CDS Elementary Mathematics 10 April 2022 Official Paper (Previous Year Paper)

Formula Used:

Sum of interior angles of a polygon of n sides = (n - 2) × 180° 

Sum of exterior angles of a polyon of n sides = 360° 

Calculation:

According to the question 

Required ratio = $\frac{(n - 2) × 180^0}{360^0}$

⇒ Required ratio = $\frac{(n - 2)}{2}$

∴ The ratio of interior and exterior angles of a polygon of n side is  $\frac{(n - 2)}{2}$.

Shortcut Trick 

If the number is in the form of an + bn and n is an odd number then 

the number an + bn must be divisible by (a + b)

So, the number 4143 + 4343 must be divisible by 41 + 43, that is 84

Alternate Method 

Concept Used:

When n is an odd number then

aⁿ + bⁿ = (a + b)(aⁿ⁻¹ b⁰ − aⁿ⁻² b +...........a. bⁿ⁻² + a⁰ bⁿ⁻¹)

Calculation:

We have 4143 + 4343

Since the given number is of the form an + bn

And 43 is an odd number 

So, 4143 + 4343 can be written as

⇒ 4143 + 4343 = (41 + 43) (4143−1 430 − 4143−2 43 +...........41. 4343−2 + 410 4343−1)

⇒ 4143 + 4343 = 84 (4143−1 430 − 4143−2 43 +...........41. 4343−2 + 410 4343−1)

∴ We can say that 4143 + 4343 is completely divisible by 84.

Given:

$x, = ,7, + ,4√ 3$

Formula Used:

a² + b² + 2ab = (a + b)²

Calculation:

We have 

x = 7 + 4√3

⇒ x = 4 + 3 + 2. 2 .√3

⇒ x = (2)² + (√3)² + 2.2.√3

⇒ x = (2 + √3)2

⇒ √x = 2 + √3      -----(i)

So, $\frac{1}{\sqrt {x}} = \frac{1}{2 + \sqrt 3}$ 

After rationalizing we get 

$\frac{1}{\sqrt x{}} = 2 - \sqrt 3$         -----(ii) 

On adding equation (i) and (ii), we get 

$\sqrt x , + ,\frac{1}{{\sqrt x }}$=  2 + √3 +  2 - √3   

∴  $\sqrt x , + ,\frac{1}{{\sqrt x }} = 4$.

Given:

The number is 461 + 462 + 463 + 464

Concept Used:

If n = a × b × c ...

Then n must be divisible by a, b, c, .....

Calculation:

We have 461 + 462 + 463 + 464

Taking 4⁶¹ common from this number, we get

⇒ 4⁶¹ (1 + 4¹ + 4² + 4³)

⇒ 461 (1 + 4 + 16 + 64)

⇒ 461 (85)

⇒ 461 × 5 × 17

So, We can write 

461 + 462 + 463 + 464 = 461 × 5 × 17

So, the number can be divisible by 5, 17 and 4ⁿ {n ≤ 61}

From option we can say that 

∴ 461 + 462 + 463 + 464 is divisible by 17.

Shortcut Trick

Given p : q = 14 : 1 and pq = 1134.

Substitute 14q for p: 14q × q = 1134 ⇒ q² = 81 ⇒ q (HCF) = 9.

Then p (LCM) = 14 × 9 = 126.

Let numbers be 9a and 9b, where a × b = 14 (LCM/HCF). Factors of 14 are (1, 14) or (2, 7).

Differences: 9(14 − 1) = 117 or 9(7 − 2) = 45. Since 45 is in the options, it is the intended answer.

∴ The correct answer is 45.

Alternate Method

Step 1: Find HCF and LCM values

⇒ p = 14q

⇒ (14q) × q = 1134

⇒ 14q² = 1134

⇒ q² = 1134 ÷ 14 = 81

⇒ q = 9 (HCF) and p = 126 (LCM)

Step 2: Find the numbers

⇒ Let the numbers be x and y. x × y = p × q = 1134.

⇒ Since HCF is 9, let x = 9a and y = 9b where a and b are co-prime.

⇒ 9a × 9b = 1134 ⇒ 81ab = 1134 ⇒ ab = 14.

Step 3: Evaluate pairs and difference

⇒ Pair 1: a=1, b=14. Numbers = (9, 126). Difference = 126 − 9 = 117.

⇒ Pair 2: a=2, b=7. Numbers = (18, 63). Difference = 63 − 18 = 45.

⇒ Comparing with given options, 45 is the only available valid difference.

∴ The correct answer is 45.

Additional Information

Product of Numbers Rule

For any two positive integers x and y, x × y = HCF(x, y) × LCM(x, y).

LCM and HCF Relationship

The HCF is always a factor of the LCM. The ratio LCM/HCF gives the product of the co-prime factors of the numbers.

Difference Property

The difference between two numbers x and y, both having HCF 'q', is always a multiple of 'q'. In this case, Difference = 9(b − a).

Shortcut Trick

The given expression follows the algebraic pattern (a³ − b³) ÷ (a² + ab + b²).

This expands and simplifies directly to (a − b).

Here, a = 5.4 and b = 0.4 (since 0.4³ = 0.064).

Required value = 5.4 − 0.4 = 5.

∴ The correct answer is 5.

