Quantitative Aptitude PYQs

Formula Used:

CI = P {1 + \(R \over 100\)}^T - P 

Where P = principal, R = rate and T = time

Calculation:

Given that

When P = x, then CI = y     

⇒ y = x {1 +\(R \over 100\) }T - x   

⇒ \(x + y \over x\) = {1 +\(R \over 100\) }T      ----(1)

When P = y, then CI = z   

z = y {1 +\(R \over 100\) }T  - y    

\(z + y \over y\) = {1 +\(R \over 100\) }T          ----(2)​​

From equation (1) & (2)

\(x + y \over x\) = \(z + y \over y\)

⇒ xy + y² = xz + xy

⇒ y² = xz.

∴ The correct answer is y² = zx.

Shortcut Trick

Innermost square area = 1 unit² ⇒ Diagonal d₁ = √2 units.

Distance from center to corner (R) of S₁ = √2/2 = 1/√2.

Distance between corners is 1, so R_n = 1/√2 + (n − 1).

Diagonal d_n = 2 × R_n = 2(1/√2 + n − 1) = √2 + 2n − 2. (Statement I matches)

Area A_n = d_n²/2 is a quadratic function of n; thus, the area difference A_n − A_n−1 depends on n. (Statement II is false)

∴ The correct answer is I only.

Alternate Method

Given:

Area of innermost square (S₁) = 1 unit²

Distance between corners of S_n and S_n−1 = 1 unit

Formula Used:

Diagonal of square (d) = √2 × side

Area of square = d² / 2

  Center (O) S1 S2 dist = 1 R<sub>n</sub> = R<sub>n-1</sub> + 1

Calculations:

For Statement I:

⇒ Area S₁ = 1 ⇒ side s₁ = 1 ⇒ diagonal d₁ = √2

⇒ Radius (center to corner) R₁ = d₁/2 = √2/2 = 1/√2

⇒ R_n = R₁ + (n − 1) × 1 = 1/√2 + n − 1

⇒ d_n = 2 × R_n = 2 × (1/√2 + n − 1) = √2 + 2n − 2

⇒ d_n = 2n + √2 − 2 (Statement I is Correct)

For Statement II:

⇒ Area A_n = d_n² / 2 = (2n + √2 − 2)² / 2

⇒ A_n = [ 4n² + 2 + 4 + 4√2n − 8n − 4√2 ] / 2

⇒ A_n = 2n² + (2√2 − 4)n + (3 − 2√2)

⇒ Area difference ΔA = A_n − A_n−1 = [2n² − 2(n−1)²] + (2√2 − 4)[n − (n−1)]

⇒ ΔA = 2(2n − 1) + 2√2 − 4 = 4n + 2√2 − 6

⇒ Since ΔA contains the term 'n', it is not independent of n. (Statement II is Incorrect)

∴ The correct answer is I only.

Additional Information

Properties of Concentric Squares

If the distance between corresponding sides is 'x', the side length increases by 2x. However, if the distance between corresponding corners is 'x', the diagonal length increases by 2x.

Area and Diagonal Relationship

The area of a square is half the square of its diagonal (A = d²/2). If the diagonal grows in an arithmetic progression, the area grows in a quadratic progression.

Linear Change in Difference

For any sequence where the general term is a quadratic (an² + bn + c), the difference between consecutive terms is always a linear function of n (2an + b − a).

Given:

AC + AB = 3AD     ---(1)

AC - AD = 4 units  ---(2)

Formula Used:

Pythagoras theorem,

AC² = AB² + BC²

Calculation:

Simplifying both equations, 

AC = (4 + AD), putting AC value in eqn (1)

⇒ (4 + AD) + AB = 3AD

⇒ AB = 3AD - AD 

⇒ AB = 2AD - 4

By Pythagoras Theorem, 

⇒ AC² = AB² + AD² 

⇒ (4 + AD)² = (2AD - 4)² + AD²

⇒ 16 + AD² + 8AD = (4AD² + 16 - 16AD) + AD²

⇒ 8AD = 4AD² - 16AD 

⇒ 4AD² = 24AD

⇒ AD = 6

∵ AB = 2AD - 4 

⇒ AB = (2 × 6) - 4 = 8

Hence, the Area of triangle ABC = 1/2 × AB × AD

∴ Area of triangle ABC = 1/2 × 6 × 8 = 24 sq. unit

Given:

