CDS Elementary Mathematics 14 Nov 2021 Official Paper (Previous Year Paper)

Given:

The number is 26381

Concept used:

Factor:- A number or quantity that when multiplied with another produces a given number or expression.

In simple words, The prime factor is finding which prime numbers multiply together to make the original number.

Example: The prime factors of 15 are 3 and 5 (because 3 ×  5 = 15, and 3 and 5 are prime numbers)

Calculation:

Prime factorisation of 26381

⇒ 23 × 31 × 37

If we check all the options we can see only option 1 and option 4 is giving unit digit 1 when numbers are multiplied

But, 29 × 17 × 37 = 18241

∴ The answer is option 4 which is 23, 31, 37

Given:

Polynomial is (x – 1) (x – 2) (x – 4) – 90

Concept used:

If (x – a) is a factor of f(x) then f(a) = 0

Factoring a polynomial is the process of decomposing a polynomial into a product of two or more polynomials of smaller degree.

For example, f(x) = x² + 5x + 6 

f(x) = (x + 3) (x + 2)

Calculation:

Let f(x) = (x – 1) (x – 2) (x – 4) – 90

With help of options we can get,

f(7) = (7 - 1)(7 - 2)(7 - 4) – 90

⇒ 6 × 5 × 3 – 90

⇒  90 – 90

⇒ 0

⇒ f(7) = 0

∴ x – 7 is a factor of the polynomial (x – 1) (x – 2) (x – 4) – 90.

Given:

The number is (23 – 4√15)

Formula used:

(a - b)² = a² + b² - 2ab

Calculation:

23 - 4√15

⇒ 3 + 20 – 4 √15

⇒ (√3)² + 4 × 5 – 2 × √3 × (2 √5)

⇒ (√3)² + 4 × (√5)² – 2 × √3 × (2 √5)

⇒ (√3)² + ( 2√5 )² – 2 × √3 × (2 √5)

⇒ (√3 – 2√5)²

Square root = √3 – 2√5

∴ The square root of (23 - 4√15) is (√3 – 2√5).

Given:

Number is 371000 

Concept used:

Divisibility Rule of 9:

If the sum of digits of the number is divisible by 9, then the number itself is divisible by 9

Calculation:

(37)1000

⇒ (36 + 1)¹⁰⁰⁰

Since 36 is divisible by 9

Remainder from 36 = 0

Now,

1¹⁰⁰⁰ divided by 9 will always give you the remainder 1

∴ The required remainder is 1

Given:

LCM + HCF = 1484

LCM – HCF = 1428

Concept used:

a × b = LCM of (a, b) × HCF of (a, b)

Where a and b are two numbers

Calculation:

LCM + HCF = 1484

LCM – HCF = 1428

⇒ 2LCM = 2912

⇒ LCM = 1456

⇒ HCF = 28

Let the other numbers be x

112 × x = LCM × HCF

⇒ 112x = 1456 × 28

⇒ x = (1456 × 28)/112

⇒ x = 364

∴ The other number is 364.

Given:

Initially number of men = 20

Time allotted = 10 days

Formula used:

M₁ D₁/ W₁ = M₂ D₂/ W₂

Here,

M = Number of Men, D = Number of Days, W = Work

Calculation:

According to the question

(20 × 5)/(1/4) = (M₂ × 5)/(3/4)

⇒ 20 × 3 = M₂

⇒ M₂ = 60

More men needed = 60 – 20 = 40

∴ 40 more men they need to employ to finish the work on time.

Given:

Condition is x is a negative real number

Calculation:

Case 1.

If we multiply a natural number with a negative real number we will always get a negative number 

eg. 1 × -2 = -2 

Hence, Case 1 is wrong

Case 2.

If we take x = – 1

(– 1)² + (– 1)

⇒ 1 – 1 = 0 which is not ≥ 0

Hence, Case 2 is also wrong

Case 3.

If we take x = – 1

2 × (– 1) < (– 1) < –(– 1)

–2 < – 1 < 1

Hence, Case 3 is correct

Case 4.

If we take real number x = -3^1/4 then we get 

x² = 3^1/2 which is not a rational number

Hence, Case 4 is also wrong

∴ 1, 2, and 4 are not correct

Given:

2x2 + xy - 3y2

Concept used:

The standard form for linear equations in two variables

ax + by + c = 0

Calculation:

2x² + xy – 3y² 

⇒ 2x² + 3xy – 2xy – 3y² 

⇒ x(2x + 3y) – y(2x + 3y)

⇒ (x – y)(2x + 3y)

Here we have two linear factors are x - y and 2x + 3y 

Now, we have to find the sum of factors 

(x – y) + (2x + 3y)

⇒ 3x + 2y

∴ The required sum of the linear factors is 3x + 2y.

Conecpt used:

When D = b² – 4ac < 0, Roots are not real

When D = b2 – 4ac = 0, Roots are real and equal 

When D = b2 – 4ac > 0, Roots are real and unequal

Calculation:

From Option 1st

Given equation is 2x2 + 16x + 3 = 0

a = 2, b = 16, c = 3

D = (16)² – 4 × 2 × 3

⇒ 256 – 24 = 232 > 0 

Hence, the above equation has real roots.

From Option 2nd

Given equation is 2x2 + 10x - 1 = 0

a = 2, b = 10, c = 1

D = (10)2 – 4 × 2 × 1

⇒ 100 – 8 = 92 > 0 

Hence, the above equation has real roots.

From Option 3rd

Given equation is x2 - 8x + 1 = 0

a = 1, b = 8, c = 1

D = (8)2 – 4 × 1 × 1

⇒ 64 – 4 = 60 > 0

Hence, the above equation has real roots.

From Option 4th

Given equation is 4x2 + 9x + 6 = 0

a = 4,b = 9, c = 6

D = (9)2 – 4 × 4 × 6

⇒ 81 – 96

⇒ – 15 < 0

Hence, the above equation has not real roots.

