CDS Elementary Mathematics 1 Sep 2024 Official Paper (Previous Year Paper)

Calculation:

Given number is

4x² + x

This can be written as

⇒ 4x² + x + ($1 \over 4$)² - ($1 \over 4$)²

⇒ (2x + $1 \over 4$)² - $1 \over 16$

Minimum value of (2x + $1 \over 4$)²  = 0.  

The minimum value of the entire expression is (- $1 \over 16$).

To find the value of x that gives this minimum:

Set (2x + $1 \over 4$)2 = 0

⇒ (2x + $1 \over 4$) = 0

⇒ x = - $1 \over 8$ = - 0.125

∴ The real number is calculated as - 0.125

Alternate MethodCalculation:

According to the question,

⇒ f (x) = x + 4 × x2

For the critical point which is "x ", f ' (x) = 0

Differentiating we get,

⇒ f ' (x) = 1 + 4 × (2x) = 0

⇒ f ' (x) = 1 + 8x = 0 

⇒ 8x = - 1

⇒ x = - 1 / 8 = - 0.125.

∴ The real number is calculated as - 0.125.

Given:

m, n are positive natural numbers, also m > n

 m² - n² = 72 

Formula Used:

(a² -b²) = (a + b) (a - b)

Calculation:

According to the question,

⇒ m2 - n2 = 72

Where,

All the factor pairs of 72 are (1, 72) (2, 36) (3, 24) (4, 18) (6, 12) (8, 9) 

However, there are only three pairs of natural numbers (m, n) that satisfy the equation  m2 - n2 = 72, and m > n.

⇒ (19, 17), (11, 7), and (9, 3).

∴ The number of pairs that satisfy the above conditions is 3.

Given:

N is a 5-digit number.

When N is divided by 6, it leaves a remainder of 2.

When N is divided by 12, it leaves a remainder of 8.

When N is divided by 15, it leaves a remainder of 11.

When N is divided by 24, it leaves a remainder of 20.

Formula Used:

Chinese Remainder Theorem

Calculation:

We know:

$N ≡ 2 ,(\text{mod}, 6)$

Means N + 4 will be divisible by 6

$N ≡ 8 ,(\text{mod}, 12)$

Means N + 4 will be divisible by 8

$N ≡ 11 ,(\text{mod}, 15)$

Means N + 4 will be divisible by 15

$N ≡ 20 ,(\text{mod}, 24)$

Means N + 4 will be divisible by 24

Now, to find the common solution, we observe:

Since 6, 12, 15, and 24 have common factors, we need to find the least common multiple (LCM) of these numbers.

$LCM(6, 12, 15, 24) = 120$

99999/120 = 833.325

Largest 5 digit number divisible by 6, 12, 15, 24 is:

833 × 120 = 99960

since N + 4 is divisible by all these numbers,

N + 4 = 99960

N = 99960 - 4 = 99956

∴ The correct answer is 99956.

Given:

111²²² + 222³³³ + 333⁴⁴⁴ 

Concept Used:

Every unit digit's power repeats every fourth time except 4 and 6, which repeats every second time.

E.g., 23²⁵,

We first divide 25 by 4, we get the remainder = 1 

Now, unit place digit 3¹ ÷ 4, we get 3 as the remainder.

Calculation:

Similarly,

111²²² :

222 ÷ 4, we get remainder = 2, 

So unit place digit 1² = 1

222³³³ :

333 ÷ 4, we get remainder 1,

So unit place digit 2¹ = 2

333⁴⁴⁴ :

444 ÷ 4, we get remainder = 0 or say 4,

So unit place digit 3⁴ = 1

By summing all the unit place digits = 1 + 2 + 1 = 4

∴ When 4 is divided by 5 we finally get the remainder = 4

Shortcut Trick

Let's consider the last 3 digits of the numbers

⇒ 345 × 789 = 272205

∴ The last three digits of the multiplication is 2, 0, 5

Calculation:

When x = 1, y= 2, z = 3 then

(x2 + y2 + z2) = 1² + 2² + 3² = 14

⇒ p ∝ (x2 + y2 + z2)

⇒ 70 ∝ 14

⇒ 70 = k (14) 

⇒ k = 5 

When, x = -1, y = 1, z = 5, then 

(x2 + y2 + z2) = (1 + 1 + 25) = 27

p = k (x2 + y2 + z2)  

⇒ p = 5 × (27) = 135

∴ The right answer is 135.

Given:

N is a number which is divided by 11 and leaves a remainder of 5

Calculation:

LCM of  6, 12, 15, 18 is 180

According to the question,

180K + 5 is divided by 11

Put the value of K which is less than 11 

So, putting the value of K is 7

⇒ 180 × 7 + 5 = 1265, this is divisible by 11

∴ The answer is 1200 < N < 1300

Concept Used: 

${1 \over √n + √(n-1)} = {1 \over √ n + √ (n-1)} \times {(√n - √(n-1)) \over(√n - √(n-1))}$

${1 \over √n + √(n-1)} = {√n - √(n-1) \over (n - (n-1))} = √ n - √ (n-1)$

Calculation:

${1 \over √10 + √9} = { (√10 - √9) \over (√ 10 + √9)(√10 - √9)}= {√ 10 - √9 \over (10 - 9 )}$ = √10 - √9

Similarly, 

${1 \over √11 + √10} =$ √11 - √10, and ${1 \over √12 + √11} = √12 - √11$ 

putting the values in the equation, 

⇒ (√10 - √9) + (√11 - √10) + (√12 -√11) + ................ + (√196 - √195) 

⇒ (-√9 + √196)

⇒  (-3 + 14) = 11

∴ The value of this expression is 11

Shortcut Trick

Apply Alligation Method on the time taken by each train to cross a man.

