CDS Elementary Mathematics 2 Feb 2020 Official Paper (Previous Year Paper)

Given:

2 × 3 × 5 × 7 × 11 + 1

Concept Used:

Prime number: A number that is divisible by either 1 or itself is called a prime number.

Composite number: A number that is divisible by number\s other than 1 or itself is called a composite number.

Calculation:

Given number = 2 × 3 × 5 × 7 × 11 + 1 = 2310 + 1 = 2311

The number 2311 is only divisible by 1 and itself,

∴ The given number is a prime number

Given: 

HCF of the numbers in a pair = 12

Concept Used:

Mean = (First term + last term)/2

Calculation:

Let, the numbers be ‘12p’ and ‘12q’

Product of numbers = 2160

⇒ 12p × 12q = 2160

⇒ pq = 15

This is only possible when the values of p and q are either 1 and 15 or 3 and 5

Hence, the two pairs of numbers are (12, 180) and (36, 60)

⇒ Mean of first pair = x = (12 + 180)/2 = 96

⇒ Mean of second pair = y = (36 + 60)/2 = 48

Mean of x and y = (96 + 48)/2 = 72

∴ Mean of x and y is 72.

Given:

(54)¹⁰

Concept Used:

loga xn = n loga x

loga xy = loga x + loga y

Calculation:

The number of digits in any number ‘n’ is equal to ${\rm{lo}}{{\rm{g}}_{10}}{\rm{n}}$+ 1, where X is a floor function and it gives the largest integer less than or equal to x

Now,

log₁₀ (54)¹⁰

= 10 log₁₀ 54                                                    [∵ log_a xⁿ = n log_a x]

= 10 log₁₀ (2 × 3³)

= 10 log₁₀ 2 + 3 × 10 log₁₀ 3                             [∵ log_a xy = log_a x + log_a y]

= (10 × 0.301) + (30 × 0.477)

= 3.01 + 14.31

= 17.32

No. of digits in (54)¹⁰ = 17.32+ 1 = 17 + 1 = 18

∴ No. of digits in (54)¹⁰ is 18.

Given: 

${{\rm{x}}^{\sqrt {\rm{x}} }} = \sqrt[{\rm{n}}]{{{{\rm{x}}^{\rm{x}}}}}$

Concept Used:

log a^b = b⋅log a

Calculation:

Taking logarithm on both sides, we get,

⇒ √x log x = (1/n) log x^x

⇒ √x log x = (x/n) log x

⇒ n √x log x = x log x

⇒ n √x log x – x log x = 0

⇒ (√x log x)(n – √x) = 0

Equating first term to 0,

⇒ √x log x = 0

⇒ log x = 0 = log 1

⇒ x = 1

Equating second term to 0,

⇒ n – √x = 0

⇒ √x = n

⇒ x = n²

∴ {1, n²} is a set of solution of the given equation

Given: 

n(Physics) = 50, n(Mathematics) = 75, n(Physics ∩ Mathematics) = 35

Formula Used:

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Calculation:

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ n(Physics ∪ Mathematics) = 50 + 75 – 35 = 90

⇒ No. of students who registered either for Physics, or Mathematics or both = 90

Total no. of students = 250

No. of students who registered neither for Physics nor for Mathematics = 250 – 90 = 160

∴ No. of students who registered neither for Physics nor for Mathematic is 160.

Given: 

Number = 10ⁿ – 1

Calculation:

Putting n = 1, number = 10 – 1 = 9

Putting n = 2, number = 100 – 1 = 99

Putting n = 3, number = 1000 – 1 = 999

This means that the given number has ‘n’ digits each of which is equal to 9,

⇒ Sum of digits of number = 9n

But, the sum of digits = 3798

So, n = 3798/9 = 422

∴ Value of n is 422

Calculation:

We can write,

⇒ 2222 < 4096

⇒ 2222 < 2¹²

⇒ (2222)² < (2¹²)²

⇒ (2222)² < 2²⁴

Hence, 2222² < 2²²²²

Similarly,

⇒ 222 < 256

⇒ 222 < 2⁸

⇒ (222)²² < (2⁸)²²

⇒ (222)²² < 2¹⁷⁶

Hence, (222)²² < 2²²²²

Also,

⇒ 22 < 32

⇒ 22 < 2⁵

⇒ (22)²²² < (2⁵)²²²

⇒ (22)²²² < 2¹¹¹⁰

Hence, (22)²²² < 2²²²²

∴ Among the given numbers, 2²²²² is the largest number

Given:

m is the number of prime numbers between 0 and 50 

n is the number of prime numbers between 50 and 100

Concept Used:

Prime number: A number that is divisible by either 1 or itself is called a prime number.

Calculation:

Prime numbers between 0 and 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

⇒ m = 15

Prime numbers between 50 and 100 are 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

⇒ n = 10

So, (m – n) = 15 – 10 = 5

∴ The value of m - n is 5

Given:

$\frac{{\rm{a}}}{{\rm{b}}} = \frac{1}{3},\frac{{\rm{b}}}{{\rm{c}}} = 2,\frac{{\rm{c}}}{{\rm{d}}} = \frac{1}{2},\frac{{\rm{d}}}{{\rm{e}}} = 3$ and $\frac{{\rm{e}}}{{\rm{f}}} = \frac{1}{4}$

Calculation:

We can write,

$\frac{{abc}}{{def}}$

$= \frac{a}{b} \times \frac{{{b^2}}}{{{c^2}}} \times \frac{{{c^3}}}{{{d^3}}} \times \frac{{{d^2}}}{{{e^2}}} \times \frac{e}{f}$

$= \frac{1}{3} \times {\left( 2 \right)^2} \times {\left( {\frac{1}{2}} \right)^3} \times {\left( 3 \right)^2} \times \frac{1}{4}$

$= \frac{3}{8}$

∴ The value of $\frac{{abc}}{{def}}$ is $\frac{3}{8}$.