Alternate Method

Given:

Expression: (5.4³ − 0.064) ÷ (5.4² + 2.16 + 0.16)

Formula Used:

a³ − b³ = (a − b)(a² + ab + b²)

Calculations:

⇒ Let a = 5.4 and b = 0.4

⇒ 0.064 = (0.4)³ = b³

⇒ 2.16 = 5.4 × 0.4 = ab

⇒ 0.16 = (0.4)² = b²

⇒ Now substitute a and b into the expression:

⇒ Expression = (a³ − b³) ÷ (a² + ab + b²)

⇒ Expression = [(a − b)(a² + ab + b²)] ÷ (a² + ab + b²)

⇒ The term (a² + ab + b²) cancels out from the numerator and denominator.

⇒ Expression = a − b

⇒ Substitute the values of a and b back into the result:

⇒ Result = 5.4 − 0.4 = 5

∴ The correct answer is 5.

Additional Information

Sum of Cubes Identity

a³ + b³ = (a + b)(a² − ab + b²). When an expression has a plus sign in the numerator matching this form, it simplifies to just a + b.

Cube of a Binomial

(a + b)³ = a³ + b³ + 3ab(a + b) and (a − b)³ = a³ − b³ − 3ab(a − b).

Difference of Squares

a² − b² = (a − b)(a + b). This foundational identity is highly useful for directly canceling out terms in division problems.

Given:

If the man walks @ 3 km/hr, he reaches 40 min early

If the man walks @ 2 km/hr, he reaches 40 min late

Formula Used:

Total distacne = Speed × Time 

Caluclation:

Let the distance between residence and office be D 

The usual speed is S km/hr and usual time be T hr

Now, According to the question 

D = 3 (T - 40/60)      -----(ii)

And D = 2 (T + 40/60)    ----(iii)

From (ii) and (iii),we get 

3 (T - 40/60) = 2 (T + 40/60)

⇒ 3T - 2 = 2T + 4/3

⇒ T = (2 + 4/3) hr

⇒ T = 10/3 hr

From (ii)

D = 3 [(10/3) - (40/60)] km

⇒ D = 3 [(10/3) - (2/3)] km

⇒ D = 3 (8/3) km

⇒ D = 8 km

∴ The distance between residence and office is 8 km.

Given:

$\frac{1}{{5\sqrt 4 , + ,4\sqrt 5 }}, + ,\frac{1}{{6\sqrt 5 , + ,5\sqrt 6 }}, + ,\frac{1}{{7\sqrt 6 , + ,6\sqrt 7 }}, + ,\frac{1}{{8\sqrt 7 , + ,7\sqrt 8 }}, + ,\frac{1}{{9\sqrt 8 , + ,8\sqrt 9 }}$

Concept used:

(a + b)(a - b) = a2 - b2

Calculation:

Consider the first term only, i.e. $\frac{1}{{5\sqrt 4 , + ,4\sqrt 5 }},$

Rationalize the denominator

⇒ $\frac{1}{{5\sqrt 4 , + ,4\sqrt 5 }},\times \frac{5\sqrt 4 , - ,4\sqrt 5 }{{5\sqrt 4 , - ,4\sqrt 5 }},$

⇒ $\frac{5\sqrt 4 , - ,4\sqrt 5}{{(5\sqrt 4)^2 , - ,(4\sqrt 5)^2 }},$

⇒ $\frac{5\sqrt 4 , - ,4\sqrt 5}{{100-80 }}, = \frac{5\sqrt 4}{20} - \frac{4\sqrt5}{20}$

⇒ $\frac{1}{\sqrt 4} - \frac{1}{\sqrt 5}$

After doing a similar operation in other terms, we will get

$\frac{1}{\sqrt 4} - \frac{1}{\sqrt 5}+\frac{1}{\sqrt 5} - \frac{1}{\sqrt 6}+\frac{1}{\sqrt 6} - \frac{1}{\sqrt 7}+\frac{1}{\sqrt 7} - \frac{1}{\sqrt 8}+\frac{1}{\sqrt 8} - \frac{1}{\sqrt 9}$

⇒ $\frac{1}{\sqrt 4} - \frac{1}{\sqrt 9}$

⇒ $\frac{1}{2} - \frac{1}{3}$

⇒ $\frac{3;-;2}{6}$

⇒ $\frac{1}{6}$

∴ Required value of the expression is $\frac{1}{6}$

Given:

x = 9999

Formula Used:

a² - b² = (a - b) (a + b)

Calculation:

To find $\frac{{4{x^3}, - ,x}}{{\left( {2x, + ,1} \right)\left( {6x, - ,3} \right)}}$

⇒ $\frac{{x \ (4x^2 - 1)}} {{\left( {2x, + ,1} \right)\left( {6x, - ,3} \right)}}$

⇒ $\frac{{x \ [ \ (2x)^2 \ - \ (1)^2 \ ]}} {{\left( {2x, + ,1} \right)\left( {6x, - ,3} \right)}}$

⇒ $\frac{{x \ (2x - 1) (2x + 1)}} {{3\left( {2x, + ,1} \right)\left( {2x, - ,1} \right)}}$

⇒ x/3

⇒ 9999/3 = 3333

∴ The value of the expression at x = 9999 is 3333.

Given:

$\left( {x, + ,\sqrt {1,, + ,,{x^2}} } \right)\left( {y, + ,\sqrt {1,, + ,,{y^2}} } \right), = ,1$

Concept used:

A² - B² = (A + B) (A - B)

(A + B)² = A² + B² + 2AB

(A - B)2 = A2 + B2 - 2AB

Shortcut TrickGiven that

$\left( {x, + ,\sqrt {1,, + ,,{x^2}} } \right)\left( {y, + ,\sqrt {1,, + ,,{y^2}} } \right), = ,1$

Put x = 0 and y = 0 (Real number)

⇒ $\left( {0, + ,\sqrt {1,, + ,,{0^2}} } \right)\left( {0, + ,\sqrt {1,, + ,,{0^2}} } \right), = ,1$

⇒ 1 × 1 = 1  satisfied.