Time for X = 24s, Time for Y = 18s, Combined relative time = 20s.

Ratio of speeds (V_X : V_Y) = |18 − 20| : |24 − 20|

Ratio = 2 : 4 = 1 : 2

∴ The correct answer is 1 : 2.

Alternate Method

Given: Time for Train X (t₁) = 24 s, Time for Train Y (t₂) = 18 s, Crossing time (t) = 20 s.

Let speed of Train X be v₁ and Train Y be v₂.

⇒ Length of Train X (L₁) = speed × time = v₁ × 24

⇒ Length of Train Y (L₂) = speed × time = v₂ × 18

When crossing in opposite directions, Relative Speed = v₁ + v₂.

⇒ Time = (L₁ + L₂) ÷ (v₁ + v₂)

⇒ 20 = (24v₁ + 18v₂) ÷ (v₁ + v₂)

⇒ 20v₁ + 20v₂ = 24v₁ + 18v₂

⇒ 20v₂ − 18v₂ = 24v₁ − 20v₁

⇒ 2v₂ = 4v₁

⇒ v₁ / v₂ = 2 / 4 = 1 / 2

∴ The correct answer is 1 : 2.

Additional Information

Relative Speed Principle

When two objects move in opposite directions, relative speed is the sum (S₁ + S₂). In the same direction, it is the difference |S₁ − S₂|.

Crossing Objects

When a train crosses a stationary point object (man, pole), distance = length of train. When it crosses another train/platform, distance = sum of lengths.

Alligation Application

The ratio of the speeds is inversely proportional to the deviation of individual crossing times from the combined crossing time.

Concept Used:

For the quadratic equation ax² + bx + c = 0

Sum of roots α + β = -b/a

Product of roots α.β = c/a

Calculation:

p, q be the roots of x2 + mx - n = 0

By using the above concept

⇒ p + q = -m

⇒ p = - m - q        ----(1)

⇒ pq = -n             -----(2)

Similarly,

m, n are the roots of x2 + px - q = 0 

⇒ m + n = -p        ------(3)

⇒ mn = -q            ------(4)

Now, Put the value of eq (1) in eq (3):

m + n = -p

⇒ m + n = -(-m - q)

⇒ m + n = m + q

⇒ n = q

From equation (4):

mn = - q

⇒ m (q) = - q

⇒ m = -1

From equation (2):

pq = -n

pq = - q

p = -1

Put the value of p = -1 and m = -1 in the equation (1):

p = - m - q 

⇒ -1 = - (-1) - q

⇒ -1 = 1 - q

⇒ q = 1 + 1 = 2

Since, n = q = 2

Now, check statements

Statement (I):  m (m + n) = -1

-1 (-1 + 2) = -1

-1 (1) = -1

-1 = -1 (true)

Statement (II): p + q = 1

• 1+ 2 = 1

1 = 1 (true)

Hence, both the options are correct.

Shortcut Trick The sinθ value lies between 0 ≤ sinθ ≤ 1

So, the max value of sinθ = 1 

Putting this to 8sinθ - 4sin²θ 

⇒  8 × 1 - 4 × 1² 

⇒  8 - 4 = 4

∴ The maximum value of  8sinθ - 4sin²θ is 4.

Alternate Method Let sinθ = x

Then, y =  8sinθ - 4sin²θ = 8x - 4x²

differentiate both sides, in order to find critical values

$\frac {dy}{dx}$ = 8 - 4 × (2x) = 0

⇒ 8 - 8x = 0 

⇒ x = 1    (put this value in above polynomial)

Then, y = 8 - 4 = 4 

∴ The maximum value of  8sinθ - 4sin²θ is 4.

Given:

The expression is (1 + tanα tanβ)² + (tanα - tanβ)².

Concept:

We will expand the given expression using basic trigonometric identities and simplifications.

Calculation:

⇒ (1 + tanα tanβ)² + (tanα - tanβ)²

⇒ 1 + 2 tanα tanβ + tan²α tan²β + tan²α - 2 tanα tanβ + tan²β

⇒ 1 + tan2α tan2β + tan2α + tan2β

⇒  tan²α(1 + tan2β) + (1 + tan2β)

⇒  (1 + tan²α)(1 + tan2β)

⇒ sec²α sec²β

∴ The correct answer is sec²α sec²β.

Given:

tan 50° - cot 50° = ?

cot 25° - tan 25° = ?

Formula used:

In 1st quadrant, all value are + ve

1. tan θ = sin θ / cos θ
2. cot θ = cos θ / sin θ

Calculations:

tan 50° - cot 50°

⇒ (sin 50° / cos 50°) - (cos 50° / sin 50°)

⇒ (sin²50° - cos²50°) / (sin 50° × cos 50°)

⇒ - (cos²50° - sin²50°) / (sin 50° × cos 50°)

⇒ - cos 100° / (sin 50° × cos 50°)

Since, - cos 100° = +ve, sin 50° × cos 50° = + ve

Thus, the overall value is Positive ⇒ True

cot 25° - tan 25°

⇒ (cos 25° / sin 25°) - (sin 25° / cos 25°)

⇒ (cos² 25° - sin² 25°) / (sin 25° × cos 25°)

⇒ cos 50° / (sin 25° × cos 25°)

Since, cos 50° = +ve, sin 25° × cos 25° = + ve

Thus, the overall value is Positive ⇒ False

Hence, the correct answer is option (1)