Given:

Expression: 3x + 3x + 1 + 3x + 2 

Calculation:

3^x + 3^{x + 1} + 3^{x + 2} = 3^x + (3^x)(3) + (3^x)(3²) = 3^x (1 + 3 + 9) = 3^x (13)

Since, x is any natural number, we can have x = 1, 2, 3 , ... so on

For x = 1, we  get ,  3x (13) = 3 × 13 = 39.

This could be the largest. But, in the options, we can see that there are numbers greater than 39. So we check for the next natural number.

So for x = 2

3x (13) = 32 × 13 = 117

We know,

3x + 3x + 1 + 3x + 2 = 3x + (3x)(3) + (3x)(32) = 3x (1 + 3 + 9) = 3x (13)

Now, we can see that the resultant expression is a multiple of 3 and 13. 

So, let us check from the options which is the largest multiple of both 3 and 13. 

As we can see, 117 is the largest multiple of 3 and 13.

∴ The largest divisor of the given expression from the given options is  117.

Given:

A two-digit number is 9 more than four times of the number obtained by interchanging its digits.

The product of digits in the two-digit number is 8

Calculation:

Let the ten’s digit and the unit digit of the number be ‘x’ and ‘y’ respectively

Given, product of digits = xy = 8      ---- (1)

⇒ Number = 10x + y

On interchanging the digits, number obtained = 10y + x

⇒ 10x + y = 9 + 4(10y + x)

⇒ 10x + y = 9 + 40y + 4x

⇒ 6x – 39y – 9 = 0

⇒ 2x – 13y – 3 = 0

Multiplying by ‘x’,

⇒ 2x² – 13xy – 3x = 0

Substituting from (1),

⇒ 2x² – 13(8) – 3x = 0

⇒ 2x² – 3x – 104 = 0

⇒ 2x² – 16x + 13x – 104 = 0

⇒ (x – 8)(2x + 13) = 0

⇒ x = 8, – 13/2 (neglect negative & fractions)

⇒ x = 8

⇒ y = 8/8 = 1

Required number = 10x + y = 80 + 1 = 81

∴ The required number is 81.

Given:

α and β are the roots of the quadratic equation x2 + kx – 15 = 0,

Formula Used:

For any quadratic equation of form (ax² + bx + c),

Sum of roots = – b/a

Product of roots = c/a

Calculation:

α and β are the roots of the quadratic equation x² + kx – 15 = 0,

⇒ α + β = – k

⇒ k = – (α + β)      ---- (1)

Also, αβ = – 15      ---- (2)

But, (α – β) = 8

Substituting from (2),

⇒ α + 15/α = 8

⇒ α² – 8α + 15 = 0

⇒ (α – 3)(α – 5) = 0

⇒ α = 3, 5

Substituting in (2),

⇒ β = – 5, – 3

Substituting in (1),

When α = 3, β = – 5, k = – (3 – 5) = 2

When α = 5, β = – 3, k = – (5 – 3) = – 2

 ∴ Positive value of k is 2

Given:

(x – 1) and (x + 2) divide the polynomial (x3 + 4x2 + px + q),

Concept Used:

According to the factor theorem, if (x – a) divides the polynomial f(x), then remainder = f(a) = 0

Calculation:

(x – 1) and (x + 2) divide the polynomial (x³ + 4x² + px + q),

⇒ f(1) = (1)³ + 4(1)² + p(1) + q = 0

⇒ 1 + 4 + p + q = 0

⇒ p + q = – 5      ---- (1)

Similarly,

⇒ f(– 2) = (– 2)³ + 4(– 2)² + p(– 2) + q = 0

⇒ – 8 + 16 – 2p + q = 0

⇒ 2p – q = 8      ---- (2)

Adding (1) and (2), we get,

⇒ p + q + 2p – q = – 5 + 8

⇒ 3p = 3

⇒ p = 1

⇒ q = – 5 – 1 = – 6

∴ The values of p and q are 1, – 6

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:

⇒ (x² + 5x + 6) = (x + 2)(x + 3)

⇒ (x² + 8x + 15) = (x + 3)(x + 5)

Using the above concept,

⇒ HCF = (x + 3)

According to the question, HCF = (x + k)

⇒ (x + 3) = (x + k)

∴ k = 3

Given: 

5^{x + 1} – 5^{x – 1} = 600

Concept Used:

a^{b + c} = a^b × a^c

Calculation:

5x + 1 – 5x – 1 = 600

⇒ (5^x × 5) – (5^x × 1/5) = 600

⇒ 5^x (5 – 1/5) = 600

⇒ 5^x = (5/24) × 600

⇒ 5^x = 125 = 5³

⇒ x = 3

10^2x = 10²⁽³⁾ = 10⁶ = 1000000

∴ The value of 10^2x is 1000000.