Area of bigger circle = \(\frac{π(2+√{3})^2}{3}\) cm²

Concept used:

circumradius of the equilateral triangle = side/√3

Calculation:

Radius of bigger circle (R) = r + circumradius of the equilateral triangle

So,

Circumradius of the equilateral triangle = side/√3 = 2r/√3

Thus,

R = r + 2r/√3 = r(1 + 2/√3) = r(√3 + 2)/√3 = r(2 + √3)/√3

So, Area of bigger circle = π [r(2 + √3)/√3]²

⇒ π [r²(2 + √3)²/3]

According to the question,

π [r2(2 + √3)2/3] = \(\frac{π(2+√{3})^2}{3}\)

⇒ r² = 1

⇒ r = 1 cm

∴ The radius of the smaller circles is 1 cm each.

Calculation:

Taking CD = h

By Pythagoras Theorem,

In  ∆ADC

⇒ h2 + x2 = 202

⇒ h2 + x2 = 400     ---(1)

In ∆BCD

⇒ h2 + (21 - x)2 = 132

⇒ h2 + x2 + 441 - 42x = 169   ---(2)

Putting the value of (h2 + x2) in 2^nd equation,

We get,

⇒ 400 + 441 - 42x = 169

⇒ 841 -169 = 42x

⇒ 672 = 42x

AD = x = 672/42 = 16 cm, 

BD = (21 - x) = 21 - 16 = 5 cm

Now, from equation (1)

⇒ h2 + 16² = 400

⇒ h2 = 400 - 256 = 144

⇒ h = 12 cm

Now, Area of ∆BCD = \(\frac{1}{2} \times BD \times CD\)

Area of ∆BCD = \(\frac{1}{2} \times 5 \times 12\) = 30 cm²

Similarly, Area of ∆ADC = \(\frac{1}{2} \times 16 \times 12\) = 96 cm²

Given:

Container A has a milk to water ratio of 1:3.

Container B has a milk to water ratio of m:n.

The mixtures from containers A and B are mixed in the ratio 2:3 to get 20 litres of a mixture with milk to water ratio of 3:7.

Calculations:

New mixture volume is 20 liters in a 2:3 ratio from containers A and B

Volume taken from container A = (20 × 2)/(2 + 3) = 8 liters

Volume taken from container B = (20 × 3)/(2 + 3) = 12 liters

Milk from A = (1/4) × 8 = 2 liters

Water from A = 8 - 2 = 6 liters

Milk from B = (m/(m + n)) × 12 = 12m/(m + n)

Water from B = (n/(m + n)) × 12 = 12n/(m + n)

Since new ratio of milk and water is 3:7

Sum of milk from A and B = 2 + 12m/(m + n)

Sum of water from A and B = 6 + 12n/(m + n)

New ratio of milk and water = 3/7

⇒ \(\frac{2\ +\frac{12m}{(m + n)}}{6\ +\frac{12n}{(m + n)}}\) = 3/7

⇒ \(\frac{2(m + n) + 12m}{6(m + n) + 12n} = \frac{3}{7}\)

⇒ \(\frac{14m + 2n}{6m + 18n} = \frac{3}{7}\)

⇒ 80m = 40n

⇒ m/n = 40/80 = 1/2

∴ The correct answer is 1/2.

Given:

The sum of the volumes of the cone, hemisphere, and cylinder equals the volume of a sphere of radius ( R ).

Formula used:

Volume of the cone: \(V_{\text{cone}} = \frac{1}{3} \pi r^2 h\)

Volume of the hemisphere: \(V_{\text{hemisphere}} = \frac{2}{3} \pi r^3\)

Volume of the cylinder: \(V_{\text{cylinder}} = \pi r^2 h\)

Volume of the sphere is given by:  \(V_{\text{sphere}} = \frac{4}{3} \pi R^3\)

Calculation:

According to the problem, the sum of the volumes of the cone, hemisphere, and cylinder is equal to the volume of a sphere:

\(V_{\text{cone}} + V_{\text{hemisphere}} + V_{\text{cylinder}} = V_{\text{sphere}} \)

Now, substitute the volumes:

\(\frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3 + \pi r^2 h = \frac{4}{3} \pi R^3\)

\(\left( \frac{1}{3} \pi r^2 h + \pi r^2 h \right) + \frac{2}{3} \pi r^3 = \frac{4}{3} \pi R^3 \)

\(\pi r^2 \left( \frac{1}{3} h + h \right) + \frac{2}{3} \pi r^3 = \frac{4}{3} \pi R^3\)