Given:

$tan(α + β) = √3$

$tan(α - β) = 1/√3$

Formula used:

tan(α + β) = tan(π/3) = √3

tan(α - β) = tan(π/6) = 1/√3

cot(2β) = 1 / tan(2β)

Calculation:

From the given information, we can find the values of α and β:

α + β = π/3

α - β = π/6

Solving these equations simultaneously:

Adding the two equations:

⇒ (α + β) + (α - β) = π/3 + π/6

⇒ 2α = π/2

⇒ α = π/4

Now, substitute α = π/4 in α + β = π/3:

⇒ π/4 + β = π/3

⇒ β = π/3 - π/4

⇒ β = π/12

Now, calculate tan(α) and cot(2β):

tan(α) = tan(π/4) = 1

cot(2β) = 1 / tan(2β)

tan(2β) = tan(2 × π/12) = tan(π/6) = 1/√3

⇒ cot(2β) = 1 / (1/√3) = √3

Finally, calculate tan(α) × cot(2β):

⇒ tan(α) × cot(2β) = 1 × √3 = √3

∴ The correct answer is Option 3.

Given:

sin²θ × cos²θ × (sec²θ + cosec²θ) = ?

Formula used:

1. sec²θ = 1 + tan²θ
2. cosec²θ = 1 + cot²θ
3. sin²θ + cos²θ = 1
4. sec²θ + cosec²θ = 1/sin²θ + 1/cos²θ

Calculations:

sec²θ + cosec²θ = (1/sin²θ) + (1/cos²θ)

⇒ sec²θ + cosec²θ = (sin²θ + cos²θ) / (sin²θ × cos²θ)

⇒ sec²θ + cosec²θ = 1 / (sin²θ × cos²θ)

Now, sin²θ × cos²θ × (sec²θ + cosec²θ)

⇒ sin²θ × cos²θ × [1 / (sin²θ × cos²θ)]

⇒ sin²θ × cos²θ cancels with the denominator.

⇒ The result = 1.

∴ The correct answer is option (2): 1.

Concept:

Calculation:

We have,

$64^{\sin ^2 θ}+64^{\cos ^2 θ}=16$

Given that, $0 \leqslant θ \leqslant \frac{\pi}{2}$, 

As we know, 64^1/2 + 64^1/2 = 8 + 8 = 16,

⇒ Sin²θ = Cos²θ = 1/2

Taking square root on both sides, 

⇒ Sin θ = Cos θ = $\frac {1}{\sqrt 2}$

⇒ θ = 45° 

Then tan 45° = cot 45° = 1 

⇒ tan 45° + cot 45° = 1 + 1 = 2 

∴ The right answer is 2

Given:

If cosecθ - cotθ = m and secθ - tanθ = n

Concept used: 

If cosecθ - cotθ = a, then cosecθ + cotθ = 1/a

if secθ - tanθ = b, then secθ + tanθ = 1/b

Formula Used:

cosec²θ - cot²θ = 1

sec²θ - tan²θ = 1

Calculation:

cosecθ - cotθ = m          --(i)

So, cosecθ + cotθ = 1/m  --(ii)

Add equation (i) and (ii),

2 cosec θ = m + 1/m

⇒ cosec θ = 1/2 (m + 1/m)

Now,

secθ - tanθ = n    --(iii)

So, secθ + tanθ = 1/n   --(iv)

Add equations (iii) and (iv),

2 secθ = n + 1/n

⇒ sec θ = 1/2 ( n + 1/n)

So,

cosec θ + sec θ = 1/2 ( m + 1/m) + 1/2(n + 1/n)

⇒ 1/2 ( m + n + 1/m + 1/n)

∴ The correct answer is option (1).

Given:

Height of the point X above the surface of the river = h

Angles of depression of points P and Q = α and β respectively

α and β are complementary angles

Formula Used:

Width of the river = 2h(tanα + cotα)

Calculation:

Since α and β are complementary, β = 90° - α

Let the width of the river be W

From point X, tanα = h / d₁ and tanβ = h / d₂

Here, d₁ and d₂ are the distances from X to points P and Q respectively.

Since β = 90° - α, tanβ = cotα

So, tanα = h / d₁ and cotα = h / d₂

Width of the river W = d₁ + d₂

⇒ d₁ = h / tanα and d₂ = h / cotα

⇒ W = h / tanα + h / cotα

⇒ W = h(tanα + cotα) / tanα.cotα

⇒ tanα.cotα = 1

⇒ W = h(tanα + cotα)

⇒ W = h($\frac{\sin\alpha }{\cos \alpha} + \frac{cos \alpha}{\sin\alpha}$)

⇒ W = h($\frac{\sin^2\alpha + \cos^2\alpha}{\sin \alpha \cos \alpha}$)

⇒ W = h($\frac{1}{\sin \alpha \cos \alpha}$)

⇒ W = hsecα.cosecα

∴ The correct answer is option 4.