∴ Required answer is option 4

Given:

Sum of the roots = 7

Product of the roots = 12

Concept used:

x² – (sum of roots)x + product of roots = 0

Calculation:

Let the roots of the given quadratic equation are α and β 

According to the question

α + β = 7   ----(i)

and α.β = 12    ----(ii)

On solving the above equations, we get

α = 4, β = 3

According to the question

New roots are α' = 4/2 = 2, β' = 3 × 2 = 6

⇒ α' + β' = 2 + 6 = 8

⇒ α'.β' = 2 × 6 = 12

So, Resulting quadratic equation is

x2 – (α' + β')x + α'.β' = 0

⇒ x² – 8x + 12 = 0

∴ The resulting quadratic equation is x2 – 8x + 12 = 0

Concept used:

For real and distinct solutions, D = b² – 4ac > 0

Calculation:

Given equation is x2 – kx + 2 = 0

According to the equation,

a = 1, b = – k, c = 2

For real and distnict roots, 

D = b² – 4ac > 0

⇒ (– k)² – 4 × 1 × 2 > 0

⇒ k² – 8 > 0

⇒ k2 – (2√2)² > 0

We know that, a² - b² = (a + b)(a - b)

⇒ (k - 2√2) (k + 2√2) > 0

This can be possible in two ways,

Case:1 (k - 2√2) > 0 and (k + 2√2) > 0

⇒ k > 2√2 and k > -2√2

⇒ k > 2√2

Case:2 (k - 2√2) < 0 and (k + 2√2) < 0

⇒ k < 2√2 and k < -2√2

⇒ k < - 2√2

∴ k > 2√2 or k < – 2√2

Given:

α + β + γ = αβ + βγ + γα     ----(i)

Calculation:

(1 – α)(1 – β)(1 – γ)

⇒ (1 – α – β – αβ)(1 – γ)

⇒ (1 – α – β – γ + αβ + αγ + βγ – αβγ)

⇒ {1 – (α + β + γ ) + αβ + αγ + βγ – αβγ}

From (i), we get

⇒ {1 – (αβ + αγ + βγ) + αβ + αγ + βγ – αβγ}

⇒ (1 – αβ – αγ – βγ  + αβ + αγ + βγ – αβγ)

⇒ 1 – αβγ

∴ Required answer is (1 – αβγ)

Given:

log10x + log10x2 = 2 log10x + 1

Formula used:

log a + log b = log ab

bloga = log a^b

Calculation:

log10x + log10x2 = 2 log10x + 1

⇒ log₁₀x³ = log₁₀x² + log₁₀10

⇒ log10x3 = log1010x2

On comparing both sides

x³ = 10x²

⇒ x = 10

∴ The value of x is 10.

Given:

LCM of two primes p and q = 2231

p > q

Concept used:

Product of two numbers = HCF × LCM

Calculation:

p and q are prime numbers

So, HCF of p and q = 1

p × q = 2231

By using the unit digit method,

1 can be got by 7 × 3, 1 × 1, 9 × 9

2231 is close to 2300

Now,

97 × 23 = 2231

So, p = 97 and q = 23

Now, p – q = 97 – 23 = 74

∴ The value of p - q is 74.

Given:

Three runners take 20, 30, and 35 minutes to complete one round.

Concept used:

LCM method to find the meeting time

Calculation:

LCM of 20 minutes, 30 minutes, and 35 minutes = 420 minutes

420 minutes = 7 hours

∴ They next meet at the starting point after 7 hours

Given:

Cost of 5 pencils, 6 notebooks and 7 erasers = Rs. 250

Cost of 6 pencils, 4 notebooks and 2 erasers = Rs. 180

Calculation:

Let cost of 1 pencil, 1 notebook and 1 eraser be p, n and e respectively

According to the question:

5p + 6n + 7e = 250     

By multiplying with 6:

30p + 36n + 42e = 1500     ----eq.1

Also, 6p + 4n + 2e = 180

By multiplying by 5:

30p + 20n + 10e = 900     ----eq.2

By subtracting eq.2 from eq.1

16n + 32e = 600

⇒ 2n + 4e = 75

∴ Cost of 2 notebooks and 4 erasers is Rs. 75

Concept:

The number of zeroes is counted by the power of 10 and (5 × 2) 

Calculation:

150 × 249 × 348 × .... × 50¹

It contains 10⁴¹, 20³¹, 30²¹, 40¹¹, 50¹

⇒ 10¹⁰⁵

It also contains 5⁴⁶, 15³⁶, 25²⁶, 35¹⁶, 45⁶, 5 (of 50) and 2⁴⁹, 12³⁹, 22²⁹, 32¹⁹, 42⁹

⇒ 546, 1536, 552, 3516, 456, 5 (of 50) and 249, 4⁴⁷, 1239, 2229, 3219, 429  there will be many number in which 2 exist.

⇒ (5 × 2)¹⁵⁷; here only power of 5s will be counted as power of 2s are in greater number

∴ Number of zeroes in the product of 150 × 249 × 348 × .... × 50¹ = 105 + 157 = 262.

Given:

p and q are roots of x2 - 60x + 899 = 0

p > q

Calculation:

x2 - 60x + 899 = 0

⇒ (x - 31) (x - 29) = 0

⇒ roots of the equation(p, q) are 31 and 29

⇒ p = 31, q = 29

Checking option 1):

31 - 29 - 1 = 1, hence it is false

Checking option 2:

31 - 58 + 27 = 0, hence it is correct.

∴ Option 2 is correct.