Hence, value (x + y)² at x = 0 and y = 0

⇒ (0 + 0)² = 0

∴ (x + y)² = 0 

Alternate Method

$\left( {x, + ,\sqrt {1,, + ,,{x^2}} } \right)\left( {y, + ,\sqrt {1,, + ,,{y^2}} } \right), = ,1$

⇒ $({x\ +\ \sqrt{1\ +\ x^2}})$ = $1 \over ({y\ +\ \sqrt{1\ +\ y^2}})$

⇒ $({x\ +\ \sqrt{1\ +\ x^2}})$ = $({\sqrt{1\ +\ y^2}\ -\ y}) \over ({\sqrt{1\ +\ y^2}\ +\ y}) × ({\sqrt{1\ +\ y^2}\ - y})$

⇒ $({x\ +\ \sqrt{1\ +\ x^2}})$ = $({\sqrt{1\ +\ y^2}\ -\ y}) \over {(\sqrt{1\ +\ y^2})^2\ -\ y^2}$

⇒ $({x\ +\ \sqrt{1\ +\ x^2}})$ = $({\sqrt{1\ +\ y^2}\ -\ y}) \over {(1\ +\ y^2)\ -\ y^2}$

⇒ $({x\ +\ \sqrt{1\ +\ x^2}})$ = $({\sqrt{1\ +\ y^2}\ -\ y}) \over 1$

⇒ $({x\ +\ \sqrt{1\ +\ x^2}})$ = $({\sqrt{1\ +\ y^2}\ -\ y})$

⇒ $x\ +\ y$ = ${\sqrt{1\ +\ y^2}}\ -\ \sqrt{1\ +\ x^2}$

⇒ $(x\ +\ y)^2$ = $({\sqrt{1\ +\ y^2}}\ -\ \sqrt{1\ +\ x^2})^2$ (Squaring both side)

⇒ ${x^2\ +\ y^2\ + 2xy}$ = ${1\ + y^2\ +\ 1\ +\ x^2\ - 2 {\sqrt{1\ +\ y^2}\ {\sqrt{1\ +\ x^2}}}}$

⇒ ${xy}$ = ${1\ -\ {\sqrt{1\ +\ y^2}\ {\sqrt{1\ +\ x^2}}}}$

⇒ ${\sqrt{1\ +\ y^2}\ {\sqrt{1\ +\ x^2}}}$ = ${1\ -\ xy}$

⇒ $({\sqrt{1\ +\ y^2}\ {\sqrt{1\ +\ x^2}}})^2$ = $({1\ -\ xy})^2$ (Squaring both side)

⇒ $(1\ +\ y^2)\ (1\ +\ x^2)$ = ${1\ +\ x^2y^2\ -2xy}$

⇒ ${x^2y^2\ +\ x^2\ +\ y^2\ +\ 1}$ = ${1\ +\ x^2y^2\ -2xy}$

⇒ ${x^2\ +\ y^2\ +\ 2xy}$ = $0$

⇒ $(x\ +\ y)^2$ = 0

∴ The value of (x + y)2 is 0.

Concept:

Using trigonometric ratios of complementary angles:-

• sin(90° - θ) = cosθ
• cos(90° - θ) = sinθ
• tan(90° - θ) = cotθ
• cot(90° - θ) = tanθ
• cotθ = $\frac{1}{tanθ}$

Calculation:

 $\frac{{2\sin ,68^\circ }}{{\cos 22^\circ }}, - ,\frac{{2\cot ,15^\circ }}{{5\tan 75^\circ }} - \frac{{3\tan 20^\circ \tan 40^\circ \tan 45^\circ \tan 50^\circ \tan 70^\circ }}{5}$

⇒  $\frac{2sin68°}{cos(90° - 68°)} - \frac{2cot15°}{5tan(90° - 15°)} - \frac{3tan(90° - 70°) tan(90° - 50°) tan45° tan50° tan70°}{5}$

⇒ $\frac{2sin68°}{sin68°} - \frac{2cot15°}{5cot15°} - \frac{3 tan45° cot70° cot50° tan50° tan70°}{5}$

⇒ $\frac{2sin68°}{sin68°} - \frac{2cot15°}{5cot15°} - \frac{3 tan45°. \frac{1}{tan70°}. \frac{1}{tan50°}. tan50°. tan70°}{5}$

⇒ $\frac{2sin68°}{sin68°} - \frac{2cot15°}{5cot15°} - \frac{3 tan45°}{5}$

⇒ $2 - \frac{2}{5} - \frac{3\times 1}{5}$    [∵ tan 45° = 1]

⇒ $2 - \frac{5}{5}$

⇒ 2 - 1 = 1

∴ The value of $\frac{{2\sin ,68^\circ }}{{\cos 22^\circ }}, - ,\frac{{2\cot ,15^\circ }}{{5\tan 75^\circ }} - \frac{{3\tan 20^\circ \tan 40^\circ \tan 45^\circ \tan 50^\circ \tan 70^\circ }}{5}$  is 1.