\(\pi r^2 \left( \frac{4}{3} h \right) + \frac{2}{3} \pi r^3 = \frac{4}{3} \pi R^3\)

\(\frac{4}{3} \pi r^2 h + \frac{2}{3} \pi r^3 = \frac{4}{3} \pi R^3 \)

Multiply by 3:

\(4\pi r^2 h + 2\pi r^3 = 4\pi R^3 \)

Using height (h = r), it simplifies:

\(4\pi r^3 + 2\pi r^3 = 4\pi R^3 \)

\( 6\pi r^3 = 4\pi R^3 \)

\(\frac{R^3}{ r^3} = \frac{6}{4} = \frac{3}{2}=1.5 \)

The correct answer is 1.5

Given:

(x3 + px2 + qx + r)  is an integer for all integral values of x.

Formula used:

Consider the polynomial when ( x ) is substituted by various integer values.

Calculation:

Statement I: p must be an integer

Consider the polynomial evaluated at (x = 0):

03 + p.02 + q.0 + r = r

Since r is an integer, we move to (x = 1):

13 + p.12 + q.1 + r = 1 + p + q + r

Since this must also be an integer, it follows that (1 + p + q + r) must be an integer.

Given that r and 1 are integers, (p + q) must also be an integer.

Now evaluate the polynomial at ( x = -1 ):

(-1)3 + p(-1)2 + q(-1) + r = -1 + p - q + r

Since this must also be an integer, (-1 + p - q + r) is an integer.

Given that ( r ) and ( -1 ) are integers, (p - q) must be an integer.

From the above expressions, we have ( p + q ) and ( p - q ) as integers. If we add these two expressions, we get:

[ (p + q) + (p - q) = 2p

Therefore, (2p) is an integer, implying p must be an integer.

Statement II: q must be an integer

Since p is established to be an integer, and both ( p + q ) and ( p - q ) must be integers (from the previous discussion),

it follows that q must be an integer to satisfy both conditions ( p + q ) and ( p - q ).

Statement III: r must be an integer

Consider again the evaluation at ( x = 0 ):

03 + p.02 + q.0 + r = r

Since any polynomial evaluated at (x = 0) gives ( r ), and it is given that the polynomial must yield an integer for all integer values of (x), r must also be an integer.

Therefore, all three statements are correct.

Given:

XYZ is a 3-digit number, with X, Y, and Z distinct non-zero digits.

XYZ - YXZ = 90

Concept Used:

General Form of 3 digit number XYZ = 100 X + 10 Y + Z

Calculation:

⇒ XYZ = 100X + 10Y + Z 

⇒ YXZ = 100Y + 10X + Z

Since, XYZ - YXZ = 90

⇒ (100X + 10Y + Z) - (100Y + 10X +Z) = 90

⇒ 90 (X - Y) = 90 

⇒ (X - Y) = 1 ⇒ X > Y

But X can be maximum at 8, then Possible values of (X, Y) are (9,8), (8,7), (7,6), (6,5), (5,4), (4,3), (3,2), (2,1)

Total possible values = 8

∴ The right answer is 8

Calculation:

Minute hand coincide with the second hand every minute as the second completes a rotation of 60 secs.

which means 60 times in 60 minutes.

⇒ 1 hour = 60 times

⇒ 2 hour = 120 times (2.01 pm to 4.01pm)

∴ The right answer is 120 times.

Concept Used:

HCF (2^m − 1, 2ⁿ − 1) = 2 ^HCF(m,n) − 1

Where, 

HCF = Greatest number that can evenly divide all of them.

e.g., (8, 12) 

Prime factors of 8 = 2 × 2 × 2 = 2³ 

Prime factors of 12 = 2 × 2 × 3 = 2² × 3

HCF = 2 × 2 = 2²

Calculation:

According to the question,

HCF of (236 - 1 and 245 - 1) can be

⇒ 2 ^{HCF (36, 45)} - 1

So,

Prime factors of 36 = 2² × 3²

Prime factors of 45 = 3² × 5

Then, HCF of 36 and 45 = 3² = 9

Now,

⇒ 2⁹ - 1 = (2⁵ × 2⁴) - 1 = (32 × 16) - 1 = 512 -1 = 511

∴ The correct answer is 511.