Given:

A number divides 12288, 28200 and 44333 so as to leave the same remainder

Calculation:

Let the number be ‘x’ and the remainder be ‘r’

⇒ (12288 – r), (28200 – r) and (44333 – r) are exactly divisible by ‘x’

As we know, if two numbers are divisible by a number, then their difference is also divisible by the same number

⇒ Difference of first two numbers = 28200 – r – 12288 + r = 15912

⇒ Difference of last two numbers = 44333 – r – 28200 + r = 16133

Hence, 15912 and are also divisible by ‘x’

⇒ 15912 = 2³ × 3² × 13 × 17

⇒ 16133 = 13 × 17 × 73

⇒ HCF of 15912 and 16133 = 13 × 17 = 221

∴ The largest number that divides 12288, 28200 and 44333 leaving the same remainder is 221

Given:

x² + 9y² = 6xy

Formula Used:

(a2 – 2ab + b2) = (a – b)2

Calculation:

x2 + 9y2 = 6xy

⇒ x² – 6xy + 9y² = 0                         (∵ (a² – 2ab + b²) = (a – b)²)

⇒ (x – 3y)² = 0

⇒ (x – 3y) = 0

⇒ x = 3y

∴ y ∶ x is 1 ∶ 3

Given: 

mⁿ = 1331

Calculation:

As we know, 11³ = 1331

Comparing, we get,

⇒ m = 11, n = 3

⇒ (m – 1)^{n – 1} = (11 – 1)^{3 – 1} = (10)² = 100

∴ The value of (m - 1)^{n - 1} is 100

Given:

$\frac{1}{{{{\rm{a}}^{{\rm{m}} - {\rm{n}}}} - 1}} + \frac{1}{{{{\rm{a}}^{{\rm{n}} - {\rm{m}}}} - 1}}$

Concept Used:

x^-1 = 1/x

Calculation:

Let, x = a^{m – n}

⇒ a^{n – m} = a^{– (m – n)} = x^{– 1} = 1/x

Hence, we can write,

$\frac{1}{{{a^{m - n}} - 1}} + \frac{1}{{{a^{n - m}} - 1}}$

$\frac{1}{{x - 1}} + \frac{1}{{\frac{1}{x} - 1}}$

<br>

$= - \frac{1}{{1 - x}} + \frac{x}{{1 - x;}}$

$= - \left( {\frac{{1 - x}}{{1 - x}}} \right)$

= -1

∴ The value of $\frac{1}{{{{\rm{a}}^{{\rm{m}} - {\rm{n}}}} - 1}} + \frac{1}{{{{\rm{a}}^{{\rm{n}} - {\rm{m}}}} - 1}}$ is - 1.

Given: 

${\rm{x}} = \sqrt 2 ,{\rm{;y}} = \sqrt[3]{3}$ and ${\rm{z}} = \sqrt[6]{6}$

Calculation:

x = √2 = (2)^1/2

⇒ x² = 2

Taking cube of both sides,

⇒ (x²)³ = (2)³

⇒ x⁶ = 8

⇒ x = ⁶√8

Similarly,

⇒ y = ³√3 = (3)^1/3

⇒ y³ = 3

Taking square on both sides,

⇒ (y³)² = (3)²

⇒ y⁶ = 9

⇒ y = ⁶√9

Now, we can write,

⇒ 3 < 8 < 9

⇒ ⁶√3 < ⁶√8 < ⁶√9

∴ z < x < y

x = √2 ,  y = 3√3 , z = ⁶√6

Taking power 6 to all three, we get

⇒ x⁶ = (2^1/2)⁶ , y⁶ = (3^1/3)⁶, z⁶ = (6^1/6)⁶

⇒ x6 = 8, y6 = 9, z6 = 6

∴ z < x < y

Given: 

log x = 1.25 and y = x^{log x}

Concept Used:

log ab = b log a

Calculation:

y = xlog x

Taking logarithm on both sides,

⇒ log y = log (x^{log x} )     

⇒ log y = (log x) log x       [∵ log a^b = b log a]

But, log x = 1.2500

⇒ log y = 1.25 × 1.25

∴ Value of log y is 1.5625.

Given:

f(x) is divided by (x – α)(x – β)

Formula Used:

According to the remainder theorem, when any function f(x) is divided by (x – a), then the remainder is equal to f(a)

Dividend = Divisor × Quotient + Remainder      ----(1)

⇒ f(x) = (x – a) × q(x) + f(a)

Calculation:

When f(x) is divided by (x – α), remainder = f(α)

⇒ f(x) = (x – α) × q₁(x) + f(α)

⇒ f(x)/(x – α) = q₁(x) + f(α)/(x – α)      ----(2)

Similarly,

When f(x) is divided by (x – β), remainder = f(β)

⇒ f(x) = (x – β) × q₂(x) + f(β)

⇒ f(x)/(x – β) = q₂(x) + f(β)/(x – β)      ----(3)

Subtracting (3) from (2), we get,

$\Rightarrow \frac{{f\left( x \right)}}{{x - \alpha }} - \frac{{f\left( x \right)}}{{x - \beta }} = {q_1}\left( x \right) - {q_2}\left( x \right) + \frac{{f\left( \alpha \right)}}{{x - \alpha }} - \frac{{f\left( \beta \right)}}{{x - \beta }};$

$\Rightarrow \frac{{\left( {x - \beta - x + \alpha } \right)f\left( x \right)}}{{\left( {x - \alpha } \right)\left( {x - ;\beta } \right)}} = {q_1}\left( x \right) - {q_2}\left( x \right) + \frac{{\left( {x - \beta } \right)f\left( \alpha \right) - \left( {x - \alpha } \right)f\left( \beta \right)}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}}$

$\Rightarrow \frac{{\left( {\alpha - ;\beta } \right)f\left( x \right)}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}} = {q_1}\left( x \right) - {q_2}\left( x \right) + \frac{{(x - \beta f\left( \alpha \right) - \left( {x - ;\alpha } \right)f\left( \beta \right)}}{{\left( {x - \alpha } \right)\left( {x - ;\beta } \right)}}$

<br>

$\Rightarrow f\left( x \right) = \frac{{\left[ {{q_1}\left( x \right) - {q_2}\left( x \right)} \right]}}{{\left( {\alpha - \beta } \right)}}\left( {{\rm{x;}}-{\rm{;\alpha }}} \right)\left( {{\rm{x;}}-{\rm{;\beta }}} \right) + \frac{{\left( {{\rm{x}} - {\rm{\beta }}} \right)f\left( \alpha \right) - \left( {x - \alpha } \right)f\left( \beta \right)}}{{\left( {{\rm{\alpha }} - {\rm{;\beta }}} \right)}}$