Concept Used:

In a right-angled triangle ABC, If a perpendicular BD is drawn on the hypotenuse AC from the right-angled vertex then 

BD² = AD × CD 

Calculation:

According to the question 

AD = 9 units and 

CD = 16 units 

And we have to find the value of BD 

Now, According to the concept used

BD2 = AD × CD 

⇒ BD2 = 9 × 16 = 144

⇒ BD2 = 12²

⇒ BD = 12 units 

∴ The required length of the perpendicular is 12 units.

Given:

43x × 47y = (2021)2, x ≠ 0, y ≠ 0

To find $\frac{{4xy, + ,x, + ,y}}{{2xy, - ,x, - ,y}}$

Calculation:

We have 43x × 47y = (2021)2

On factorization of 2021 we get 

2021 = 43 × 47

So, we can write the equation (i) as

43x × 47y = (43 × 47)2

⇒ 43x × 47y = 432 × 472 

On comparing we get 

x = 2 and y = 2

Now we have to find the value of $\frac{{4xy, + ,x, + ,y}}{{2xy, - ,x, - ,y}}$

⇒ $\frac{4 \times 2 \times 2 + 2 + 2}{2 \times 2 \times 2- 2 - 2}$ = $\frac{20}{4} = 5$  

∴ The required value is 5.

Given: 

$\frac{1}{{{\rm{a}},{\rm{ + }},\frac{1}{{{\rm{b}},{\rm{ + }}\frac{1}{{{\rm{c}},{\rm{ + }},\frac{1}{{\rm{d}}}}}}}}} = ,,\frac{{17}}{{60}}$

Calculation:

We can write the RHS $\frac{17}{60}$ as 

⇒ $\frac{1}{\frac{60}{17}}$ = $\frac{1}{3+ \frac{9}{17}}$ = $\frac{1}{3+ \frac{1}{\frac{17}{9}}}$

⇒ $\frac{1}{3+ \frac{1}{1 + \frac{8}{9}}}$ = $\frac{1}{3+ \frac{1}{1 + \frac{1}{\frac{9}{8}}}}$ = $\frac{1}{3+ \frac{1}{1 + \frac{1}{1 + \frac{1}{8}}}}$

Now, On comparing 

$\frac{1}{{{\rm{a}},{\rm{ + }},\frac{1}{{{\rm{b}},{\rm{ + }}\frac{1}{{{\rm{c}},{\rm{ + }},\frac{1}{{\rm{d}}}}}}}}}$ by $\frac{1}{3+ \frac{1}{1 + \frac{1}{1 + \frac{1}{8}}}}$

We get a = 3, b = 1, c = 1 and d = 8

So, abcd = 3 × 1 × 1 × 8 = 24

∴ The product of a, b, c, d is 24.

Given:

x2 = 17x + y and y2 = x + 17y, x ≠ y

Formula Used:

a² - b² = (a - b) (a + b)

Calculation:

We have 

x2 = 17x + y      -----(i)

y2 = x + 17y       -----(ii)

On applying (i) - (ii), we get 

x2 - y2 = 16x - 16y

⇒ (x - y) (x + y)  = 16(x - y)

⇒ (x + y) = 16      ----(iii)

Now, On applying (i) + (ii), we get 

x2 + y2 = 18x + 18y

⇒ x2 + y2 = 18(x + y)

⇒ x 2 + y2 = 18 × 16      [from (i)]

⇒ x 2 + y2 = 288    ----(iv)

Now, we have to find the value of $\sqrt {{x^2}, + ,{y^2}, + ,1}$

So, from equation (iv), we get 

⇒ $\sqrt {288 + 1} = \sqrt{289} = 17$

∴ The required value of $\sqrt {{x^2}, + ,{y^2}, + ,1}$ is 17.

Given :

194480 + n = m4

Concept used :

When we square a number, on squaring the unit digit can only be 0, 1, 4, 5, 6, 9

Calculations :

We know that,

On squaring,  the unit digit can only be 0, 1, 4, 5, 6, 9

Thus, a number of squares never have 2, 3, 7, and 8 as unit digit

Equation (1) can be written as 

194480 + n = (m²)²       ----- (1)

When we again square the number  0, 1, 4, 5, 6, 9, we get unit digit as 0, 1, 6, 5 only.

Putting n = 1 in equation (1), we get

⇒ 194480 + 1 = (m2)² 

⇒ (√194481) = m²

⇒ √ 441 = m

⇒ m = 21

∴ The least value of n is 1.

Given:

$\frac{{x - y}}{{x√ y , + ,y√ x }}, = ,\frac{1}{{√ x }};\left( {x, > ,0,y, > ,0} \right)$

Formula Used:

a² - b² = (a + b) (a - b)

Calculation:

We have$\frac{{x - y}}{{x√ y , + ,y√ x }}, = ,\frac{1}{{√ x }}$

⇒ $\frac{{(√ x)^2 - (√ y)^2}}{{√ x √ x√ y , + , √ y √ y√ x }}, = ,\frac{1}{{√ x }}$

⇒ $\frac{(√ x + √ y) (√ x - √ y)}{{√ x .√ y \ (√ x , + , √ y) }}, = ,\frac{1}{{√ x }}$

⇒ $\frac{√ x - √ y}{√ x . √ y} = \frac{1}{√ x}$

⇒ √x - √y = √y 

⇒ √x = 2√y 

On squaring both side, we get 

⇒ x = 4y 

⇒ x/y = 4

∴ The required value of x/y is 4.