Given:

∠ABC = 60°, AB = 6cm and BC = 8 cm

Formula Used:

Area of triangle = 1/2 × AB × BC × sin∠ ABC

Calculations:

Since AB = 6 cm and BC = 8 cm 

∠ ABC = 60° 

⇒ Sin∠ ABC = sin 60° = √ 3/2

⇒ Area = 1/2 × AB × BC × sin 60°

⇒ Area = (1/2) × 6 × 8 × sin 60° 

⇒ Area = 1/2 × 6 × 8 × (√3)/2= 12√3

∴ The right answer is 12√3

Concept used:

ax2 + bx + c = 0, then roots p + q = -b/a, pq = c/a

Calculation:

x² - sin²θx - cos²θ = 0 , when compared to ax² + bx + c = 0,

we get a = 1, b = -sin²θ, c = - cos²θ 

⇒ p + q = -b/a = -(-sin²θ / 1) = sin²θ 

⇒ pq = (-cos²θ / 1) = - cos²θ 

⇒ p² + q² = (p + q)² - 2pq = (sin²θ)² - 2 × (-cos²θ )

⇒ p2 + q2 = (p + q)2 - 2pq = sin⁴θ + 2 cos²θ 

For minimum value, put θ = 0° , 90°, 45° 

At θ = 0°, sin4θ + 2 cos2θ = 0 + 2(1)² = 2

At θ = 90°, sin4θ + 2 cos2θ = (1)⁴ + 2(0) = 1

At θ = 45°, sin4θ + 2 cos2θ = (1/√2)⁴ + 2(1/√2)² = 5/4

we get minimum value at θ = 90° at which p² + q² = 1

∴ The correct answer is 1.

Calculation:

The arithmetic mean of n numbers = M

Sum of n numbers = n × M = nM

Sum of (n - 1) numbers = k 

Then,

n^th number = Sum of n numbers - sum of (n -1) numbers                               

n^th number = nM - k 

∴ The correct answer is nM - k

Formula used:

GM = ⁿ√ x₁ × x₂ × x₃ × x_n =  (x1 × x2 × x3 × xn)^1/n

Calculation:

GM = (3 × 9 × 27 × 81 × 243 × 729 × 2187)^1/7

⇒  (3¹ × 3² × 3³ × 3⁴ × 3⁵ × 3⁶ × 3⁷)^1/7

⇒  (3^28/7) = 3⁴

∴ The answer is 81.

Given:

A person purchases one kg of tea powder

• A's place cost of sugar 1 kg = ₹ 1000
• B's place cost of sugar 2 kg = ₹ 1000
• C's place cost of sugar 4 kg = ₹ 1000
• D's place cost of sugar 5 kg = ₹ 1000

​

​Calculation:

Total purchase = 4 kg   (A + B + C + D)

cost of 1 kg tea powder in A's place = 1000

cost of 1 kg tea powder in B's place = 1000 / 2 = 500

cost of 1 kg tea powder in C' place = 1000 / 4 = 250

cost of 1 kg tea powder in D' place = 1000 / 5 = 200

Total cost = 1000 + 500 + 250 + 200 = ₹ 1950

According to the above calculation,

At ₹ 1950 = 4 kg

At ₹ 1000 = (4 / 1950) × 1000 = 2.05 kg

∴ On average, he purchases 2.05 kg of sugar for every ₹1000 spent.

Calculation:

All single-digit prime no = 2, 3, 5, 7

⇒ Largest no. = 7532

⇒ Smallest no = 2357

Sum:

⇒ 7532 + 2357 = 9889

∴ The sum of the largest and the smallest 4-digit numbers made by using single digit prime numbers is 9889.

Shortcut Trick

Convert the base 3 to a power close to the divisor 28: 3³ = 27.

Rewrite the expression: 3²⁵⁵ = (3³)⁸⁵ = 27⁸⁵.

Since 27 is 1 less than 28, the remainder is (−1)⁸⁵ = −1.

Final positive remainder = 28 − 1 = 27.

∴ The correct answer is 27.

Alternate Method

Given: Dividend = 3²⁵⁵, Divisor = 28

Formula Used: Remainder of (a − 1)ⁿ ÷ a is (−1)ⁿ

⇒ 3²⁵⁵ = (3³)⁸⁵

⇒ 3³ = 27

⇒ Remainder [3²⁵⁵ ÷ 28] = Remainder [27⁸⁵ ÷ 28]

⇒ 27 ≡ −1 (mod 28)

⇒ Remainder = (−1)⁸⁵ = −1

⇒ Positive Remainder = 28 + (−1) = 27

∴ The correct answer is 27.

Additional Information

Remainder Theorem for Powers

If a number (a + k)ⁿ is divided by a, the remainder is kⁿ. In this case, (28 − 1)⁸⁵ divided by 28 leaves (−1)⁸⁵.

Negative Remainder Property

A negative remainder −r is equivalent to a positive remainder (Divisor − r). It is a useful intermediate step to simplify calculations.

Odd and Even Powers of −1

(−1)ⁿ = 1 if n is even, and (−1)ⁿ = −1 if n is odd. This rule determines the sign of the intermediate remainder.

Given:

(10097 + 10054 + x + 1) is divisible by 9

x, (0 < x < 8)

Concept:

The divisibility rule for 9 states that a number is divisible by 9 if the sum of its digits is

either 0 or a multiple of 9.

Calculation:

If 10⁴ = 10000, the sum of all digits is 1

10⁶ = 1000000 also the sum of all digits is 1

sum of all digit of (10097 + 10054 + x + 1) is x + 3

According to the divisibility rule for 9,

If we add 6 to x + 3 so that their sum is 9, then x + 3 will be divisible by 9 with 0

remainder.

∴ The value of x is 6.

Given:

AB.DC = AC.BD → $\frac{AB}{BD}=\frac{AC}{DC}$

∠ BAD = α ,  ∠ CAD = β 

Concept Used:

By Angle Bisector Theorem:

A line that bisects an angle divides the opposite side into two segments that are proportional to the other two sides of the triangle.

i.e., $\frac{AB}{BD}=\frac{AC}{DC}$

Calculation:

According to the above theorem, 

⇒ If $\frac{AB}{BD}=\frac{AC}{DC}$ then we can say that D bisects ∠ BAC.