Comparing with eq. (1), we can write,

When f(x) is divided by (x – α)(x – β),

∴ Reminder is$\rm \frac{{\left( {{\rm{x}} - {\rm{\beta }}} \right){\rm{f}}\left( {\rm{\alpha }} \right) - \left( {{\rm{x}} - {\rm{\alpha }}} \right){\rm{f}}\left( {\rm{\beta }} \right)}}{{\left( {{\rm{\alpha }} - {\rm{\beta }}} \right)}}$

Given: 

Area of square = 2401x⁴ + 196x² + 4

Formula Used: 

Area of square = (Side)²

Calculation:

Area of square = 2401x4 + 196x2 + 4

⇒ Area of square = (49x²)² + (2 × 49x² × 2) + (2)²

∵ (a² + 2ab + b²) = (a + b)²

⇒ Area of square = (49x² + 2)²

⇒ Area of square = (Side)² = (49x² + 2)²

∴ Side of square is 49x² + 2

Given:

x ∝ yz

Calculation:

x ∝ yz

⇒ $\frac{x}{z}$ ∝ y

⇒ $\frac{z}{x}$ ∝ $\frac{1}{y}$

∴ y varies inversely as $\frac{z}{x}$

Given:

The points P and Q represents real number $0.7\bar 3$ and $0.5\bar 6$ on the number line

Calculation:

Let, x = $0.7\bar 3$= 0.733333…

⇒ 100x = 73.3333…

⇒ 10x = 7.3333…

⇒ 100x – 10x = 73.3333… – 7.3333…

⇒ 90x = 66

⇒ x = 66/90 = 11/15

Similarly,

Let, y = $0.5\bar 6$ = 0.566666…

⇒ 100y = 56.6666…

⇒ 10y = 5.6666…

⇒ 100y – 10y = 56.6666… – 5.6666…

⇒ 90y = 51

⇒ y = 51/90 = 17/30

⇒ x – y = (11/15) – (17/30) = 1/6

∴ Distance between point P and Q is 1/6

Given:

5x + 2y = 7xy

10x + 3y = 8xy

Calculation:

5x + 2y = 7xy        ----(1)

10x + 3y = 8xy       ----(2)

Multiply equation 1 by 2 and subtract equation 2 from it, we get

⇒ y = 6xy

⇒ x = 1/6

Now, from equation 1:

⇒ 5/6 + 2y = 7y/6

⇒ 5y/6 = -5/6

⇒ y = -1

∴ $\left( {\frac{1}{6},; - 1} \right)$ is the required point.

Shortcut Trick

Calculate successive differences step-by-step using multiplication factors 1.2 and 1.1.

Diff 2011 = 5000 × 1.2 − 2000 × 1.1 = 6000 − 2200 = 3800

Diff 2012 = 6000 × 1.2 − 2200 × 1.1 = 7200 − 2420 = 4780

Diff 2013 = 7200 × 1.2 − 2420 × 1.1 = 8640 − 2662 = 5978

In 2013, the price difference 5978 clearly exceeds 5000.

∴ The correct answer is 2013.

Alternate Method

Given:

Initial Price of X in 2010 = ₹5000, rate of increase = 20%

Initial Price of Y in 2010 = ₹2000, rate of increase = 10%

Formula Used:

Price after n years = Initial Price × (1 + Rate/100)ⁿ

Price Difference Progression

Year Price X Price Y Difference

2010 ₹5000 ₹2000 ₹3000

2011 ₹6000 ₹2200 ₹3800

2012 ₹7200 ₹2420 ₹4780

2013 ₹8640 ₹2662 ₹5978

Calculations:

⇒ In 2010: Difference = 5000 − 2000 = ₹3000

⇒ In 2011: X = 5000 × 1.2 = ₹6000, Y = 2000 × 1.1 = ₹2200

⇒ Difference in 2011 = 6000 − 2200 = ₹3800

⇒ In 2012: X = 6000 × 1.2 = ₹7200, Y = 2200 × 1.1 = ₹2420

⇒ Difference in 2012 = 7200 − 2420 = ₹4780

⇒ In 2013: X = 7200 × 1.2 = ₹8640, Y = 2420 × 1.1 = ₹2662

⇒ Difference in 2013 = 8640 − 2662 = ₹5978

⇒ As ₹5978 > ₹5000, the difference exceeds ₹5000 for the first time in 2013.

∴ The correct answer is 2013.

Additional Information

Successive Percentage Increase

When a value increases continuously over years, the formula is: Final Value = Initial Value × (1 + R/100)ⁿ, which uses the exact same principles as the Compound Interest formula.

Multiplying Factor Method

To increase a given number by R%, multiply it directly by the factor (1 + R/100). For example, a 20% continuous yearly increase means multiplying by 1.2 repeatedly each year.

Percentage Change Calculation

The overall percentage change between two values is calculated as: % Change = (Final Value − Initial Value) ÷ Initial Value × 100.

Given:

The ratio of speeds of X and Y is 5 : 6.

Y allows X a start of 70 m in a 1.2 km race

Formula:

Distance = Speed × Time

Calculation:

Suppose the speeds of X and Y are 50 m/s and 60 m/s respectively.

Y allows X a start of 70 m in a 1.2 km race;

That means X covers 1130 m and Y covers 1200 m.

Time in which X covers 1130 m = 1130/50 = 22.6 sec

And

Time in which Y covers 1200 m = 1200/60 = 20 sec

It is evident that Y wins the race and X takes 2.6 sec more than Y to finish the race.