Given:

Roots of x2 - 4x - log10N = 0 are real

Formula used:

Roots of quadratic equation ax² + bx + c = 0 are real 

if D ≥ 0  i.e b² - 4ac ≥ 0 

And lf logaN = x

Then N = a^x

Calculation:

We have the quadratic equation x2 - 4x - log10N = 0

On comparing this equation with ax2 + bx + c = 0, we get 

a = 1, b = -4 and c = - log10N

Now, According to the question 

D = b2 - 4ac  ≥ 0

⇒ (- 4)² - 4 × 1 × (- log10N) ≥ 0

⇒ 16 + 4log10N ≥ 0

⇒ 4log10N ≥ -16

⇒ log10N ≥ - 4

⇒ N ≥ 10^-4

⇒ N ≥ 0.0001

∴ The minimum value of N is 0.0001.

Given:

5x-1 = (2.5)log105

Formula Used:

If a^x = n then x = log_aⁿ

log_a^b = log_e^b/log_e^a

Calculation:

We have 5x-1 = (2.5)log105

⇒ (2.5)log105 = 5x-1 

⇒ log₁₀⁵  = log_2.5^5x-1 

⇒ log105  = (x - 1) log2.55

⇒ (x - 1) = (log105)/(log2.55)

⇒ (x - 1) = log₁₀^2.5

⇒ x = log102.5 _{+ 1}

⇒ x = log102.5 ₊ log₁₀¹⁰

⇒ x = log1010 × 2.5

⇒ x = log1025

⇒ x = log105²

⇒ x = 2log10​5

∴ The value of x is 2log10​5.

Shortcut TrickWe have $96 - 64a^3 + \frac{8}{a^6} - \frac{48}{a^3 } - t^3 = 0$

Only one equation contains two variable 

So, We can solve this question by value putting method 

Let a = 1 we can't take a = 0  since 1/0 = ∞ 

Now, On putting a = 1 in the equation we get 

96 - 64 + 8 - 48 - t³ = 0

⇒ -8 - t³ = 0

⇒ t = -2 

Now, we have to find the value of a2t + 4a3

On putting a = 1 and t = -2 in the above equation, we get 

1² × (-2) + 4 × 1³ = 2

∴ The value of a2t + 4a3 is 2.

 

Alternate Method 

Given:

$96 - 64a^3 + \frac{8}{a^6} - \frac{48}{a^3 } - t^3 = 0$

Formula Used:

(a - b)³ = a³ - b³ - 3a²b + 3ab²

Calculation:

We have $96 - 64a^3 + \frac{8}{a^6} - \frac{48}{a^3 } - t^3 = 0$

$⇒ 3.(\frac{2}{a^2}) (4a)^2 - (4a)^3 + (\frac{2}{a^2})^3 - 3.(\frac{2}{a^2})^2.(4a) - t^3 = 0$

$⇒ (\frac{2}{a^2})^3 - (4a)^3- 3.(\frac{2}{a^2})^2.(4a) + 3.(\frac{2}{a^2}) (4a)^2- t^3 = 0$

$⇒ [\frac{2}{a^2} -4a]^3- t^3 = 0$

$⇒ t^3 = [\frac{2}{a^2} -4a]^3$

$⇒ t = [\frac{2}{a^2} -4a]$

Now, We have to find the value of a2t + 4a3

⇒ $a^2[\frac{2}{a^2} -4a] + 4a^3$

⇒ 2 - 4a³ + 4a³

⇒ 2

∴ The value of a2t + 4a3 is 2.

Given:

$A + B = \rm \frac{x^2 - 8}{x + 2} \ \ and \ A - B = \frac{-x^2 + 2x + 4}{x + 2}$

Calculation:

We have 

$A + B = \rm \frac{x^2 - 8}{x + 2} \ \ ---(i)$

$A - B = \frac{-x^2 + 2x + 4}{x + 2} \space ---(ii)$

Subtract equation (ii) from equation (i), we get 

$2B = \rm \frac{x^2 - 8}{x + 2} - \frac{[-x^2 + 2x + 4]}{x + 2}$

$2B = \frac{x^2-8+x^2 - 2x - 4}{x + 2}$

$2B = \frac{2(x^2-x - 6)}{x + 2}$

$B = \frac{(x^2-x - 6)}{x + 2}$

⇒ B = (x - 3) (x + 2)/ (x + 2)

⇒ B = (x - 3)

Now, we have to check which option is similar to (x - 3)

On solving option (i) we get (x - 2)/(x + 2) ≠ (x - 3)

On solving option (ii) we get (x + 2)/(x - 2) ≠ (x - 3)

On solving option (iii) we get (2x - 1)(x - 3)/(2x - 1) = (x - 3)

∴ The 3rd option is correct opiton.

Calculation:

P(x) = $\frac{x^4}{(x^2 - y^2) (x^2 - z^2) }+ \frac{y^4}{(y^2 - x^2) (y^2 - z^2)} + \frac{z^4}{(z^2 - x^2)(z^2 - y^2)}$

To simplify this expression, put any value except the value in which p(x) become indeterminate.

Put z = 0

⇒ P(x) = $\frac{x^4}{(x^2 - y^2) x^2 }+ \frac{y^4}{(y^2 - x^2) y^2 }$

⇒ P(x) = $\frac{x^2}{(x^2 - y^2) }- \frac{y^2}{(x^2 - y^2) }$

⇒ P(x) = $\frac{(x^2-y^2)}{(x^2 - y^2) }$

⇒ P(x) = 1

Given:

(2ab - b2) : (6a2 - ab) = 1 : 6

Calculation:

We have 

(2ab - b2) : (6a2 - ab) = 1 : 6

⇒ 6 × (2ab - b2) = 1 × (6a2 - ab)

⇒ 12ab - 6b² = 6a² - ab

⇒ 6a2 - 13ab  + 6b2 = 0

⇒ 6a2 - 9ab - 4ab + 6b2 = 0

⇒ 3a (2a - 3b) - 2b (2a - 3b) = 0 

⇒ (2a - 3b) (3a - 2b) = 0 

⇒ (2a - 3b) = 0 or (3a - 2b) = 0 

⇒ 2a = 3b or 3a = 2b

Case:1   2a = 3b

⇒ a : b = 3 :  2

⇒ (a + b) : (a - b) = (3 + 2) : (3 - 2) 

⇒ (a + b) : (a - b) = 5

Case:2   3a = 2b 

⇒ a : b = 2 : 3

⇒ (a + b) : (a - b) = (2 + 3) : (2 - 3) 

⇒ (a + b) : (a - b) = -5

∴ The value of (a + b) : (a - b) is -5 or 5.