Given:

Three equal circles of radius 4 cm touch each other 

Formula Used:

Area of circle = πr2

Area of equilateral triangle = √3a²/4

Where, r = radius of circle and 

a = side of an equilateral triangle 

Calculation:

According to the question 

We can draw figure like this 

Now, From figure, we can say that 

AB = BC = CA = 8 cm

So, ΔABC is an equilateral triangle 

And we know that angles of equilateral triangle is 60° 

Now, Area of shaded region = Area of ΔABC - Area of sectors formed in three circles 

⇒ Area of shaded region = $\frac{\sqrt3}{4} (8)^2 - 3 \times \frac{60^0}{360^0} \pi (4)^2$

⇒ $16{\sqrt3}- 8 \pi$  square cm

∴ The area of shaded region is $16{\sqrt3}- 8 \pi$  square cm.

Given:

A car travels from A to B at a speed of 40 km/hr

Travels back from B to A at a speed of 30 km/hr

And again goes from A to B at a speed of 60 km/hr

Formula Used:

Speed = Distance/Time

Average speed = Total distance/Total time 

Calculation:

Let the distance between A and B is d km 

According to the question 

Total distance = d + d + d = 3d 

Total time = $\frac{d}{40} + \frac{d}{30} + \frac{d}{60}$

Now, Average speed = $\frac{3d}{\frac{d}{40} + \frac{d}{30} + \frac{d}{60}}$

⇒ $\frac{3d}{\frac{3d + 4d + 2d}{120}}$ = $\frac{3d \times 120}{9d}$ = 40 km/hr

∴ The average speed of car in the whole journey is 40 km/hr.

Given:

Given number is 9410

Concept used:

Long division method to find the square root

Calculation:

So, the number should be 9409 which is 97²

The smallest natural number which must be subtracted is 9410 - 9409 = 1

∴ The required answer is 1

Alternate Method ⇒ 95² < 9410 < 100²

⇒ 97² = 9409

∴ The perfect square = 9410 - 1 = 9409

Shortcut Trick Convert 27 to base 3: 275 = (33)5 = 315 Expression: 315 + 313 Factor out the lowest power: 313(32 + 1) Simplify the bracket: 313 × (9 + 1) = 313 × 10 The result is a multiple of 10, thus it is divisible by 10. ∴ The correct answer is 10. Alternate Method Given: 275 + 313 Formula Used: (am)n = am × n am + an = an(am−n + 1), where m > n Calculation: ⇒ 275 + 313 ⇒ (33)5 + 313 ⇒ 315 + 313 ⇒ 313(315−13 + 1) ⇒ 313(32 + 1) ⇒ 313(9 + 1) ⇒ 313 × 10 ⇒ Since 10 is a factor of the resulting expression, it must be divisible by 10. ∴ The correct answer is 10. Additional Information Laws of Indices Key rules include am × an = am+n and (am)n = am×n, which allow simplification of large exponential terms to a common base. Divisibility Rule of 10 A number is divisible by 10 if its units digit is 0. Any integer multiplied by 10 results in a number divisible by 10. Factoring Exponential Expressions When adding or subtracting powers of the same base, always factor out the term with the lowest exponent to simplify the constant factor.

Given :

p = 22n + 2 + m and q = 24n - m

n is even natural number

Formula used :

(M^ab) = (M^a)^b       

(M^{a + b}) = (M^a.M^b)      

Calculations :

If p is divisible by 5, we have

⇒ p/5 = (22n + 2 + m)/5

Using the formula avove

⇒ ((2²)ⁿ × 2² + m)/5

⇒ (4ⁿ × 4 + m)/5

When 4 is divided by 5 it leaves - 1 as a negative remainder, so

⇒ (-1ⁿ × 4 + m)/5

n is an even positive number 

⇒  (1 × 4 + m)/5

⇒ (4 + m)/5

The least number required for m is 1

⇒ (4 + 1)/5 =  5/5 = 1

Also, we have q = 24n - m

⇒ q = 16ⁿ - m

Since n is an even positive integer. Therefore,

(16 - m) will always be one factor of q.  

If m = 1  then q = 16 - 1 = 15

Which is divisible by 5.

∴ The least value of m such that p, as well as q, is divisible by 5 is 1.

Calculation:

Given number is x; x > 0

According to the question 

x² = x + 12

⇒ x2 - x - 12 = 0

⇒ x2 - 4x + 3x - 12 = 0

⇒ x(x - 4) + 3(x - 4) = 0

⇒ (x - 4) (x + 3) = 0

⇒ x = 4 or x = - 3

Since x > 0 

∴ The correct value of x is 4.

Given:

$\frac{1}{{1, + ,\sqrt 2 }}, + ,\frac{1}{{\sqrt 2 , + ,\sqrt 3 }}, + ,\frac{1}{{\sqrt 3 , + ,\sqrt 4 }}, + ,..., + ,\frac{1}{{\sqrt {2020} , + ,\sqrt {2021} }}$

Concept used:

(a + b)(a - b) = a2 - b2

Calculation:

Consider the first term only, i.e. $\frac{1}{{1, + ,\sqrt 2 }},$

Rationalize the denominator

⇒ $\frac{1}{{1, + ,\sqrt 2 }},\times \frac{\sqrt 2 , - ,1 }{{\sqrt 2 , - ,1 }},$

⇒ $\frac{\sqrt 2 , - ,1 }{{(\sqrt 2)^2 , - ,(1)^2 }},$

⇒ $\sqrt2-1$

After doing a similar operation in other terms, we will get

$\sqrt2-1+\sqrt3-\sqrt2+\sqrt4-\sqrt3;.......+\sqrt{2021} - \sqrt{2020}$

⇒ $\sqrt {2021} - 1$

∴ Required value of the expression is $\sqrt {2021} - 1$

Given:

$x,, + ,,\frac{1}{x} = ,\frac{5}{2}$     ----(i)

Formula Used:

(a + b)² = a² + b² + 2ab

Calculation:

We have to find the value of $\frac{{5x}}{{7{x^2}, - ,3x, + ,7}}$

Taking x common from the denominator, we get 

$\frac{{5x}}{{x(7{x}, - ,3, + ,\frac{7}{x})}}$

⇒ $\frac{{5}}{{(7{x}, , + ,\frac{7}{x} - ,3)}}$

⇒ $\frac{{5}}{{(7[{x}, , + ,\frac{1}{x}] - ,3)}}$

From (i)

⇒ $\frac{5}{7 \ \times \frac{5}{2} - 3} = \frac{10}{35 - 6}$

⇒ $\frac{10}{29}$

∴ The required value of the expression is $\frac{10}{29}$.

Given:

a + b = 2, $\frac{1}{{\rm{a}}}, + ,\frac{1}{{\rm{b}}}, = ,2$ 

Formual Used:

a3 + b3 = (a + b) (a² + b² - ab)

Calculation:

We have a + b = 2    ----(i)

And  $\frac{1}{{\rm{a}}}, + ,\frac{1}{{\rm{b}}}, = ,2$ 

⇒ a + b = 2ab

⇒ 2 = 2ab      [from (i)]

⇒ ab = 1    ----(ii)

Squaring the equation (i), we get 

a2 + b2 + 2ab = 4

⇒ a2 + b2 + 2 × 1 = 4

⇒ a2 + b2  = 2       ----(iii) 

According to the formula used 

a3 + b3 = (a + b) (a2 + b2 - ab)

And from equation (i), (ii) and (iii), we get

⇒ a3 + b3 = 2 (2 - 1)

⇒ a3 + b3 = 2

∴ The required value of  a3 + b3 is 2.

Given:

8 men can do a piece of work in 24 days

12 women can do a piece of work in 24 days

Concept Used:

Work = Efficiency × Time 

Calculation:

According to the question,

8 Men = 12 Women

⇒ Men/Women = 12/8

⇒ Men/Women = 3/2

Total work = Men × day 

⇒ 8 Men × 24 

⇒ 8 × 3 × 24 = 576 unit

Work be done by 8 men and 12 women

⇒ 576/(8 × 3 + 12 × 2)

⇒ 12 days.

∴ The required time to complete the work by 8 Men and 12 Women together is 12 days

Alternate Method

Since 8 men can do a piece of work in 24 days

And 12 women can do a piece of work in 24 days

Efficiency of 8 Men = Efficiency of 12 Women = x (let)

Now, Total work = x × 24 = 24x units    ----(i)

Now, (8 Men + 12 Women) together work

So, Combined efficiency = x + x = 2x    ----(ii)

From (i) and (ii), we get 

Total time to complete the work by (8 M + 12 W) = 24x ÷ 2x = 12 days 

∴ The required time to complete the work by 8 Men and 12 Women together is 12 days.

Given:

A car takes p minutes to travel 350 km with an average speed of u km/hr.

Another car takes q minutes to travel 350 km with an average speed of v km/hr.

u - v = 5         ----(i)

q - p = 140       ----(ii)

Concept Used:

Distance = Speed × Time 

If the distance covered is constant then 

Product of speed and time is also constant 

Calculation:

According to the question 

The total distance covered by both the car is constant 

u × p =  v × q

⇒ v/u = p/q

Using Componendo rule we get 

⇒ (u - v)/u = (q - p)/q

⇒ 5/u = 140/q

⇒ q = 28u     ----(iii)

Now, According to the question

u × p × (1/60) = 350

⇒ u × p = 21000       ----(iv)

And Also v × q × (1/60) = 350

⇒ v × q = 21000          ----(v)

From (i), v = u - 5          ----(vi)

From (ii), p = q - 140       ----(vii)

From (iv) and (vii), we get 

⇒ u × (q - 140) = 21000  

⇒ uq = 21000 + 140u  

Now, from (iii), we get 

⇒ uq = 21000 + 140u  

⇒ u × 28u = 21000 + 140u

⇒ 28u² - 140u - 21000 = 0

⇒ u2 - 5u - 750 = 0

⇒ u2 - 30u + 25u - 750 = 0

⇒ u(u - 30) + 25(u - 30) = 0

⇒ (u - 30) (u + 25) = 0 

u = 30, u = -25  [not possible]

∴ The value of u is 30 km/hr.

Concept Used:

1 Week = 7 Days

1 Day = 24 Hours 

1 Hour = 60 Minutes 

Calculation:

We have to find total minutes in x weeks and x days

x weeks x days = (x × 7) days + x days

⇒ 8x days = 8x × 24 hours 

⇒ 192x hours = 192x × 60 minutes

⇒ 11520x minutes 

∴ The total minutes in x weeks and x days is 11520x minutes.

Given:

A is the arithmetic mean of p and q 

G is the geometric mean of p and q

And p > q 

Concept Used:

For andy two numbers

A.M ≥ G.M

Calculation:

For any two positive numbers p and q

A ≥ G

So, options 1) and 3) both can be the answer 

But for option 3)

We have to take p = q 

Now, if we put p = q in A ≥ G

i.e in (p + q)/2 ≥ √pq, we get 

2p/2 ≥ √p²

⇒ p ≥ p

i.e p = p 

But According to the question we have

p > q 

∴ The only 1) option will be correct.