∴ Distance by which Y wins the race = 2.6 × 50 = 130 m

Given:

A train takes two hours less for a journey of 300 km

Formula Used:

Distance = Speed × Time

Calculation:

Suppose the usual speed of the train = x km/hr

According to the question:

300/x – 300/(x + 5) = 2

⇒ 300x + 1500 – 300x = 2x² + 10x

⇒ 2x² + 10x – 1500 = 0

⇒ 2x² + 60x – 50x – 1500 = 0

⇒ 2x (x + 30) – 50(x + 30) = 0

⇒ (x + 30) (2x – 50) = 0

⇒ x = 25      (As negative value of x is not possible)

∴ Usual speed of train is 25 km/hr

Given:

6 men and 8 women can do a piece of work in 10 days

13 men and 24 women can do the same work in 4 day

Calculation:

Suppose M and W are the efficiencies (daily work done) of 1 man and 1 woman respectively.

According to the question,

(6M + 8W) × 10 = (13M + 24W) × 4

⇒ 60M + 80W = 52M + 96W

⇒ 8M = 16W

⇒ M = 2W

⇒ M/W = 2/1

∴ M : W is 2 : 1

Given:

The number of students is increased by 2 in each row, then the number of rows decrease by 3

The number of rows decreases by 3. If the number of students is increased by 4 in each row

Concept Used:

By method of elimination of 2 variable

Calculation:

Suppose total number of rows is A and number of students in each row is B;

∴ Total students = A × B

According to the question:

(A – 3) × (B + 2) = A × B

⇒ 2A – 3B = 6      ---- (1)

And

(A – 5) × (B + 4) = A × B

⇒ 4A – 5B = 20      ---- (2)

Multiply equation 1 by 2 and subtract equation 2 from it:

⇒ B = 8

Now from equation 1:

⇒ 2A = 6 + 24

⇒ A = 15

Then  A × B = 120

∴ Total number of students in the class is120

Given:

Time = 2 year

Difference of 1% in interest rates, SI increases by Rs. 24

Formula used:

Simple interest = (P × R × T)/100

Calculation:

Suppose the principal is Rs. P

Time = 2 years

From the question, we can say that due to a difference of 1% in interest rates, SI increases by Rs. 24.

So,

24 = (P × 2 × 1)/100

⇒ P = 1200

∴ The principal amount is 1200

Given:

The populations of two villages are 1525 and 2600 respectively

The ratio of male to female population in the first village is 27 : 34

The ratio of male to female population in the second village is 6 : 7

Concept Used:

If the ratio of male and female is a : b

Then male = a/(a + b)

Calculation:

Population of 1^st village = 1525

Ratio of male to female population in the first village is 27 : 34

So,

Number of males in first village = 1525 × (27/61) = 25 × 27 = 675

Number of females in first village = 1525 – 675 = 850

And

Population of 2^nd village = 2600

Ratio of male to female population in the 2^nd village is 6 : 7

So,

Number of males in 2^nd village = 2600 × 6/13 = 200 × 6 = 1200

Number of females in 2^nd village = 2600 – 1200 = 1400

Hence,

Ratio of male to female population of these two villages taken together

= (675 + 1200) : (850 + 1400)

= 1875 : 2250

= 5 : 6

∴ The ratio of male to female population of these two villages 5 : 6 taken together

Given:

The ratio of number of girls to that of boys is 3 : 4. The average height of students in the class is 4.6 feet.

The average height of the boys in the class is 4.8 fee

Concept Used:`

The ratio of a and b is a₁ : b₁, total average is P, and average of a and b is c and d respectively.

Total average (P) =  (a₁ × c + b₁ × d)/(a₁ + b₁)

Calculation:

Suppose the number of girls & boys in the class are 3a and 4a;

The average height of all students is 4.6 feet and average height of the boys is 4.8 feet.

Suppose average height of the girls is X feet;

So,

4a × 4.8 + 3a × X = 4.6 × (3a + 4a)

19.2 + 3X = 32.2

3X = 13

X = 4.33

∴ Average height of the girls in the class is more than 4.3 feet but less than 4.4 feet.

Given:

The data 3, 5, 9, 4, 6, 11, 18

Formula Used:

Median = [(n + 1)/2]^th terms

Calculation:

Data in ascending order = 3, 4, 5, 6, 9, 11, 18

Total numbers (n) = 7 (Odd)

Median = [(n + 1)/2]^th terms

Median = [(7 + 1)/2]^th terms

Median = 4^th term

And 4^th term is 6

∴ Median of 4^th term is 6

Given:

The ratio of the angles of the sectors is 1 : 2 : 3

Concept Used:

The sum of angles of in a pie chart = 360°

Calculation:

The ratio of the angles of the sectors is 1 : 2 : 3

We know the sum of angles of in a pie chart = 360°

So, largest angle = [3/(1 + 2 + 3)] × 360° = 360°/2 = 180°

∴ The largest angle is 180°.

Given:

Maximum marks are converted from 250 to 50.

 The highest marks scored were 170 and the lowest marks were 70

Calculation:

Maximum marks are converted from 250 to 50.

Highest marks 170 will be converted to: 170 × (50/250) = 34

Lowest marks 70 will be converted to: 70 × (50/250) = 14

Difference = 34 – 14 = 20

∴ The difference is 20.

Given:

Formula used:

If a, b and c is the numbers of student in class

% of a = [a/(a + b + c)] × 100

Calculation:

Percentage of trained convicts:

In jail A: [45/(86 + 45 + 25)] × 100 = [45/156] × 100 = 28.8%

In jail B: [903/(1305 + 903 + 461)] × 100 = [903/2669] × 100 = 33.8%

In jail C: [940/(2019 + 940 + 474)] × 100 = [940/3433] × 100 = 27.3%

In jail D: [869/(1166 + 869 + 416)] × 100 = [869/2451] × 100 = 35.4%

In jail E: [544/(954 + 544 + 254)] × 100 = [544/1752] × 100 = 31.05%

In jail F: [464/(1198 + 464 + 174)] × 100 = [464/1836] × 100 = 25.2%

∴ Jails with highest and smallest percentage of trained convicts are D and F respectively.