Formula used:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 

Calculation:

Given that,

$\frac{x}{b + c -a } = \frac{y}{b - c - a} = \frac{z }{a + b -c} = k$

⇒ x = k(b + c - a)  ----(1)

⇒ y = k(b - c - a)  ----(2)

⇒ z = k(a + b - c)  ----(3)

Now, let

P = x2 + y2 + z2 - 2xy - 2yz + 2zx 

As, term zx is positive and other terms xy & yz are negative, then x and z must be negative.

⇒ P = (y - x - z)²

⇒ P = [k(b - c - a) - k(b + c - a) - k(a + b - c)]²

⇒ P = [kb - kc - ka - kb - kc + ka - ka - kb + kc]²

⇒ p = [- ka - kb - kc]²

⇒ P = k²(a + b + c)²

Given:

Ratio of three numbers = 2 : 3 : 5

Sum of squares of the numbers = 1368

Calculation:

Let the numbers be 2x, 3x and 5x 

According to the question:

4x² + 9x² + 25x² = 1368

⇒ x² = 36

⇒ x = 6, - 6

According to the question the numbers are positive 

So, We will not taking negative sign

⇒ 2x = 12, 3x = 18, 5x = 30

∴ Sum of all numbers = 60

Calculation:

Statement:1    $\frac{a^2 - b^2 }{a^2 + b^2 } > \frac{a - b}{a + b}$ where a > b > 0

Let a = 2 and b = 1

⇒ $\frac{2^2 - 1^2 }{2^2 + 1^2 } > \frac{2 - 1}{2 + 1}$

⇒ $\frac{3 }{5 } > \frac{1}{3}$ which is correct.

Hence, statement 1 is correct.

Statemnt:2   $\frac{a^3 + b^3 }{a^2 + b^2 } > \frac{a^2 + b^2}{a + b}$ only when a > b > 0

Let a = 2 and b = 1

⇒ $\frac{2^3 + 1^3 }{2^2 + 1^2 } > \frac{2^2 + 1^2}{2 + 1}$

⇒ $\frac{9 }{5 } > \frac{5}{3}$ Which is also correct.

This inequality also holds for a = 1 and b = 2 i.e.

Statement 2 is correct for b > a > 0 as well.

But according to statement 2, inequality will be correct only if a > b > 0

So, statement 2 is incorrect.

Hence, statement (1) is correct but (2) statement (2) is incorrect.

Mistake PointsPlease note the language of  the second statement

"only when a > b > 0"

We have proved thet, statement is correct for 

a = 2 & b = 1 (i.e. a > b > 0) and b = 2 & a = 1 (i.e. b > a > 0)

Hence, due to only word, statement 2 will become incorrect.

Given:

Work done by (3n - 1) men in (2n + 1) days = x

Work done by (3n + 1) men in (4n - 3) days = y

x : y = 6 : 11

Formula used:

(m₁ × d₁)/w₁ = (m2 × d₂)/w₂

m = number of men

d = number of days

w = work done

Calculation:

Let x be 6a and y be 11a

According to the question:

(3n - 1)(2n + 1)/6a = (3n + 1)(4n - 3)/11a

⇒ 11(6n² + n - 1) = 6(12n² - 5n - 3)

⇒ 66n² + 11n - 11 = 72n² - 30n - 18

⇒ 6n² - 41n - 7 = 0

⇒ (6n + 1)(n - 7) = 0

⇒ n = - 1/6, 7

∴ n = 7

Given:

$\frac{ay - bx}{c}= \frac{cx - az}{b}= \frac{bz - cy}{a}$

Calculation:

In the above equation multiple by c, b, and a in the consecutive equations

$⇒ \frac{ay - bx}{c} \times \frac {c}{c} = \frac{cx - az}{b} \times \frac {b}{b}= \frac{bz - cy}{a} \times \frac {a}{a}$

$⇒ \frac{acy - bcx}{c^2} = \frac{bcx - abz}{b^2} = \frac{abz - acy}{a^2}$

$⇒ \frac{acy - bcx}{c^2} = \frac{bcx - abz}{b^2} = \frac{abz - acy}{a^2} = k (let) ----(i)$

On equating equations with k, we get 

$⇒ {acy - bcx} = c^2 k \space \space ----(ii)$

$⇒ {bcx - abz} = b^2 k \space \space ----(iii)$

$⇒ {abz - acy} = a^2 k \space \space ----(iv)$

On adding equation (ii), (iii) and (iv), we get 

⇒ 0 = k(a2 + b2 + c2)

⇒ k = 0    ----(v)

From equation (i) and (v), we get

acy = bcx

⇒ x/y = a/b    ----(vi)

bcx = abz

⇒ x/z = a/c    ----(vii)

abz = acy

⇒ y/z = b/c     ----(viii)

From (vi), (vii) and (viii), we get

x = ar, y = br and z = cr  [let r be the ratio]

Now, we have to check the given conditions

Condition 1. is true from equation (vi)

Condition 2. is $\frac{x + y+ z}{a + b + c} = \frac{z}{c}$

⇒ (ar + br + cr)/(a + b + c) = r   [LHS]

⇒ cr/c = r  [RHS]

Condition 2. is also true 

∴ Both the conditions 1 and 2 are true.