⇒ Then, ∠ BAD = ∠ CAD which is α = β 

∴ The correct relation between α and β is α = β

Given:

N = 12345678AB which is divisible by 9

Concept Used:

Any number whose summation of the digits is divisible by 9, then the number itself can be divisible by 9.

Calculation:

Statement I:

All digits 1+2+3+4+5+6+7+8 = 36 which is divisible by 9.

Then,

⇒ (A + B) will be divisible by 9.

⇒ Statement I is correct.

Statement II:

For A & B to be divisible by 9, 

A & B must be one odd & one even. 

Such as: (1,8), (2,7), (3,6), (4,5), (9,0)

⇒ Statement II is false.

Hence, The correct answer is option 1.

Given:

x³ + $1 \over x³$ = 65/8

y³ + $1 \over y³$ = 730/27

Calculation:

⇒ x³ + $1 \over x³$³ = 65/8 = 8$\frac{1}{8}$

⇒ x³ + $1 \over x³$ = 8 + 1/8

We get x³ = 8 

⇒ x = 2

Also,

⇒ y³ + $1 \over y³$ = 730/27 = 27$\frac{1}{27}$

⇒ y³ + $1 \over y³$ = 27 + $1 \over 27$

We get, y³ = 27

⇒ y = 3

Now, 

⇒ xy = 2 × 3 = 6

∴ The correct answer is 6.

Given:

11x + 5y is a prime number where x, y are natural numbers.

Formula Used:

Find the minimum value of (x + y) such that 11x + 5y is a prime number.

Calculation:

Let's test the values of x and y to find the minimum sum (x + y).

For x = 1:

11 × 1 + 5y ⇒ 11 + 5y

11 + 5y should be a prime number.

For y = 1:

11 + 5 × 1 ⇒ 11 + 5 = 16 (not a prime number)

For y = 2:

11 + 5 × 2 ⇒ 11 + 10 = 21 (not a prime number)

For y = 3:

11 + 5 × 3 ⇒ 11 + 15 = 26 (not a prime number)

For y = 4:

11 + 5 × 4 ⇒ 11 + 20 = 31 (prime number)

Thus, the minimum value of (x + y) when x = 1 and y = 4:

x + y = 1 + 4 = 5

The minimum value of (x + y) is 5.

Given:

A 4-digit number N which has e×actly 15 distinct divisors.

Concept:

Let, N = P² × Q³

⇒ Its number of factors = (2 + 1) (3 + 1) = 3 × 4 = 12

Calculation:

According to the question,

Let the 4-digit number N = (pa) (qb)

⇒ Its number of factors = (a + 1) (b + 1)

As per the question, (a + 1) (b + 1) = 15

⇒ 15 can be written as (3 ×  5) or (1 × 15)

 

But, N= q¹⁴ cannot satisfy the 4-digit Number condition.

∴  The correct number of factors is 45.

Given:

p, q, and r are the lengths of the sides of a right-angled triangle.

Calculation:

Let's consider a right-angle triangle of sides (3, 4, 5)

i.e. p = 3, q = 4, r = 5

So,

(p - q - r) (q - r - p) (r - p - q)

⇒ (3 - 4 - 5)(4 - 5 - 3)(5 - 3 - 4)

⇒ (3 - 9)(4 - 8)(5 - 7)

⇒ (-6)(-4)(-2) 

After multiplication, we get a negative term.

Thus, (p - q - r) (q - r - p) (r - p - q) is always negative only.

∴ The correct answer is option (2).

Given:

Expression: $\frac{\left(a^8+a^4+1\right)\left(b^8+b^4+1\right)}{a^4 b^4}$, where a>0a>0

, b>0b>0

.

Formula used:

AM-GM inequality: For positive numbers $(x_1, x_2, \dots, x_n$, $(x_1 + x_2 + \dots + x_n)/n \geq \sqrt[n]{x_1x_2\dots x_n}$.

Calculations:

Let $x = a^4$ and $y = b$, so the expression becomes:

⇒ $\frac{(x^2+x+1)(y^2+y+1)}{xy}$

Apply AM-GM on $(x^2, x, 1)$:

⇒ $(x^2 + x + 1 \geq 3\sqrt[3]{x^2 \cdot x \cdot 1} = 3\sqrt[3]{x^3} = 3x.$

Equality holds when x = 1.

Similarly, $(y^2 + y + 1 \geq 3y)$.

Now substitute into the expression:

⇒ $\frac{(x^2+x+1)(y^2+y+1)}{xy} \geq \frac{(3x)(3y)}{xy}.$

⇒ $\frac{(x^2+x+1)(y^2+y+1)}{xy} \geq 9.$

The minimum value is achieved when x = y = 1, i.e., a = b = 1.

⇒ Minimum value = 9.

∴ The correct answer is option (3): 9.

Given:

• Total no of students = 200
• students prefer both tea and coffee = n
• students prefer coffee = 2n
• students prefer tea = 3n
• students prefer neither tea nor coffee = 4n

 

Calculation:

Students prefer only Tea = students prefer Tea - students prefer both tea and coffee 

⇒ 3n - n = 2n

Students prefer only coffee = students prefer coffee - students prefer both tea and coffee

⇒ 2n - n = n

Students prefer only Tea + only coffee + both + neither tea nor coffee = 200

⇒ ​2n + n + n + 4n = 200

⇒ 8n = 200

⇒ n =  $\frac{200}{8}$

⇒ n = 25

∴ The correct answer is 25.