Given:

Formula used:

If a, b and c is the numbers of student in class

% of a = [a/(a + b + c)] × 100

Calculation:

Placement rate of trained convicts:

In jail A: [25/45] × 100 = 55.55%

In jail B: [461/903] × 100 = 51.05%

In jail C: [(474/940) × 100] = 50.42%

In jail D: [416/869] × 100 = 47.87%

In jail E: [(254/544) × 100] = 46.7%

In jail F: [174/464] × 100 = 37.5%

∴ Jail with highest placement rate of trained convicts is jail A.

Given:

Formula used:

If a, b and c is the numbers of student in clss

% of a = [a/(a + b + c)] × 100

Calculation:

Placement rate of trained convicts:

In jail A: [25/45] × 100 = 55.55%

In jail B: [461/903] × 100 = 51.05%

In jail C: [474/940] × 100 = 50.42%

In jail D: [416/869] × 100 = 47.87%

In jail E: [254/544] × 100 = 46.69%

In jail F: [174/464] × 100 = 37.50%

∴ From Jails A, B and C, more than half of the trained convicts are offered jobs.

Given:

Three-digit numbers (all digits are different) which are divisible by 7 and also divisible by 7 on reversing the order of the digits

Concept Used:

For divisibility of 7 is the last term is 1, 3, 7 and 9

Calculation:

Suppose the possible number is ‘abc’ then ‘cba’ will also be divisible by 7.

We can write:

100a + 10b + c = 7x

100c + 10b + a = 7y

Subtracting both equations:

99a – 99c = 7(x – y)

99(a – c) = 7(x – y)

Here, ‘99(a – c)’ must be divisible by 7.

Since, 99 is not divisible by 7, (a – c) will be divisible by 7.

Hence, only possible values of (a – c) are 0 and 7.

If (a – c) = 0 then a = c [Which is not possible as all digits are different]

If (a – c) = 7

Possible values of a = 9, 8 & c = 2, 1 [c = 0 is not possible as it will give 2-digit number]

Hence, multiples of 7 in the form of ‘abc’ with condition a = 9, 8 & c = 2, 1 are:

168, 259 & their reverse i.e. 861, 952.

∴ Total four numbers are possible. 

Confusion Points In the question it is mentioned that the digits should not be repeated.

Given:

x and y satisfy the equation 5x + 9y = 7

Formula Used:

In A.P. 

Last term(l) = First term(a) + [Number of terms(n) - 1]⋅Common difference(d)

l = a + (n - 1)d

Then n = [(l - a)/d] + 1

Calculation:

We need to find those integer values of x which will give integer values of y when we put them in the equation.

Put x = -1 y = 12/9 (Not an integer)

Put x = -2 y = 17/9 (Not an integer)

Put x = -4 ⇒ y = 3 (Integer)

Put x = -5 y = 22/9 (Not an integer)

Put x = -13 ⇒ y = 8 (Integer)

Put x = -22 ⇒ y = 13 (Integer)

So here we can observe that values of x (starting with -1) with gap of 9 are giving the integer values for y.

So basically, we need to find the all the multiples of 9 between -500 and 500.

AP will be: -495, -486, …, 486, 495.

Number of terms = [495 + 495]/9 + 1 = 110 + 1 = 111

∴ 111 integral values of x and y will satisfy the equation 5x + 9y = 7.

Calculation:

S = XYZ + YZX + ZXY

S = 100X + 10Y + Z + 100Y + 10Z + X + 100Z + 10X + Y

S = 111X + 111Y + 111Z

S = 111 (X + Y + Z)

We can observe here:

111 is divisible by 3 and 37.

Hence, S will be divisible by (X + Y + Z), 3 and 37 always. 

∴ 1 and 3 are correct

 

On Checking statement 2

take X = 1, Y = 2 & Z = 4 

We get S = 111 (1 + 2 + 4) = 777

Which is not divisible by 9.

Hence, S will not be always divisible by 9.

∴ 2 is incorrect.

Mistake Points 

At first, we have to understand here that 

We can't take x = 0 

If we take any one of them equal to zero then we will not get a three-digit number 

like if we take x = 0, y = 2 and z = 4 

then XYZ = 024 which is a two-digit number 

Some cases also generate divisible by 9, but in all cases not satisfy the condition. the Question said always divisible by 9, but it is not possible for all cases. Like S = 111(1 + 2 + 4) = 777, or S = 111(2 + 3 + 0) = 555 both number is not divisible by 9.

Satisfy cases, S = 111(1 + 2 + 3) = 666, S = 111(2 + 3 + 4) = 999 both number also divisible. 

Confusion PointsX, Y and Z are not given equal to each other, hence we cannot consider this special case to answer the question.

Given:

Average speeds of X and Y are in the ratio 4 : 5

X takes 45 minutes more than Y to reach the destination

Formula Used:

Speed = Distance/Time

Calculation:

Hence, Ratio of time taken by X and Y to reach the destination = 5 : 4

Suppose time taken by X and Y to reach the destination are 5x and 4x respectively.

X takes 45 minutes more than Y to reach the destination.

(5x – 4x) = x = 45

So, 4x = 45 × 4 = 180

∴ Time taken by Y to reach the destination is 180 minutes.

Given: 

The sum is S and product is P.

Formula Used:

Harmonic mean of a and b = 2/(1/a + 1/b) = 2ab/(a + b)

Calculation:

The sum is S and product is P.