Given:

Area of the rectangular garden = 375 m²

65 m wire can fence 3 sides of the garden

Formula used:

Area of a rectangle = lb

l = length of the rectangle

b = breadth of the rectangle

Calculation:

We have, lb = 375      ----(i)

According to the question,

There are two possibilities 

2b + l = 65      ----(ii)

or 2l + b = 65      ----(iii)

Solving (i) and (ii), we get 

⇒ 2b + 375/b = 65

⇒ 2b² - 65b + 375 = 0

⇒ (2b - 15) (b - 25) = 0

⇒ b = 7.5, 25 and l = 50, 15

Pair (50, 7.5) is possible here, as length > breadth

So, The perimeter of rectangle = 2 (50 + 7.5) = 115m

Solving equation (i) and (iii), we get 

⇒ l = 7.5, 25 and b = 50, 15

Pair (25, 15) is possible here, as length > breadth

So, The perimeter of rectangle = 2 (25 + 15) = 80m

∴ Form option we can say that 80m is the perimeter of the garden.

Given:

One root of ax2 - 4ax + 15 = 0 is 3/2

Calculation:

If 3/2 is root of ax2 - 4ax + 15 = 0; then it will satisfy the equation

⇒ 9a/4 - 6a + 15 = 0

⇒ -15a/4 = -15

⇒ a = 4

Equation becomes 4x2 - 16x + 15 = 0

⇒ 4x2 - 10x - 6x + 15 = 0

⇒ (2x - 5)(2x - 3) = 0

⇒ x = 3/2, 5/2

Now, Sum of squares of roots = (3/2)² + (5/2)²

⇒ 34/4 = 17/2

∴ The value of sum of square of roots is 17/2.

Given:

Product of the digits of a 2-digit number = 8

If 63 added to the number, its digits interchange

Calculation:

Let the number be 10x + y

According to the question

xy = 8    ----(i)

Also, 10x + y + 63 = 10y + x

⇒ 9y - 9x = 63 

⇒ y = x + 7      ----(ii)

From equation (i) and (ii), we have

⇒ 8/x = x + 7

⇒ x² + 7x - 8 = 0

⇒ (x + 8) (x - 1) = 0

⇒ x = 1, - 8; negative term rejected

⇒ y = 8

Hence, the number is 18

∴ Sum of the digits = 1 + 8 = 9

Given:

Speed of motor boat in still water(b) = 30 kmph

Total downstream distance = 60 km

Total upstream distance = 60 km

Total time 9/2 hours

Formula used:

d = b + w

u = b - w

w = speed of stream, d = downstream speed

u = upstream speed

Calculation:

According to the question:

60/(30 + w) + 60/(30 - w) = 9/2

⇒ 3600 = 9/2(900 - w²)

⇒ (900 - w2) = 800

⇒ w2 = 100

⇒ w = 10

∴ Speed of the stream = 10 kmph

Given:

Present age of father  = Sum of present ages of 4 children

10 year hence,

Sum of ages of children = 1.6 × age of the father

Calculation:

Let the present age of father be f and present sum of four children be c

According to the question

f = c    ----(i) and 

1.6(f + 10) = c + 40    ----(ii)

Form (i) and (ii), we get

⇒ 1.6f + 16 = f + 40

⇒ f = 40

∴ Present age of father = 40 years

Given:

Sum of numerator and denominator = 10

If numerator increased by 3 and denominator decreased by 1, fraction becomes 1

Calculation:

Let n be the numerator and (10 - n) be the denominator

⇒ Fraction is n/(10 - n)

According to the question:

(n + 3)/(10 - n - 1) = 1

⇒ n + 3 = 9 - n

⇒ n = 3

⇒ Denominator = 10 - n = 7

∴ Difference between numerator and denominator = 7 - 3 = 4

Given:

Equations 7x + ky = 27 and kx + 7y = 19 have unique solution

Formula used:

a₁x + b₁​y + c₁ ​= 0 

a₂x + b₂y + c₂​ = 0

If both equations have unique solution,

then a₁/a₂ ≠ b₁/b₂

Calculation:

For equations to have unique solution:

7/k ≠ k/7

⇒ k² ≠ 49

∴  k ≠ 7

Given:

Principal(P) doubles itself in 5 years when compounded annually

Formula used:

Amount(A) = P[1 + (R/100)]^T

R = Rate of interest

T = time period

Calculation:

According to the question:

2P = P[1 + (R/100)]⁵

⇒ 2 = [1 + (R/100)]5

⇒ 2^1/5 = 1 + (R/100) ----(i)

Let the required time period be T years

⇒ 4P = P[1 + (R/100)]^T ----(ii)

From (i) and (ii), we get

4 = [2^1/5]^T

⇒ 2^1/5 = 4^1/T

⇒ 21/5 = 22/T

⇒ 2/T = 1/5

⇒ T = 10

∴ It will become 4 times in 10 years

Formula Used:

The angle between the hour hand and minutes hand

θ = (30H - 11 M/2)

Calculation:

H = 3 and angle = 0°

When minute & hour hands coincide, then θ = 0

⇒ 0 = 30 × 3 - 11 M/2

⇒ 0 = 90 - 11 M/2

⇒ 90 = 11 M/2

⇒ 180 = 11 M

⇒ M = $16\frac{4}{11}$

⇒ M = $16\frac{4}{11}$ Minutes

∴  Both hour hand and minute hand will coincide past 3 O’clock between 16 to 17 minutes.

Given:

Sheela can stitch a suit in 2 days

Meena can stitch a suit in 3/2 days

Formula used:

Work = Efficiencey × Time

Calculation:

Efficiency of Sheela = 1/2 

Efficiency of Meena = 2/3

Combined efficiency = 1/2 + 2/3 = 7/6 suits per day

Now, They can stitch 7/6 suit in 1 day 

So they can stitch 1 suit in 6/7 days 

So they can stitch 35 suits in 35 × 6/7 = 30 days 

∴ Days taken for stitching 35 suits = 30 days

Given:

2x - 3y - 7 = 0

Formula used:

(a - b)³ = a³ - 3a²b + 3ab² - b³

Calculation:

2x - 3y = 7

By cubing both sides

⇒ 8x³ - 3. (2x)² (3y) + 3 (2x) (3y)² - 27y³ = 343

⇒ 8x3 - 36x2y + 54xy2 - 27y3 = 340 + 3

⇒ 8x3 - 36x2y + 54xy2 - 27y3 - 340 = 3

∴ The value of 8x3 - 36x2y + 54xy2 - 27y3 - 340 is 3.