Alternate Method 

Concept Used:

HCF (2m − 1, 2n − 1) = 2 HCF(m,n) − 1

Calculation:

HCF of (236 - 1 and 245 - 1) 

⇒ 2 HCF (36, 45) - 1

⇒ HCF (36, 45) = 9

⇒ 29 - 1

⇒ 512 - 1 = 511

∴ The HCF is 511.

Given:

Side of equilateral triangle = 14 cm

π = \(\frac{22}{7} \)

Formula used:

Height of equilateral triangle = \(\frac{√3}{2} \)× side

The volume of the cone = \(\frac{1}{3}πr^2h\)

 r = Side of triangle / 2

Concept used:

Height of cone = height of equilateral triangle

Radius of cone = half of triangle base

Calculation:

Height of equilateral triangle = \(\frac{√3}{2} \)× side

⇒  \(\frac{√3}{2} \)× 14 = 7√3 cm

Radius of cone = side of triangle / 2

⇒ r = 14 / 2

⇒ r  = 7 cm

The volume of the cone = \(\frac{1}{3}πr^2h\)

⇒ (1 / 3) × (22 / 7) ×  7 × 7 × 7√3

⇒ 1078 / √3 cubic cm

∴ The column of cone is 1078 / √3 cubic cm.

Given:

Radius of cone = 3 cm

Formula used:

Circumcircle of equilateral triangle = Side / √3

Area of circle = πr²

Concept used:

The circumcircle of an equilateral triangle = Radius of the circle

Calculation:

The side length s of the equilateral triangle formed by the vertices of the cones is the distance between the vertices of any two cones.

Since the cones are arranged in such a way that their bases are touching, the distance between the centers of the bases of the two cones is twice the radius of the base:

⇒ s = 2 × radius

⇒ s = 2 × 3 = 6 cm

⇒ Circumcircle of equilateral triangle = 6 / √3 = 2√3

⇒ Radius of circle = 2√3 cm

⇒ Area of circle = π(2√3)²

∴ Area of circle is 12π square cm.

Shortcut Trick:

Radius of a incircle = (p + b - h)/2

Where p = perpendicualr, b = base, h = hypotenus

h² = p² + b² = 5² + 12² = 25 + 144 = 169

So, h = √169 = 13 cm

So, Inradius = (5 + 12 - 13)/2 = (17 - 13)/2 = 4/2 = 2 cm

Hence, the radius of incircle is 2 cm.

Alternate Method

Given:

In a right-angled △ABC where ∠B=90°,

AB = 5 and BC = 12

Formula Used:

r = \( {△ \over s}\)

Where,

r = radius of the in-circle

△ = area of the triangle & s = semi perimeter

Calculation:

By Pythagoras theorem,

AC² = AB²+ BC²

⇒ AC² = 5² + 12²

⇒ AC² = 25 + 144 = 169

⇒ AC = √(169) =13 cm

Area of △ABC = \(1 \over 2\) × b × h

⇒ △ = \(1 \over 2\) × 5 × 12 = 30 cm²

s = \(AB+BC+CA \over 2 \) = \(5+12+13 \over 2\) = 15 cm

Then, r = \(30\over 15 \) = 2 cm

∴ The radius of the in-circle is 2 cm.

Given:

A sum of interior angles to exterior angles of a regular polygon of n sides is \(\frac{7}{2}\).

Formula used:

The sum of interior angle = 180° (n - 2)

Interior angle = \(\frac{180° (n - 2) }{n}\)

The sum of the exterior angles of any polygon is always 360°

Calculation:

A sum of interior angles / exterior angles = 7 : 2

⇒ \(\frac{180° (n - 2)}{360°}\) = \(\frac{7}{2}\)

⇒ n = 9

⇒ interior angle = {180 (9 - 2)} / 9

⇒ interior angle = (180 × 7) / 9

 ∴ The interior angle is 140°

Given:

199 = m² - n²

Formula used:

a² - b² = (a - b)(a + b)

Calculation:

199 is a prime no.

So, 1 and 199 is only a factor of 199

m2 - n2 = (m - n)(m + n)

⇒ m + n = 199      ---(1) 

⇒ m - n = 1

⇒ m = 1 + n

putting the value of m in equation (1)

⇒ 1 + n + n = 199

⇒ n = 99

⇒ If n = 99 then m = 100

⇒ mn = 99 × 100

∴ mn is 9900.

Concept:

Prime Number: Numbers that are perfectly divisible only either by itself or 1.

Calculation:

Numbers in the form (2ⁿ - 1) can be expressed by putting n = 1, 2, 3... are

1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047....