Then, the Harmonic mean of these two observations = 2P/S

∴ H.M. of two observations is 2P/S

Given:

Annual income of X is 20% more than that of Y;

The income of Y is less than that of X by p%.

Concept used:

If smaller value is m% less than the greater value,

then the value of m = [(greater value - smaller value)/greater value] × 100

Calculation:

Suppose annual income of Y = 100

So, annual income of X = 100 × 1.2 = 120

Income of Y is less than that of X by p%

Hence,

Required percentage (p) = [(120 – 100)/120] × 100 = $16\frac{2}{3}$%

∴ value of p is  $16\frac{2}{3}$%

Calculation:

First, we need to get the LCM of 3, 4, 5, 6, and 7.

After having lcm of it we gets 3 × 2 × 2 × 5 × 7 = 3 × 4 × 5 × 7 = 420

We see here that 2 has already it's squares but 3, 5, and 7 don't have their squares

Since we have to find the perfect square

Therefore to get the perfect square we must make 3, 5, and 7 in the form of a square.

To have the squares of these three digits we have to multiply the lcm by 3, 5 and 7

After multiplying we get 420 × 3 × 5 × 7 = 44100

Hence this will be the perfect square which will be divisible by the digits 3, 4, 5, 6, and 7.

Given:

Time in which both outlets emptied the tank together = 20 minutes

Time in which 1^st outlet emptied the tank alone = 30 minutes

Concept Used:

1/(A's time) + 1/(B's time) + ...... = 1/Total time

Calculation:

Suppose 2^ndoutlet can empty the tank alone in X minutes;

So,

1/30 + 1/X = 1/20

⇒ 1/X = 1/60

⇒ X = 60

∴ Time taken to empty the tank by second outlet alone in 60 minutes

Given:

If (x² - 1) is a factor of ax⁴ + bx³ + cx² + dx + e

Concept Used:

If (x - a) is a factor of any polynomial equation 

So. x - a = 0, x = a is satisfy that equation

Cslculation:

(x² - 1) is the factor of ax⁴ + bx³ + cx² + dx + e.

(x² - 1) = (x – 1) (x + 1)

So, x = 1, -1 will satisfy the given equation:

Putting x = -1:

a – b + c – d + e = 0

⇒ a + c + e = b + d

∴ Option (4) is corret.

Given:

$\left( {{x^8} + \frac{1}{{{x^8}}}} \right) = 47$

**Formula Used:**If (a + 1/a)2 = P

Then, (a + 1/a) = √(P + 2)

Calculation::

$\left( {{x^8} + \frac{1}{{{x^8}}}} \right) = 47$

So,

(x⁴ + 1/x⁴)² = (47 + 2)

(x4 + 1/x4)​ = 7

and

(x² + 1/x²)² = (7 + 2) 

(x2 + 1/x2) = 3

Taking cube both sides:

$\left( {{x^6} + \frac{1}{{{x^6}}}} \right)$ + 3 × 1 × (x² + 1/x²) = 27

So,

$\left( {{x^6} + \frac{1}{{{x^6}}}} \right)$= 27 – 3 × 3 = 18

∴ The value of $\left( {{x^6} + \frac{1}{{{x^6}}}} \right)$ is 18.

Given:

A wheel makes 360 revolutions in one minute

Concept Used:

In one complete revolution radian covered = 2π

Revolutions in 1 minute or 60 seconds = 360

So, revolutions in 1 second = 360/60 = 6

Calculation:

In one complete revolution radian covered = 2π

∴ In 6 revolutions radian covered = 6 × 2π = 12π

∴ Number of radians the wheel turns in one second is 12π

Given: 

25 cosec²x + sec²x

Formula Used:

Cosec²x = 1 + Cot²x

Sec²x = 1 + Tan²x

Calculation:

25 cosec2x + sec2x

⇒ 25(1 + Cot²x) + (1 + Tan²x)

⇒ 26 + (25Cot²x + Tan²x)

∴ Minimum value = 26 + 2√(25 × 1) = 26 + 10 = 36

∴ Minimum value is 36

Given:

$\alpha = {\rm{\Pi }}_{n = 1}^{99}\cot n\theta$

⇒ α = Cotθ × Cot2θ × Cot3θ × Cot4θ × ………….. × Cot 97θ × Cot 98θ × Cot 99θ

And 90° = 100θ

Formula Used:

Tanθ = Cot(90° – θ)

Calculation:

Tan θ = Cot (100θ – θ) = Cot 99θ

Tan 2θ = Cot (100θ – 2θ) = Cot 98θ

And so on.

Hence, we can write the above expression as:

⇒ α  = Cotθ × Cot2θ × Cot3θ × Cot4θ × …… Cot 49θ × Cot 50θ × Tan 49θ ………. × Tan 3θ × Tan 2θ × Tan θ

⇒ α  = Cot 50θ

Since, 100θ = 90°

So, 50θ = 45°

⇒ α = Cot 45° = 1

∴ The value of $\alpha = {\rm{\Pi }}_{n = 1}^{99}\cot n\theta$ is 1.

Given:

tan6θ = cot2θ

Formula Used:

cot θ = tan (90° - θ)

Calculation:

tan6θ = tan(90° – 2θ)

⇒ 6θ = 90° – 2θ

⇒ 8θ = 90°

⇒ 4θ = 45° [Here 6θ = 67.5° which satisfies the given condition] 

∴ The valure of sec 45° is $\sqrt 2$.