Given:

p ∝ q/r2

Formula used:

% Change = [(Final value - Initial Value)/ Initial Value] × 100

Calculation:

According to the question

p ∝ q/r² 

⇒ p = kq/r²; k = constant

Now,

q increases by 20% & r decrease by 20%

p' = $\frac{k(1+\frac{20}{100})q}{(1-\frac{20}{100}r)^2}=\frac{1.2kq}{0.64r^2}$

⇒ p' = 1.875kq/r²

Hence, 

% Change = (Final value - Initial Value)/ Initial Value

% Change = $\frac{\frac{1.875kq}{r^2}-\frac{kq}{r^2}}{\frac{kq}{r^2}}× 100$

%Change = (0.875) × 100

∴ Change in p = 87.5%

Given:

Total days of work = 30 days

Earning on present-day = Rs. 500

Losing on absent day = Rs. 100

Total amount received = Rs. 11400

Calculation:

Let he was present for p days

Then, according to the question:

500p - 100(30 - p) = 11400

⇒ 500p - 3000 + 100p = 11400

⇒ 600p = 14400

⇒ p = 24

∴ He worked for 24 days

Given:

Cost price(CP) of a chair and a table = Rs. 750

Gain at chair = 5%

Gain at table = 20%

Total gain = 16%

Formula used:

Gain = SP - CP

Gain% = (Gain/CP) × 100

SP = Selling price

Calculation:

Let CP of chair be c and that of table be t

According to the question:

Total SP = 1.16 × 750 = 870

⇒ Total SP = 870

Again according to the question

c + t = 750

Multiplying it by 6

⇒ 6c + 6t = 4500     ----(i)

Again according to the question

1.05c + 1.2t = 870

Multiplying it by 5

⇒ 5.25c + 6t = 4350     ----(ii)

By subtracting (ii) from (i), we get

⇒ 0.75c = 150

⇒ c = 200

⇒ t = 550

∴ Original cost of table = Rs. 550

Given:

A person rode one-third of a journey at 60 km/hr

One-third at 50 km/hr and the rest at 40 km/hr

Had the person ridden half of the journey at 60 km/hr and the rest at 40 km/hr,

He would have taken 4 minutes longer to complete the journey

Formula used:

Speed = distance/time

Calculation:

Let total distance be 6x

Time is taken in 1st case

t₁ = 2x/60 + 2x/50 + 2x/40

⇒ t₁ = 37x/300

Time taken in 2nd case

t₂= 3x/60 + 3x/40

⇒ t₂ = 5x/40

According to the question:

t₂  = t₁  + 4/60    (∵ 1 hr = 60 min)

⇒ 5x/40 = 37x/300 + 1/15

⇒ 5x/8 = 37x/60 + 1/3

⇒ 75x = 74x + 40

⇒ x = 40

∴ Total distance = 6x = 240 km

Given:

The person saves Rs. 1000 more each next year, while his first-year savings is Rs. 2000

Formula used:

Sn = (n/2)[2a + (n - 1)d]

S_n = sum of n terms of an Arithmetic Progression

n = number of terms

a = first term

d = common difference

Calculation:

a = 2000

d = 1000

S_n = 170000

Sn = (n/2)[2a + (n - 1)d]

⇒ 170000 = (n/2)[2 × 2000 + (n - 1)1000]

⇒ 340 = n(4 + n - 1)

⇒ n² + 3n - 340 = 0

⇒ (n - 17)(n + 20) = 0

⇒ n = 17, - 20; negative rejected

∴ He will save Rs. 170000 in 17 years

Shortcut Trick

For trigonometric expression "acos3θ + bsec3θ" 

Case I: If a ≤ b then the minimum value is a + b 

Case II: If a > b then the minimum value is 2√ab

Here in this question a = 1 and b = 1 

So, According to case I,

The minimum value is 1 + 1 = 2

 

Alternate Method 

Given: 

cos3θ + sec3θ  where 0° ≤ 0 < 90°

Concept Used:

To find the minimum or maximum value we have to find dy/dθ = 0 

Calculation:

Let y = cos3θ + sec3θ

⇒ dy/dθ = 3cos2θ (-sinθ) + 3sec2θ . secθ. tanθ 

⇒ dy/dθ = -3sinθ cos2θ + 3 sec³θ sinθ/cosθ 

⇒ dy/dθ = sinθ(-3cos3θ + 3 sec3θ)/cosθ 

⇒ dy/dθ = tanθ.(-3cos3θ + 3 sec3θ)

Now, Put dy/dθ = 0 

⇒ tanθ.(-3cos3θ + 3 sec3θ) = 0

⇒ tanθ = 0 or (-3cos3θ + 3 sec3θ) = 0 

⇒ θ = 0° or

⇒ cosθ = secθ ⇒ θ = 0° 

Now, The minimum value at θ = 0° 

y = cos³0° + sec³0° 

⇒ y = 1 + 1 = 2

∴ The minimum value of the trigonometric expression is 2.