Given:

x² + αx + β = 0

α and β are the roots of above equation

Concept Used:

For equation ax² + bx + c = 0

Sum of roots = -b/a

Product of roots = c/a 

Calculation:

The given equation is x2 + αx + β = 0

Here, a = 1, b = α, c = β 

Sum of roots (α + β) = -α /1

⇒ 2α + β = 0      -----(i)

Product of roots (α**.**β) = β/1

⇒ α = 1     ----(ii)

From (i) and (ii), we get 

β = -2      ----(iii)

Now, we have to find the value of α - β

⇒ α - β = 1 - (-2) = 3

∴ The required value of α - β is 3.

Given:

Marked Price of article = Rs. 200

Discount = 10% on Marked Price 

Gain = 20% on Cost Price 

Concept Used:

S.P = M.P [100 - D%]%

S.P = C.P [100 + P%]%

Where, S.P = Selling Price 

M.P = Marked Price, C.P = Cost Price 

D% = Discount Percent, P% = Profit Percent

Calculation:

Let the cost price of the article be Rs. x

S.P = M.P [100 - D%]%

⇒ S.P = 200 × [100 - 10]%

⇒ S.P = 200 × 90/100 = Rs. 180      ----(i)

Now, S.P = C.P [100 + P%]%

⇒ S.P = x × [100 + 20]%

⇒ S.P = x × (120/100)

⇒ S.P = 6x/5     ----(ii)

Now, From (i) and (ii), we get

6x/5 = 180 

⇒ x = 150

∴ The cost price of article is Rs. 150.

Shortcut TrickWhen Discount% and Profit% are given Then the Ratio of C.P:M.P is:

C.P/M.P = (100 - D%)/(100 + P%)

⇒ C.P/200 = (100 - 10%)/(100 + 20%)

⇒ C.P/200 = 90/120

⇒ C.P = 200 × 90/120

⇒ C.P = 150

∴ The cost price of article is Rs. 150.

Given:

A person borrowed ₹9,000 at 7% simple interest

borrowed  ₹12,000 at 8% simple interest

borrowed ₹15,000 at 9% simple interest

Time = n years 

Amount to be paid = ₹50,700

Formual Used:

Simple Interst = PRT/100

Amount = Principal + Simple Interest

Calculation:

Simple interest on the 9000 [S₁]

S₁ = (9000 × 7 × n)/100 = 630n

A₁ = 9000 + 630n     ----(i)

Where A₁ is the amount on 12000 

Simple interest on the 12000 [S2]

S2 = (12000 × 8 × n)/100 = 960n

A2 = 12000 + 960n     ----(ii)

Where A2 is the amount on 12000 

Simple interest on the 15000 [S3]

S3 = (15000 × 9 × n)/100 = 1350n

A3 = 15000 + 1350n     ----(iii)

Where A3 is the amount on 15000 

Now from (i), (ii) and (iii), we get 

9000 + 630n + 12000 + 960n + 15000 + 1350n = 50,700

⇒ 2940n + 36000 = 50700

⇒ 2940n = 14700

⇒ n = 5

∴ The required time is 5 years.

Concept:

Natural number: All the positive integers from 1 to infinity.

Even number: Whole numbers that have the digits 0, 2, 4, 6, or 8 in their one's place are even numbers.

Odd number: Any number which cannot be divided by two. 

Calculation:

Statement: 1

Let n, n+1, and n + 2 be three consecutive natural numbers.

Let, P(n) = n³ + (n + 1)³ + (n + 2)³ is divisible by 9.

For n =1, 

⇒ 1³ + 2³ + 3³

⇒ 1 + 8 + 27 = 36 is divisible by 9.

⇒ For n = 4

⇒ 4³ + 5³ + 6³

⇒ 64 + 125 + 215 = 405 is divisible by 9

Hence, statement 1 is correct. 

Statement: 2

Take 3², 5², 7² we get

⇒ 9, 25, 49 which is divisible by 8 and gives 1 as remainder.

Similarly, we can check for any other value as well.

Hence, statement 2 is correct. 

∴ Both 1 and 2 are correct.

Given :

Number of factors/divisor of 1000 (excluding 1 and 1000)

Formula used :

For N = x^a × y^b × z^c

Where x, y and z are prime numbers and a, b and c are their powers

Number of factors = (a + 1)(b + 1)(c + 1) 

Calculations :

Prime factorization of 1000

⇒ 1000 = (2 × 5)³ 

⇒ 1000 = 2³ × 5³   

Hence, number of factors = (3 + 1)(3 + 1)

⇒ 4 × 4 = 16

The total number of factors or divisors for 1000 is 16.

Now excluding 1 and 1000, we get

⇒ 16 - 2

⇒ 14

∴ The number of divisors of 1000 (excluding 1 and 1000) is 14.

Given:

Equation is  x2 - k2x + 30kx - 161x - 64 = 0 

Sum of roots = 0

Concept Used:

For quadratic equation ax² + bx + c = 0 

Sum of roots (α + β) = -b/a

Product of roots (α. β) = c/a

(α - β)² = (α + β)² - 4α.β

Calculation:

We have x2 - k2x + 30kx - 161x - 64 = 0 

⇒ x2 - (k2 - 30k + 161)x - 64 = 0 

α + β =  (k2 - 30k + 161)/1 = 0 

And α.β = -64/1 = -64

Now, We have to find α - β

So, Using formula 

(α - β)2 = (α + β)2 - 4α.β

⇒ (α - β) 2 = (0)2 - 4 × -64

⇒ (α - β) 2 = 4 × 64

⇒ α - β = √(4 × 64)

⇒ α - β = ±16 

∴ The difference between the roots is 16.