Given:

Height = 15 m

Makes angle = 30° 

Formula Used:

sin θ = perpendicular/hypotenius

sin 30° = 1/2

Calculation:

From the figure:

In triangle ABC:

Sin30° = AB/BC

⇒ 1/2 = AB/BC

⇒ BC = 2AB

Total height of the tree = AB + BC = 15 m

Hence, 3AB = 15

⇒ AB = 5 m

∴ The height from the ground to the point where tree is broken is 5 m

Given:

Two vertical towers distance =100 feet

The distance of the pole from the shorter tower = 75

The shorter tower length = 40 feet 

 A pole length = 6 feet

Formula Used:

tan θ = Perpendicular/Base

Calculation:

In triangle FGH:

GH = 40 – 6 = 34 feet

HF = DE = 75 feet

So, tanx = GH/HF = 34/75

In triangle ABF:

BF = CE = 100 + 75 = 175 feet

So, tanx = AB/BF = AB/175

Hence,

34/75 = AB/175

⇒ AB = 7/3 × 34 = 79.33 feet

∴ Height of taller tower = AC = 79.33 + 6 = 85.33 or 85 feet (Approximately)

Shortcut Trick

If the angles of elevation from two points at distances p and q from the base of a tower are complementary (θ₁ + θ₂ = 90°), the height (h) is always the geometric mean of the distances.

Given angles: 27° and 63° ⇒ Sum = 27° + 63° = 90°.

Shortcut Formula: h = √(p × q)

∴ The correct answer is √pq.

Alternate Method

Given: Distances from base are p and q. Angles are 27° and 63°.

Formula Used: tan θ = Perpendicular / Base

⇒ In first triangle: tan 27° = h / q

⇒ In second triangle: tan 63° = h / p

⇒ Since 63° = 90° − 27°, tan 63° = cot 27°

⇒ cot 27° = h / p ⇒ tan 27° = p / h

⇒ Equating tan 27°: h / q = p / h

⇒ h² = p × q

⇒ h = √pq

∴ The correct answer is √pq.

Additional Information

Complementary Angles Rule

If α + β = 90°, then tan α × tan β = 1. This property is extensively used in height and distance problems involving two points.

Geometric Mean in Trigonometry

When the product of two trigonometric ratios is constant, the unknown side (height) often results in a geometric mean √ab of the given lengths.

Trigonometric Identities

tan(90° − θ) = cot θ and cot θ = 1 / tan θ are the core identities for solving complementary angle problems.

Given:

sin2 6° + sin2 12° + sin2 18° + … + sin2 84° + sin2 90°

Formula Used:

sin (90° - θ) = cos θ

Calculation:

sin² 6° + sin² 12° + sin² 18° + … + sin² 78° + sin² 84° + sin² 90°

⇒ sin² 6° + sin² 12° + sin² 18° + … + sin² (90 – 12) + sin² (90 – 6) + sin² 90°

⇒ sin² 6° + sin² 12° + sin² 18° + … + Cos² 12° + Cos² 6° + sin² 90°

⇒ (sin² 6° + cos² 6°) + (sin² 12° + cos² 12°) + ………… (sin² 42° + cos² 42°) + sin² 90°

⇒ 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

⇒ 8

∴ The value of sin2 6° + sin2 12° + sin2 18° + … + sin2 84° + sin2 90° is 8.

Given:

$\frac{{\cos θ }}{{1 + \sin θ }} + \frac{1}{{\cot θ }}$

Formula USed:

cot θ = cos θ/sin θ 

sin² θ + cos² θ = 1

Calculation:

$\frac{{\cos θ }}{{1 + \sin θ }} + \frac{1}{{\cot θ }}$

⇒ $\frac{{\cos θ }}{{1 + \sin θ }} + \frac{{sin;θ }}{{\cos θ }}$

⇒ $\frac{{co{s^2}θ + sinθ + ;si{n^2}θ }}{{(1 + \sin θ );cosθ }}$

⇒ $\frac{{1 + sinθ }}{{(1 + \sin θ );cosθ }}$

⇒ 1/cosθ

⇒ secθ

∴ The value of $\frac{{\cos θ }}{{1 + \sin θ }} + \frac{1}{{\cot θ }}$ is sec θ.

Given:

$\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} - \frac{{\sin \theta + 1}}{{\cos \theta }}$

Calculation:

Let x = $\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}}$ and y = $\frac{{\sin \theta + 1}}{{\cos \theta }}$

x = $\frac{{\sin \theta - \cos \theta + 1}}{{(\sin \theta + \cos \theta) - 1}}\times\frac{{\sin \theta +\cos \theta + 1}}{{(\sin \theta + \cos \theta) +1}}$

x = $\frac{{(\sin \theta+ 1) - \cos \theta }}{{(\sin \theta + \cos \theta) - 1}}\times\frac{{(\sin \theta + 1)+\cos \theta }}{{(\sin \theta + \cos \theta) +1}}$

x = $\frac{{(\sin \theta+ 1)^2 - \cos^2\theta }}{{(\sin \theta + \cos \theta)^2- 1^2}}$

x = $\frac{{(\sin^2 \theta+ 1+ 2\sin \theta) - \cos^2\theta }}{{(\sin^2 \theta + \cos^2 \theta+ 2\sin \theta \cos\theta)- 1}}$

x = $\frac{{(\sin^2 \theta+ 1+ 2\sin \theta) -(1- \sin^2\theta )}}{{(1+ 2\sin \theta \cos\theta)- 1}}$

x = $\frac{{2\sin^2 \theta+2\sin \theta}}{{ 2\sin \theta \cos\theta}}$

x = $\frac{{2\sin \theta(1+\sin \theta)}}{{ 2\sin \theta \cos\theta}}$

x = $\frac{{1+\sin \theta}}{{ \cos\theta}}$

and y = $\frac{{\sin \theta + 1}}{{\cos \theta }}$

∴ x - y = $\frac{{1+\sin \theta}}{{ \cos\theta}}$ - $\frac{{\sin \theta + 1}}{{\cos \theta }}$ = 0

∴ The value of $\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} - \frac{{\sin \theta + 1}}{{\cos \theta }}$ is 0.