Given:

14 sin2θ + 10 cos2θ = 11

where 0° < θ < 90°,

Formula used:

sin2θ + cos2θ = 1

Calculation:

14 sin2θ + 10 cos2θ = 11

⇒ 4 sin2θ + 10 sin2θ + 10 cos2θ = 11

⇒ 4 sin2θ + 10 (sin2θ + cos2θ) = 11

⇒ 4 sin2θ + 10 = 11

⇒ sin2θ = 1/4

⇒ sinθ = 1/2, - 1/2

Since 0° < θ < 90°, hence negative value rejected

⇒ perpendicular/hypotenuse = 1/2

⇒ base² = 2² - 1² = 3

⇒ base = √3

⇒ tanθ + cotθ = (1/√3) + (√3)

∴ tanθ + cotθ = 4/√3

Given:

$\frac{\sin^3 θ + \cos^3 θ }{\sin θ + \cos θ} + \frac{\sin^3 θ - \cos^3 θ}{\sin θ - \cos θ}$

Formula Used:

a³ + b³ = (a + b) (a² + b² - ab)

a3 - b3 = (a - b) (a2 + b2 + ab)

sin²θ + cos²θ = 1

Calculation:

We have 

$\frac{\sin^3 θ + \cos^3 θ }{\sin θ + \cos θ} + \frac{\sin^3 θ - \cos^3 θ}{\sin θ - \cos θ}$

$⇒ \frac{(\sin θ + \cos θ)(\sin^2 θ + \cos^2 θ - sinθ. cosθ) }{(\sin θ + \cos θ)} + \frac{(\sin θ - \cos θ)(\sin^2 θ + \cos^2 θ + sinθ. cosθ)}{(\sin θ - \cos θ)}$

⇒ (sin2θ + cos2θ) - sinθ.cosθ + (sin2θ + cos2θ) + sinθ.cosθ 

⇒ 1 + 1 = 2

∴ The value of $\frac{\sin^3 θ + \cos^3 θ }{\sin θ + \cos θ} + \frac{\sin^3 θ - \cos^3 θ}{\sin θ - \cos θ}$is 2.

Shortcut Trick

Let the height of the flagstaff be h.

The ladder of 10 m reaches (h − 10) m height from the ground.

In a 30°−60°−90° triangle, the side opposite to 60° is √3 times the base x, so h = x√3.

Using Pythagoras theorem for the ladder: x² + (h − 10)² = 10².

Substitute x = h/√3 → h²/3 + h² − 20h + 100 = 100 → 4h²/3 = 20h → 4h = 60 → h = 15 m.

∴ The correct answer is 15 m.

Alternate Method

Given:

Length of ladder (BD) = 10 m

Distance from top AB = 10 m

Angle of elevation of top (A) from D = 60°

Formula Used:

In ΔACD, tan 60° = AC / CD

In ΔBCD, BC² + CD² = BD²

Calculations:

Let AC = h and CD = x

⇒ tan 60° = h / x

⇒ √3 = h / x ⇒ x = h / √3

In right ΔBCD, BC = AC − AB = h − 10

⇒ (h − 10)² + (h / √3)² = 10²

⇒ h² − 20h + 100 + h² / 3 = 100

⇒ 4h² / 3 − 20h = 0

⇒ 4h² = 60h

⇒ h = 15 m

∴ The correct answer is 15 m.

Additional Information

Trigonometric Ratios

For a right-angled triangle, tan θ = Perpendicular / Base and sin θ = Perpendicular / Hypotenuse.

Pythagoras Theorem

In any right-angled triangle, Base² + Height² = Hypotenuse².

Standard Angles

The value of tan 60° = √3, tan 30° = 1/√3, and tan 45° = 1.

Given:

Trigonometric expression is 1 + 2sin2θ cos2θ - sin4θ - cos4θ where 0° < θ < 90°

Formula Used:

2sinθ.cosθ = sin2θ 

sin²θ + cos²θ = 1

a² + b² = (a + b)² - 2ab

Calculation:

We have 1 + 2sin2θ cos2θ - sin4θ - cos4θ

⇒ 1 + 2sin2θ cos2θ - (sin4θ + cos4θ)

⇒ 1 + 2sin2θ cos2θ - [(sin2θ)² + (cos2θ)²]

⇒ 1 + 2sin2θ cos2θ - [(sin2θ + cos2θ)² - 2sin2θcos2θ]

⇒ 1 + 2sin2θ cos2θ - 1 + 2sin2θcos2θ

⇒ 4sin2θ cos2θ

⇒ (2sinθ.cosθ)²

⇒ (sin2θ)²

⇒ sin²2θ

The maximum value of sin²x = 1 at x = 90° 

∴ The maximum value of sin22θ = 1 at 2θ = 90°

i.e θ = 45° which satisfy the given condition

Given:

Height of airplane from river = 1200 m

Width of the river = 3000 m

Angles of depression on opposite points of two banks = 30º and θ respectively

Calculation:

We have height of aeroplane from river = AD = 1200 m

Width of river = BC = 3000 m

Now, In ΔABD

cot 30º = BD/1200

⇒ BD = 1200 cot 30º

⇒ BD = 1200√3 m

And, CD = BC - BD

⇒ CD = 3000 - 1200√3

⇒ CD = 600(5 - 2√3) ≈ 922 m

Now, In ΔADC

tanθ = AD/CD

⇒ tanθ = 1200/922 ≈ 1.301

We know that tan45° = 1 and tan60° = √3 = 1.732

And 1< 1.301 < 1.732

∴ 45° < θ < 60°.

Given:

$\dfrac{\cos^2 θ - 3 \cos θ + 2}{\sin^2 θ} = 1$

Caculation:

We have $\dfrac{\cos^2 θ - 3 \cos θ + 2}{\sin^2 θ} = 1$

⇒ cos²θ - 3 cosθ + 2 = sin2θ 

⇒ cos2θ - 3 cosθ + 2 = 1 - cos2θ

⇒ 2cos2θ - 3 cosθ + 1 = 0

⇒ 2cos2θ - 2 cosθ - cosθ + 1 = 0

⇒ 2 cosθ (cosθ - 1) - 1 (cosθ - 1) = 0

⇒ (cosθ - 1) (2cosθ - 1) = 0 

So, (cosθ - 1) = 0 or (2cosθ - 1) = 0 

θ = 0° or θ = 60° 

Since 0° < θ < 90° 

So, θ = 60° is the possible angle 

Now, we have to find the value of sin2θ + cosθ at θ = 60°

⇒ sin260° + cos60°

⇒ 3/4 + 1/2 = 5/4

∴ The required value is 5/4.