Given:

A piece of cloth costs Rs. 10,000

Concept used:

Total cost = total item × price of each item

Calculation:

Let at the beginning piece of cloth be x m

According to the question,

[10000/x - 10000/(x + 2)] = 250

⇒ [(10000x + 20000 - 10000x)/x(x + 2)] = 250

⇒ [20000/x(x + 2)] = 250

⇒ 20000 = 250x² + 500x

⇒ 250x2 + 500x - 20000 = 0

⇒ x2 + 2x - 80 = 0 [By dividing 250 from both side]

⇒ x2 + 10x - 8x - 80 = 0

⇒ x(x + 10) - 8(x + 10) = 0

⇒ (x + 10)(x - 8) = 0

So, x = 8 or - 10

But x cannot be negative

∴ The original length of the piece of cloth is 8 m

Given:

ax2 + bx + c = 0 is a quadratic equation 

Roots are in the ratio c : 1

i.e α : β = c : 1 or β : α = c : 1

Concept Used:

Sum of roots (α + β) = -b/a

Product of roots (α.β) = c/a

Calculation:

Given α : β = c : 1     ----(i)

According to the question 

(α + β) = -b/a      ----(ii)

(α.β) = c/a      ----(iii)

On squaring the (i) equation, we get 

α² + β² + 2α.β = $\frac{b^2}{a^2}$

Dividing the above equation by α.β, we get 

⇒ $\frac{\alpha^2}{\alpha .\beta} + \frac{\beta^2}{\alpha .\beta} + 2 = \frac{\frac{b^2}{a^2}}{\frac{c}{a}}$

⇒ $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2 = \frac{b^2}{ac}$

Now, from (i)

⇒ $c + \frac{1}{c} + 2 = \frac{b^2}{ac}$ 

Multiplying the above equation by ac both side, we get 

⇒ ac² + a + 2ac = b²

⇒ a (c2 + 1 + 2c) = b2

⇒ a (c + 1)2 = b2

∴ The correct relation is b2 = a(c + 1)2.

Given:

The sides of a triangle are 6x2 and (2x2 - 1)

The area of the triangle is 84 sq. unit

Formula Used:

In a right angles triangle 

Area of triangle = (1/2) × base × perpendicular

Calculation:

Let BC is base 

So, AB will be perpendicular 

Now, According to the formula used 

Area of triangle = 84

⇒ (1/2) × (2x2 - 1) × 6x2 = 84

⇒ (2x2 - 1) × x2 - 28 = 0

Let x² = y

So, The above equation can be written as

⇒ (2y - 1) × y - 28 = 0

⇒ 2y² - y - 28 = 0

⇒ 2y2 - 8y + 7y - 28 = 0

⇒ 2y (y  - 4) + 7 (y - 4) = 0

⇒ (y  - 4) (2y + 7) = 0

⇒ y = 4 or y = -7/2

So, x2 = 4 or x2 = -7/2 [not possible]

⇒ x = ±2

So, AB = 6 × 2² = 24 and 

BC = 2 × 22 - 1 = 7

We know that (7, 24, 25) is a triplet of a right-angled triangle 

So, The hypotenuse will be 25 

So, Perimeter = (7 + 24 + 25) unit = 56 unit

∴ The perimeter of the triangle is 56 unit.

Shortcut Trick

For such questions, the fastest method is checking the options to satisfy the given condition.

Let T be the time taken by the faster train, so the slower train takes (T + 2) hours.

Difference in speeds = 192/T − 192/(T + 2) = 16.

Check T = 4 hours from Option 2: Speeds are 192/4 = 48 km/hr and 192/6 = 32 km/hr.

Difference = 48 − 32 = 16 km/hr, which perfectly satisfies the condition.

∴ The correct answer is 4 hours.

Alternate Method

Given:

Distance between the two cities = 192 km

Time difference between trains X and Y = 2 hours

Speed difference between trains X and Y = 16 km/hr

Formula Used:

Speed = Distance ÷ Time

Calculations:

Let the time taken by the faster train (Train X) be T hours.

Then, the time taken by the slower train (Train Y) = (T + 2) hours.

⇒ Speed of the faster train = 192 / T km/hr

⇒ Speed of the slower train = 192 / (T + 2) km/hr

According to the given condition, the difference in speeds is 16 km/hr:

⇒ 192/T − 192/(T + 2) = 16

Dividing the entire equation by 16:

⇒ 12/T − 12/(T + 2) = 1

⇒ [12(T + 2) − 12T] ÷ [T(T + 2)] = 1

⇒ 24 ÷ (T² + 2T) = 1

⇒ T² + 2T = 24

⇒ T² + 2T − 24 = 0

⇒ T² + 6T − 4T − 24 = 0

⇒ T(T + 6) − 4(T + 6) = 0

⇒ (T − 4)(T + 6) = 0

Since time cannot be negative, T = 4 hours.

∴ The correct answer is 4 hours.

Additional Information

Relative Speed

When two bodies are moving in the same direction, Relative Speed = |S₁ − S₂|. When moving in opposite directions, Relative Speed = S₁ + S₂.

Average Speed

Average Speed is the total distance traveled divided by the total time taken. For two equal distances covered at speeds S₁ and S₂, Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂).

Unit Conversion

To convert km/hr to m/s, multiply the speed by 5/18. To convert m/s to km/hr, multiply by 18/5.