Given: 

AB = 9 cm, BC = 12 cm

∠ABC = θ 

Area of triangle = 36 cm²

Formula used:

Area of triangle = $\frac{1}{2}$ × Ist side × 2nd side × sin θ

Calculation:

Area of triangle = $\frac{1}{2}$ × Ist side × 2nd side × sin θ

36 =  $\frac{1}{2}$ × 9 × 12 × sin θ

⇒ sin θ = $\frac{2}{3}$

We know that

cos²θ = 1 - sin²θ 

cos2θ = 1 - (2/3)2

⇒ cos2θ = 1 - 4/9 = 5/9

∴ Cos θ = $\frac{\sqrt 5}{3}$

Given:

Let n be a natural number.

The HCF of n and n + 10 is 10.

The LCM of n and n + 10 is x (a 2-digit number).

Formula Used:

For any two numbers a and b,

HCF(a, b) × LCM(a, b) = a × b

Calculation:

Let n = 10k. Therefore, n + 10 = 10k + 10 = 10(k + 1).

Given HCF(n, n + 10) = 10, it implies k and k + 1 are co-prime (i.e., HCF(k, k+1) = 1).

We know,

HCF(n, n + 10) × LCM(n, n + 10) = n × (n + 10)

⇒ 10 × x = 10k × 10(k + 1)

⇒ 10 × x = 100k(k + 1)

⇒ x = 10k(k + 1)

Since x is a 2-digit number,

10 ≤ 10k(k + 1) ≤ 99

⇒ 1 ≤ k(k + 1) ≤ 9.9

⇒ 1 ≤ k(k + 1) ≤ 9

Checking values of k,

For k = 1, k(k + 1) = 1 × 2 = 2

For k = 2, k(k + 1) = 2 × 3 = 6

For k = 3, k(k + 1) = 3 × 4 = 12 (not possible as it exceeds 9)

Possible values of k are 1 and 2.

Corresponding values of x are:

For k = 1, x = 10 × 1 × 2 = 20

For k = 2, x = 10 × 2 × 3 = 60

Hence, there are only two values of x: 20 and 60.

The correct answer is option 2.

Calculation:

Value putting approach:

Let's put a = 2,

a4 + 2a3 + 3a2 + 2a + 1

⇒ 2⁴ + 2(2)³ + 3(2)² + 2(2) + 1 = 49

a6 - 2a3 + 1

⇒ (2)⁶ - 2(2)³ + 1 = 49

So, HCF of 49 and 49 is 49.

Now, check options:

Option (1):

a3 + 3a2 + 2a + 1

⇒ (2)³ + 3(2)² + 2(2) + 1 = 25 (false)

Option (2):

a3 + a2 + a + 1

⇒ (2)3 + (2)2 + 2 + 1 = 15 (false)

Option (3):

(a2 + a + 1)2

⇒ (2² + 2 + 1)² = (4 + 2 + 1)² = 7² = 49 (true)

∴ The correct answer is option (3).

Additional Information

Here, we can not put a = 0 and 1 because at a = 0, all values become 1 and at a = 1, the second expression becomes negative.

Given:

The quadratic equation is:

$x² - (k - 2)x + (k + 1) = 0$

Roots are equal.

Formula used:

For a quadratic equation ax² + bx + c = 0, the roots are equal if:

$D = b² - 4ac = 0$

Here, a = 1, b = -(k - 2), c = k + 1.

Calculations:

Discriminant (D) = b² - 4ac

⇒ $D = [-(k - 2)]² - 4 × 1 × (k + 1)$

⇒ $D = (k - 2)² - 4(k + 1)$

⇒ $D = k² - 4k + 4 - 4k - 4$

⇒ $D = k² - 8k$

⇒ Set D = 0:

⇒ $k² - 8k = 0$

⇒ $k(k - 8) = 0$

⇒ $k = 0$ or $k = 8$

The valid values of k are 0 and 8.

∴ The correct answer is option 2.

Given:

$\dfrac{\cos\theta - \sin\theta + 1}{\cos\theta + \sin\theta - 1} \times (\cot\theta - \csc\theta)$

Formula used:

$\cot\theta = \dfrac{\cos\theta}{\sin\theta}, cosec\theta = \dfrac{1}{\sin\theta}$

Calculations:

⇒ Rewrite $\cot\theta - cosec\theta$

⇒ $\cot\theta - cosec\theta = \dfrac{\cos\theta}{\sin\theta} - \dfrac{1}{\sin\theta} = \dfrac{\cos\theta - 1}{\sin\theta}$

⇒ $\dfrac{\cos\theta - \sin\theta + 1}{\cos\theta + \sin\theta - 1} \times \dfrac{\cos\theta - 1}{\sin\theta}$

 

⇒ $\dfrac{(\cos\theta - \sin\theta + 1)(\cos\theta - 1)}{(\cos\theta + \sin\theta - 1)(\sin\theta)}$

⇒ Factorize terms:

⇒ $\dfrac{-(\cos\theta + \sin\theta - 1)}{\cos\theta + \sin\theta - 1}$

⇒ Cancel common terms:

⇒ -1

∴ The value is -1.