Formula used:

sin²x + cos²x = 1

1 + tan²x = sec²x

Calculation:

Consider 1:

sin4θ - sin2θ = cos4θ - cos2θ

⇒ sin²θ(sin²θ - 1) = cos²θ(cos²θ - 1)

⇒ sin2θ(- cos2θ) = cos2θ(- sin2θ)

Since LHS = RHS, hence true

Consider 2:

sin4θ + cos4θ = 1 + 2sin2θ cos2θ

⇒ sin4θ + cos4θ - 2sin2θ cos2θ = 1

⇒ (sin2θ - cos²θ)² = 1

Since, LHS is not equal to RHS, hence false

Consider 3:

tan4θ + tan2θ = sec4θ - sec2θ 

⇒ tan2θ(tan2θ + 1) = sec2θ(sec2θ - 1)

⇒ tan2θ sec2θ = sec2θ tan2θ

Since, LHS = RHS, hence true

∴ Only 1 and 3 are correct.

Formula used:

sin(90 - A) = cos A

tan(90 - A) = cot A

Calculation:

sin 24° sin66° - cos24° cos66° + tan24° tan 66° - cot 24° cot 66°

⇒ sin(90° - 66°) sin(90° - 24°) - cos24° cos66° + tan(90° - 66°) tan(90° - 24°) - cot 24° cot 66°

⇒ cos 66° cos 24º - cos24° cos66° + cot 24° cot 66° - cot 24° cot 66° = 0

∴ The value of sin 24° sin66° - cos24° cos66° + tan24° tan 66° - cot 24° cot 66° is 0.

Given:

x = psinA cosB

y = psinA sinB

z = pcos A

Formula used:

sin²θ + cos²θ = 1

Calculation:

x2 + y2 + z² = (psinA cosB)² + (psinA sinB)² + (pcosA)²

⇒ x2 + y2 + z2 = p²(sin²A cos²B + sin²A sin²B + cos²A)

⇒ x2 + y2 + z2 = p2[sin2A(cos2B + sin2B) + cos2A]

⇒ x2 + y2 + z2 = p2[sin2A + cos2A]

∴ x2 + y2 + z2 = p²

Given:

x = m secA + n tanA 

y = m tanA + n secA

Formula used:

sec2x - tan2x = 1

Calculation:

x = m secA + n tanA 

⇒ x² = m² sec²A + n² tan²A + 2mn secA.tanA

y = m tanA + n secA

⇒ y² = m² tan²A + n² sec²A + 2mn tanA.secA

x2 - y² = (m2 sec2A + n2 tan2A + 2mn secA.tanA) - (m2 tan2A + n2 sec2A + 2mn secA.tanA)

⇒ x2 - y2 = m2 sec2A + n2 tan2A + 2mn secA.tanA - m2 tan2A - n2 sec2A - 2mn secA.tanA

⇒ x2 - y2 = m²(sec2A - tan2A) + n²(tan2A - sec2A)

⇒ x2 - y2 = m2(sec2A - tan2A) - n2(sec2A - tan2A )

∴ x2 - y2 = m2 - n2

Given:

tanθ = 1; 0º < θ < 90º

Formula used:

tan 45º = 1

sin 45º = 1/√2

cos 45º = 1/√2

Calculation:

We have tanθ = 1

⇒ tanθ = tan45° 

⇒ θ = 45º

Now, we have to find the value of sin2θ - 2sinθ cosθ at θ = 45º

⇒ sin245º - 2 sin45º cos45º

⇒ (1/2) - 2(1/√2)(1/√2)

∴ sin2θ - 2sinθ cosθ = - 1/2

Formula used:

sin²x + cos²x = 1

Calculation:

$\dfrac{2 \sin^3 θ - \sin θ}{\cos θ - 2 \cos ^3 θ }$ = $\dfrac{sin θ}{cos θ}[\dfrac{2sin^2θ - 1}{1 - 2cos^2θ}]$

⇒ $\dfrac{2 \sin^3 θ - \sin θ}{\cos θ - 2 \cos ^3 θ }$ = $tanθ[\dfrac{2sin^2θ - sin^2θ - cos^2θ}{sin^2θ + cos^2θ - 2cos^2θ}]$

⇒ $\dfrac{2 \sin^3 θ - \sin θ}{\cos θ - 2 \cos ^3 θ }$ = $tanθ[\dfrac{sin^2θ - cos^2θ}{sin^2θ - cos^2θ}]$

∴ $\dfrac{2 \sin^3 θ - \sin θ}{\cos θ - 2 \cos ^3 θ }$ = tanθ

Given:

A, B and C are interior angles of the triangle ABC

Formula used:

sin(90º - x) = cos x

tan(90º - x) = cot x

Calculation:

Since, A, B and C are interior angles of the triangle ABC

Then, A + B + C = 180º

⇒ B + C = 180º - A

By multiplying both sides by 1/2

⇒ (B + C)/2 = 90º - A/2

Taking both sides sin

⇒ sin(B + C)/2 = sin(90º - A/2)

⇒ sin(B + C)/2 = cos A/2

Similarly, it can be proved that tan(B + C)/2 = cot A/2

Now, we have to find the value of $\tan \left( \frac{B + C}{2} \right) + \sin \left( \frac{B + C}{2} \right) - \cot \left( \frac{A}{2} \right) - \cos \left( \frac{A}{2} \right)$ 

⇒ cot A/2 + cos A/2 - cot A/2 - cos A/2 = 0

∴ $\tan \left( \frac{B + C}{2} \right) + \sin \left( \frac{B + C}{2} \right) - \cot \left( \frac{A}{2} \right) - \cos \left( \frac{A}{2} \right)$ = 0.