According to the Question, we have to take up to 2000 prime numbers.

So, only 3, 7, 31, 127 satisfies, i.e., four numbers

∴ The Prime numbers of the form (2ⁿ - 1) and less than 2000 are 4.

Shortcut Trick 

Given:

If the students opt for either only one language or all three languages

⇒ d + e + f = g

Calculation:

According to the question,

⇒ 160 = E + H + S - d - e - f - g

⇒ 160 = 130 + 120 +110 - 2 × g     ----(d+e+f = g)

⇒ 160 = 360 - 2 × g

⇒ 2 × g = 200

⇒ g = 100

∴ The number of students who studied all subjects is 100.

Alternate Method

Given:

Let E, H and S denote the students who opted for English, Hindi, and Sanskrit

Total no. of students = 160

n(E) = 130, n(H) = 120 and n(S) = 110

Formula Used:

According to the Inclusion-Exclusion Principle of Set Theory,

n(E∪H∪S) = n(E) + n(H) + n(S) − n(E∩H) − n(H∩S) − n(E∩S) + n(E∩H∩S)

where, n(E∪H∪S) = total no. of the students

n(E∩H) = Students opted for both English and Hindi

n(H∩S) = Students opted for both Hindi and Sanskrit

n(E∩S) =Students opted for both English and Sanskrit

 n(E∩H∩S) =Students opted for all subjects

According to the question,

If the students opt for either only one language or all three languages

⇒ { n(E∩H) + n(H∩S) + n(E∩S) } =  3 × n(E∩H∩S)

∵ In E∩H, H∩S, E∩S consists of 3 E∩H∩S

Calculation:

n(E∪H∪S) = n(E) + n(H) + n(S) − (3 × n(E∩H∩S)) - n(E∩H∩S)

⇒ 160 = 130 + 120 + 110 - (2 × n(E∩H∩S))

⇒ 160 = 360 - (2 × n(E∩H∩S))

⇒ 2 × n(E∩H∩S) = 200

⇒ n(E∩H∩S) =100

∴ The number of students who studied all subjects is 100.

Calculation:

As per the question, 

We need to find the last digit of 5^a + 7^b + 11^c + 13^d

When each digit is divided by 10:

The last digit of 5^a is always 5.

Similarly, the last digit of 11^c is always 1.

The last digit of 13^d depends on the cycle of 3:

• 13¹ → 3 mod 10
• 13² → 9 mod 10
• 13³ → 7 mod 10
• 13⁴→ 1 mod 10

     then repeat after every 4 powers

• 71 → 7 mod 10
• 7² → 9 mod 10
• 7³ → 3 mod 10
• 7⁴ → 1 mod 10

    then repeats after every 4 powers

We need to consider all combinations of the last digits of  5a + 7b + 11c + 13d

• 5^a always contributes 5.
• 11^c always contributes 1.
• 7b can contribute 7, 9, 3, or 1.
• 13^d can contribute 3, 9, 7, or 1.

We sum the contributions from each term and find the possible remainders when divided by 10:

• 5 + 1 + 3 + 7 = 16 → 6 mod 10
• 5 + 1 + 3 + 9 = 18 → 8 mod 10
• 5 + 1 + 3 + 1 = 10 → 0 mod 10
• 5 + 1 + 9 + 7 = 22 → 2 mod 10
• 5 + 1 + 9 + 9 = 24 → 4 mod 10
• 5 + 1 + 9 + 1 = 16 → 6 mod 10
• 5 + 1 + 7 + 7 = 20 → 0 mod 10
• 5 + 1 + 7 + 9 = 22 → 2 mod 10
• 5 + 1 + 7 + 1 = 14 → 4 mod 10
• 5 + 1 + 1 + 7 = 14 → 4 mod 10
• 5 + 1 + 1 + 9 = 16 → 6 mod 10
• 5 + 1 + 1 + 1 =  8 →  8 mod 10

The distinct remainder modulo 10 from the above sums are: 0, 2, 4, 6, 8.

∴ The number of distinct remainders of S when divided by 10 is 5.

Formula used:

sin 300= \(\frac{1}{2}\) = \(\frac{P}{H}\) 

Calculation:

In Δ ACD

∠ ACD = 180° - (90° + 60°) = 30° 

Now,

In Δ ABC,

sin C = p/h

⇒ sin 30 = AB/6

⇒ 1/2 = AB/6

⇒ AB = 6/2 = 3 cm

∴ The length of AB is 3 cm.