

Calculation:

Statement 1:

x8y9 < 0

As x⁸ will be always positive 

⇒ y⁹ < 0    (product of +ve and -ve is negative) 

⇒ y < 0

Statement: 2

x9y10 < 0

As y¹⁰ will be always positive 

⇒ x9 < 0    (product of +ve and -ve is negative) 

⇒ x < 0

From 1 & 2, we can say x < 0 & y < 0

Since, product of -ve and -ve is positive

⇒ xy > 0

∴ Both Statement-1 and Statement-2 are sufficient to answer the question

Given:

a = 3b, 4b = 5c, 6c = 7d

Calculation:

a = 3b, 4b = 5c, 6c = 7d

⇒ $\frac{a}{{b}}$ = $\frac{3}{{1}}$, $\frac{b}{{c}}$ = $\frac{5}{{4}}$, $\frac{c}{{d}}$ = $\frac{7}{{6}}$

⇒ ($\frac{a}{{b}}$ × $\frac{b}{{c}}$ × $\frac{c}{{d}}$) = ($\frac{3}{{1}}$ × $\frac{5}{{4}}$ × $\frac{7}{{6}}$)

⇒ $\frac{a}{{d}}$ = $\frac{35}{{8}}$

Now, the value of $\rm\frac{d+a}{d-a}$

⇒ $(\rm\frac{8+35}{8-35})$ = $-\rm\frac{43}{27}$

∴ The required value is $-\rm\frac{43}{27}$.

Given:

The given polynomial = (k - 3) x2 - kx - 1

Concept used:

If ax² + bx + c = 0 has no real roots, then

Discriminant (D) = b²- 4ac < 0

Calculation:

Given quadratic equation is (k - 3) x2 - kx - 1

According to the above concept,

⇒ b² - 4ac < 0

⇒ (-k)² - 4(k - 3)(-1) < 0

⇒ k² + 4k - 12 < 0

⇒ k² + 6k - 2k - 12 < 0

⇒ k(k + 6) - 2(k + 6) < 0 

⇒ (k - 2)(k + 6) < 0

∴ The required value is -6 < k < 2.

Given:

bc + cd = 2bd

a + c = 2b

Shortcut Trick

Put a = 1, b = 2 or c = 3

⇒ bc + cd = 2bd

⇒ (2 × 3) + 3d = 2 × 2 × d

⇒ 6 + 3d = 4d

⇒ d = 6

From option 3:

ad − bc = 1 × 6 - 2 × 3 = 0

∴ Option 3 is right.

Alternate Method

According to the question

bc + cd = 2bd     -----(1)

a + c = 2b

Putting the value of 2b in equation (1) then,

⇒ bc + cd = (a + c) × d

⇒ bc + cd = ad + cd

⇒ bc = ad

⇒ ad − bc = 0

∴ The required value is ad − bc = 0

Given:

Distance covered by flight = 2800 km

Average speed reduced by 100 km/hr

Time of flight increased by 30 min = 1/2 hour

Formula Used:

Distance = Speed × Time

Shortcut Trick

We have,

⇒ $\frac{Distance}{Speed} = Time$

⇒ $\frac{2800}{(x - 100)} - \frac{2800}{x} = \frac{1}{2}$

Put x = 800

⇒ $\frac{2800}{700} - \frac{2800}{800} = time$

⇒ $4 - \frac{7}{2} = time$

⇒ 0.5 = $\frac{1}{2}$ = time

∴ The original average speed of the aircraft is 800 km/hr.

Alternate MethodLet the original average speed (S₁) = x

Average speed reduced by 100(S₂)  = x - 100

Total speed ΔS = S₂ - S₁ = x - x - 100 = 100

⇒ D = $\frac{S_1 \ \times S_2}{\Delta S} \times \Delta t$

⇒ $2800 = \frac{x(x - 100)}{100} \times \frac{1}{2}$

⇒ 560000 = x(x - 100)

⇒ x² - 100x - 560000 = 0

⇒ x² -  800x + 700x - 560000 = 0

⇒ x(x - 800) + 700(x - 800) = 0 

⇒ (x - 800)(x + 700) = 0

⇒ x = 800 or x = - 700

∴ The original average speed of the aircraft is 800 km/hr.

Shortcut TrickPut a = 1, b = 0, c = 0

x = b + c = 0

y = c + a  = 1

z = a + b = 1

⇒ (x + y + z)3 – 24xyz = (0 + 1 + 1)³ - 24 × 0

⇒ (x + y + z)3 – 24xyz = 8

From option (3)

8 (a3 + b3 + c3) = 8(1³ + 0 + 0) = 1

∴ 8 (a3 + b3 + c3) is the correct answer.