Given:

$\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$

Formula used:

$\sin^3 \theta = \sin \theta \cdot \sin^2 \theta$ and $cos^3 \theta = \cos \theta \cdot \cos^2 \theta$

$sin^2 \theta + \cos^2 \theta = 1$

Calculations:

Numerator: $sin \theta - 2 \sin^3 \theta$

⇒ $sin \theta (1 - 2 \sin^2 \theta)$

⇒ $sin \theta (1 - 2(1 - \cos^2 \theta)$

⇒ $sin \theta (2 \cos^2 \theta - 1)$

Denominator: $(2 \cos^3 \theta - \cos \theta)$

⇒ $cos \theta (2 \cos^2 \theta - 1))$

Now simplify the expression:

⇒ $\frac{\sin \theta (2 \cos^2 \theta - 1)}{\cos \theta (2 \cos^2 \theta - 1)}$

⇒ $(\frac{\sin \theta}{\cos \theta})$

⇒ $tan \theta$

∴ The correct answer is tanθ.

Given:

The mean weight of all 150 students = 60 kg.

The mean weight of boys = 70 kg.

Mean weight of girls = 55 kg.

Calculation:

Let the no of boys be x

Let the no of girls be y

⇒ 70x + 55y = 9000        --------------(i)

⇒ ( x + y = 150 ) × 60

⇒ 60x + 60y = 9000        --------------(ii)

By comparing equation (I) and (ii)

⇒ 70x + 55y = 60x + 60y

⇒ x / y = 1 / 2

∴ The ratio of number of boys to the number of girls is 1 : 2.

<br>

Alternate Method By allegation method

Given:

Length of tower A = 23 m

Length of tower B = 11 m

Distance between tower A and tower B = 9 m

Formula used:

Pythagoras theorem = P² + b² = H²

Calculation:

             

In

PQT

H² = P2 + b2

⇒ H2 = 12² + 9²

⇒ H = 15 m = 1500 cm

If the monkey covered 1500 cm in 5 min(300 sec)

Speed = 1500 / 300 = 5 cm/sec

∴ Speed of the monkey is 5 cm/sec

Given:

The radius of circle = r

The circle is divided into eight identical parts by cutting by planes passing through the same diameter.

Formula used:

Area of sphere = 4 π r²

Area of circle = π r2

Concept used:

The surface area of each final piece = Area of circle + ( Area of sphere / 8 )

Calculation:

The surface area of each final piece = Area of circle + ( Area of sphere / 8 )

⇒ πr² + (4πr2 / 8)

⇒ πr2 + πr2 / 2

⇒ 3πr2 / 2

∴ The surface area of each final piece is 3πr2 / 2.

Given:

Radius of circular track = 50 m = 164 foot

Radius of wheel = 1 foot

Formula used:

1 meter = 164 foot

Circumference of circle = 2 π r

Calculation:

For Right wheel

Circumference of wheel = 2 × π × 1 =  2π 

If the right wheel turns 1000 times, the total Circumference

⇒ 1000 × 2π = 2000 π m

Circumference of track = 2 × π × 164

⇒ 328 π m

For Left wheel

The radius of track = 51 m = 167.3 foot

Circumference of track = 2 × π × 167.3 m

⇒ 328 π m = 2000 π m

⇒ 1 m = ( 2000 π m ) / 328 π m

 ⇒ 2 π 167.3 m = (2000 π / 328 π) ×  2 π 167.3

⇒ 1020.26 turns

∴ The left wheel turn at approx 1020.

Calculation:

We have

70 × 71 × 72 × 73 × 74 × 75 × 76 × 77 × 78 × 79

⇒ $\frac{70 × 71 × 72 × 73 × 74 × 75 × 76 × 77 × 78 × 79 }{1000}$

In the given product, the numbers 70 and 75 provide three factors of 5,

The numbers 70, 72, 74, 76, and others contribute more than enough factors of 2. Since

we only need 2³ × 5³ to divide by 1000

The product is divisible by 1000, resulting in a remainder of 0.

∴ The remainder is 0.

Formula used:

1. tan θ = P / B
2. tan (α + β) = $\frac{tan(α) + tan(β))}{1-tan(α) tan(β)}$

 

Where, P = perpendicular, B = Base

Calculation:

In  $\bigtriangleup$ PQA

According to formula,

tan α = 80 / 64

⇒ Tan α = 5 / 4

In  $\bigtriangleup$ PQB

tan β = 80 / 100

⇒ tan β = 4 / 5

According to formula no 2

⇒ tan (α + β) = $\frac{\frac{5}{4} + \frac{4}{5}}{1-\frac{5}{4}\times\frac{4}{5}}$ = (41/20) / (1 - 1) = ∞ 

Since the denominator is zero,

⇒ tan (α + β) = ∞ = tan 90° 

So. (α + β) = 90° 

 ∴ It is undefined, which means α + β = 90°

Formula used:

x^a × x^b = x^{(a + b)}

Calculation:

(1 + x3) (1 + x6) = 1 + x³ + x⁶ + x⁹

Now,

x8k+3 + x8k+6 + x8k+9 + x8k+12

⇒ x^8k × x³  + x8k × x6 + x8k × x9 + x8k × x¹²

⇒ x8k × x³ (1 +  x3 + x6 + x9)

So,

x8k × x3 (1 +  x3 + x6 + x9) ÷ ( 1 + x3 + x6 + x9)

⇒ x8k × x3 = x8k+3

∴ The correct answer is option (3).

Given:

Side of outer square = 14 cm

Formula used:

Area of Triangle = ( 1 / 2) × base × height

Calculation:

In $\bigtriangleup$ APQ

Side AP = (1 / 2) × AD 

⇒ AP = AQ = 7 cm

Area of $\bigtriangleup$ APQ = ( 1 / 2) × 7 × 7

⇒ 49 / 2

The area of the shaded region:

Area of ( $\bigtriangleup$ APQ  + $\bigtriangleup$ DPS + $\bigtriangleup$ RSC  + $\bigtriangleup$ QRB)

⇒ (49 / 2) × 4 = 98 cm²

∴ The area lying between the two squares is 98 cm²

Given:

Base = (4/3) × Height

Area = 54 cm²

Formula used:

Area of triangle = (1/2) × Base × Height

Pythagoras theorem: Hypotenuse² = Base² + Height²

Perimeter = Base + Height + Hypotenuse

Calculation:

⇒ (1/2) × (4/3) × Height × Height = 54

⇒ (2/3) × Height² = 54

⇒ Height² = 54 × (3/2) = 81

⇒ Height = √81 = 9 cm

⇒ Base = (4/3) × 9 = 12 cm

⇒ Hypotenuse² = 9² + 12² = 81 + 144 = 225

⇒ Hypotenuse = √225 = 15 cm

⇒ Perimeter = 9 + 12 + 15 = 36 cm

∴ The perimeter of the triangle is 36 cm.

Given:

Side lengths of the triangle: a = 4 units, b = 4 units, c = 6 units.

Formula Used:

Area of a triangle using Heron's formula:

$\text{Area} = √{s(s-a)(s-b)(s-c)}$

where s is the semi-perimeter of the triangle.

$s = \frac{a + b + c}{2}$

Calculation:

Semi-perimeter, s :

s = $\frac{4 + 4 + 6}{2}$

⇒ s = 14/2

⇒ s = 7 units

Using Heron's formula to find the area:

$\text{Area} = √{7(7-4)(7-4)(7-6)}$

⇒ $\text{Area} = √{7 × 3 × 3 × 1}$

⇒ $\text{Area} = √{63}$

⇒ $\text{Area} = 3√{7}$

The area of the triangle is 3√7 square units.

Given:

AB = 10 cm & BP = CP

∠ BAP = 45°  & ∠ CAP =  θ 

Formula Used:

$\left(\text { } \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\right)$

Calculation:

In △ABP,

⇒ tan45° = $BP \over AB$

⇒ BP = AB = 10 cm    ---(tan45° = 1) 

Also, In △BAC,

⇒ tan(45° + θ) = $\frac{tan45^0 + tanθ} {(1 - tan45^0.tanθ)}$

But from trigonometry, tan(45° + θ) = $\frac{BC}{AB }$

⇒ $\frac{BC}{AB } = \frac{tan45^0 + tanθ} {(1- tan45^0.tanθ)}$

from the figure, BC = 2BP &  AB = BP

⇒$2BP \over BP$= $\frac{1 + tanθ} {(1 - tanθ)}$

⇒ 2 = $\frac{1 + tanθ} {(1 - tanθ)}$

⇒ 2 ( 1- tanθ) = 1 + tanθ 

⇒ 3 tanθ = 1

⇒ tanθ = $1 \over 3$

∴ The correct answer is $1 \over 3$.

Given:

AB = 10 cm,  BP = CP 

∠ BAP = 45° & ∠ ACP = γ

Calculation:

In △ABP,

⇒ tan45° = $BP \over AB$

⇒ BP = AB = 10 cm    ---(tan45° = 1) 

 In △BAC,

⇒ tanγ = $AB \over BC$

But, AB = BP & BC = 2BP

⇒ tanγ = $BP \over 2BP$

⇒ tanγ = $1 \over 2$

∴ The correct answer is $1 \over 2$.

Given:

 

AB = 10 cm,  BP = CP 

∠ BAP = 45° 

Calculation:

Statement I:

∵ P divides BC into two equal parts, So, AP is a median

⇒ AP divides △ABC into two equal parts.

Hence, statement I follows.

Statement II:

AB = 10 cm, BC = 20cm, BP = 10cm

by Pythagoras theorem,

AC² = AB² + BC²

⇒  AC² = 100 + 400 = 500

⇒ AC = √500 = 22.36 

AP2 = AB2 + BP2

⇒  AP2 = 100 + 100 = 200

⇒ AP = √200 = 14.14

∴ The perimeter of △APC is 10 + 14.14 + 22.36 = 46.50 cm which is more than 46 cm

Hence, statement II also follows.

Statement III:

Area of △ABC = $1 \over 2$ × BC × AB

⇒ Area of △ABC = $1 \over 2$ × 20 × 10 = 100 cm²

⇒ Area of △APC = 100/2 = 50 cm²

Hence, statement III also follows.

All statements follow.

Formula Used:

Median = l + ($n/2 - cf \over f$) × h

where,

l = lower term of the class 

n = total number of students

cf = cumulative frequency value of the previous class

f = frequency value of the class

h = size of the class

Calculation:

 

Median lies on (n/2)^th observation, where,

n = 50, n/2 = 25

Comparing 25 with the "cf" value,

Then, the class lies in the interval (44.5 - 53.5), as 25 is under 37.

Also, l = 44.5

f = 15

cf = 22 

h = (53.5 - 44.5) = 9

⇒ Median = 44.5 + ($25 - 22 \over 15$) × 8

⇒ Median = 44.5 + $\frac{3}{15}$ × 9

⇒ Median = 44.5 + 9/5 = 44.5 + 1.8 = 46.3

∴ The median is calculated as 46.3.

Mistake Points

When we determine the mean, median or mode of the grouped data (interval data), then the interval must be in regular intervals. If it is an irregular interval then we have to convert it into a regular interval.

To convert into regular intervals:

Lower limit ⇒ lower limit - 0.5

Upper limit ⇒ upper limit + 0.5