Alternate Method

Formula Used:

(a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a)

Calculation:

We have,

⇒ (x + y + z)3 – 24xyz

⇒ (b + c + c + a + a + b)³ - 24(b + c)(c + a)(a + b)

⇒ [2(a + b + c)]³ - 24(b + c)(c + a)(a + b)

⇒ 8(a + b + c)³ - 24(b + c)(c + a)(a + b)

⇒ 8[a³ + b³ + c³ + 3(a + b)(b + c)(c + a)] - 24(b + c)(c + a)(a + b)

⇒ 8(a³ + b³ + c³) + 24(a + b)(b + c)(c + a) - 24(b + c)(c + a)(a + b)

⇒ 8(a³ + b³ + c³)

∴ The value of (x + y + z)3 – 24xyz is equal to 8(a³ + b³ + c³).

Given:

A person X and another person Y walk in the same direction

Average speed of X = u km/hr

Average speed of Y = v km/hr

Concept Used:

When two objects move in the same direction, then

Relative speed = S1 - S2

Formula used:

Speed = Distance/Time

Calculation

Time = Distance/Speed

⇒ t = d/(u − v)

Distance covered by X to overtake Y,

⇒ d = [(u × d/(u − v)]

⇒ d = $\rm\frac{ud}{(u-v)}$

∴ The required value is  $\rm\frac{ud}{(u-v)}$.

Shortcut TrickLet the number be 2

Since p is the difference between a number and its reciprocal 

⇒ 2 - 1/2 = 3/2 = p

⇒ p⁴ + 4p² = p²(p² + 4) = (9/4)[9/4 + 4]

⇒ p4 + 4p2 = 225/16

Since q is the difference between the square of the same number

and the square of its reciprocal

⇒ (2)² - (1/2)² = 15/4 = q

⇒ q² = 225/16

∴ The required value is q2

Alternate Method

Formula used:

1. (a² − b²) = (a − b)(a + b)
2. (a − b)² = a² + b² − 2ab
3. (a + b)² = a² + b² + 2ab

 

Calculation:

Let the number be n

⇒ n − $\frac{1}{{n}}$ = p          ------(1)

⇒  n² − $\frac{1}{{n^2}}$ = q       -----(2)

⇒ q = (n − $\frac{1}{{n}}$) (n + $\frac{1}{{n}}$)    [Using formula (1)]

⇒ q = p × (n + $\frac{1}{{n}}$)            [From equation (1)]

⇒ q/p = (n + $\frac{1}{{n}}$)             ------(3)

The value of  p4 + 4p²

⇒ p²(p² + 4)

⇒ p²(n − $\frac{1}{{n}}$)² + 4

⇒ p²(n² + $\frac{1}{{n^2}}$ − 2 + 4)       [Using formula (2)]

⇒ p²(n² + $\frac{1}{{n^2}}$+ 2)

⇒ p²(n + $\frac{1}{{n^2}}$)²                        [Using formula (3)]

⇒ p² × $\frac{q^2}{{p^2}}$ = q²                    [From equation (3)]

∴ The required value is q²

Given:

x = 2 + 2^1/2

Formula Used:

(a - b)² = a² + b² - 2ab

Calculation:

We have,

⇒ x = 2 + 2^1/2

⇒ x - 2 = 2^1/2

On squaring both the sides, we get

⇒ (x - 2)² = (2^1/2)²

⇒ (x - 2)² = 2

⇒ x² - 4x + 4 = 2

⇒ x² - 4x + 4 - 2 = 0

⇒ x² - 4x + 2 = 0

⇒ x² + 2 = 4x

On dividing by 'x', we get

⇒ $x \ + \ \frac{2}{x} \ = \ 4$

Again squaring on both the sides, we get

⇒ $x^2 \ + \ \frac{4}{x^2}\ + \ 4 = 16$

⇒ $x^2 \ + \ \frac{4}{x^2} = 12$

Again squaring on both the sides, we get

⇒ $x^4 \ + \ \frac{16}{x^4} \ +\ 8 = 144$

⇒ $x^4 \ + \ \frac{16}{x^4 } = 136$

∴ The value of x4 + 16x-4 is 136.

Given:

Time is taken by X longer than Y = 3 Hours

Distance = 30 km

Time taken by X if doubles his speed = 2 hours is less than Y 

**Formula used:**​

Let the speed of Y be u km/h & the speed of X be v km/h

According to the question

T_x = T_y + 3

30/u = 30/v + 3         ----(1)

Now, If X doubles his speed

30/2u = 30/v - 2        -----(2)

Equation (1) - (2)

(30/u - 30/2u) = (30/v + 3) - (30/v - 2)

30/2u = 5 or u  = 3 

Therefore, from equation (1)

30/3 = 30/v + 3

30/v = 7 or v = 30/7 = $4\frac{2}{7}$ km/hr

∴ The required value of Y is $4\frac{2}{7}$ km/hr.

Shortcut Trick

1% S.I interest in 10000 for 1st quarter = 100 Rs

4% S.I interest in 10000 1st quarter = 400 Rs

We know that, after the first quarter, C.I > S.I

∴ C > Rs. 400

<br>

Alternate Method

Formula used:

CI = [P(1 + R/100)T − P]

Where, P = Principal, R = Rate of interest and T = Time 

Calculation:

According to the question

Principal = Rs. 10,000

Rate = 4% per annum quarterly = (4/4)% = 1%

Time = 4 year  [1year = 4 quarters]

CI = [P(1 + R/100)T − P]

⇒ Rs. [10,000(1 + 1/100)⁴ − 10,000]

⇒ Rs. [10,000(1 + 0.01)⁴ − 10,000]

⇒ Rs. [10,000(1.01)⁴ − 10,000]

⇒ Rs. (10,000 × 1.0406 − 10,000)

⇒ Rs. (10,406 − 10,000)

⇒ Rs. 406

∴ C > Rs. 400

Given:

$\rm 6x^2 - 25x + \frac{6}{x^2} + \frac{25}{x} + 12 = 0$

Formula used:

If n - $\frac {1}{n}$ = p then n² + $\frac{1}{{n^2}}$ = p² + 2

Calculation:

$\rm 6x^2 - 25x + \frac{6}{x^2} + \frac{25}{x} + 12 = 0$

⇒ 6(x² + $\frac{1}{{x^2}}$) − 25(x − $\frac {1}{x}$) + 12 = 0

Let, x − $\frac {1}{x}$ be 'a'

From the above concept, (x² + $\frac{1}{{x^2}}$) = a² + 2

⇒ 6(a² + 2) − 25a + 12 = 0

⇒ 6a² + 12 − 25a + 12 = 0

⇒ 6a² − 25a + 24 = 0

⇒ 6a² − 16a − 9a + 24 = 0

⇒ 2a(3a − 8) − 3(3a − 8) = 0

⇒ (2a − 3) (3a − 8) = 0

⇒ a = $\frac {3}{2}$ or $\frac {8}{3}$

Since, $\frac {8}{3}$ is not given 

∴ The required value of x − $\frac {1}{x}$ = $\frac {3}{2}$.

Given:

$\rm 6x^2 - 25x + \frac{6}{x^2} + \frac{25}{x} + 12 = 0$

Formula used:

If n - $\frac {1}{n}$ = p then n2 + $\frac{1}{{n^2}}$ = p2 + 2

Calculation:

$\rm 6x^2 - 25x + \frac{6}{x^2} + \frac{25}{x} + 12 = 0$

⇒ 6(x2 + $\frac{1}{{x^2}}$) − 25(x − $\frac {1}{x}$) + 12 = 0

Let, x − $\frac {1}{x}$ be 'a'

From the above concept, (x2 + $\frac{1}{{x^2}}$) = a2 + 2

⇒ 6(a2 + 2) − 25a + 12 = 0

⇒ 6a2 + 12 − 25a + 12 = 0

⇒ 6a2 − 25a + 24 = 0

⇒ 6a2 − 16a − 9a + 24 = 0

⇒ 2a(3a − 8) − 3(3a − 8) = 0

⇒ (2a − 3) (3a − 8) = 0

⇒ a = $\frac {3}{2}$ or $\frac {8}{3}$

Case:1 a = $\frac {3}{2}$  

⇒ (x2 + $\frac {1}{x^2}$) = (a² + 2) = (3/2)² + 2

⇒ (x2 + $\frac {1}{x^2}$) = (9/4 + 2) = 17/4 (not in the option)

Case:2  a = $\frac {8}{3}$ 

⇒ (x2 + $\frac {1}{x^2}$) = (8/3)² + 2

⇒ (x2 + $\frac {1}{x^2}$)  = (64/9 + 2) = 82/9

∴ The require value of x² + $\frac {1}{x^2}$ is $\frac {82}{9}$.