Quantitative Aptitude PYQs

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.

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

Formula Used:

Mode = l + \(f_1 - f_0 \over (2f_1 - f_0 - f_2)\) × h

Where,

l = lower value of the class

f₁ = frequency

f₀ = frequency before the class

f₂ = frequency after the class

h = size of the class

Calculation:

 

Class (44.5 -53.5) is the class with the maximum number of students.

So, l = 44.5, f₁ = 15, f₀ = 10, f₂ = 8, and h = (53.5 - 44.5) = 9

⇒ Mode = 44.5 + \(15 - 10 \over (2 × 15 - 10 - 8)\) × 9

⇒ Mode = 44.5 + \(5 \over 12\) × 9

⇒ Mode = 44.5 + 15/4 = 44.5 + 3.75 = 48.25

∴ The Mode is calculated as 48.25

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

Given:

ABC is a right-angled triangle at B.

AC - AB = 2 cm, BC = 16 cm.

Formula used:

Pythagoras theorem: AC² = AB² + BC².

sinθ = Opposite / Hypotenuse, cosθ = Adjacent / Hypotenuse.

sinθ + cosθ = Expression to calculate.

Calculations:

Let AB = x and AC = x + 2.

Using Pythagoras theorem:

(x + 2)² = x² + 16²

⇒ x² + 4x + 4 = x² + 256

⇒ 4x = 252

⇒ x = 63

Thus, AB = 63 cm, AC = 65 cm, BC = 16 cm.

Now calculate sinθ and cosθ:

sinθ = BC / AC = 16 / 65

cosθ = AB / AC = 63 / 65

sinθ + cosθ = (16 / 65) + (63 / 65) = 79 / 65

∴ The correct answer is option 4: 79 / 65.

Given:

ABC is a right-angled triangle at B.

AB = 63 cm, BC = 16 cm, AC = 65 cm.

BD is the perpendicular from B to AC.

Formula used:

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

Area can also be calculated as: (1/2) × AB × BC.

Equate the two expressions for area to find BD.

Calculations:

Area using AB and BC:

⇒ Area = (1/2) × AB × BC

⇒ Area = (1/2) × 63 × 16 = 504 cm²

Area using AC and BD:

⇒ Area = (1/2) × AC × BD

⇒ 504 = (1/2) × 65 × BD

⇒ BD = 1008 / 65

⇒ BD = 1008 / 65

∴ The length of BD is \(\frac{1008}{65} \mathrm{~cm}\).

Given:

OM = 10 cm, MN = 16 cm 

Concept Used:

A perpendicular drawn from the center bisects the chord MN.

i.e., MQ = QN = \(1\over 2\) × MN = \( {1 \over 2}\) × 16 = 8 cm

Calculation:

Using Pythagoras theorem,

⇒ OM2 = OQ2 + MQ2

⇒ OQ2 = OM2 - MQ2

⇒ OQ2 = 102 - 82

⇒  OQ2 = 100 - 36 

⇒ OQ2 = 36

⇒ OQ = 6 cm

∴ The length of OQ is 6 cm

Given:

OM = 10 cm, MN = 16 cm

Concept Used:

A perpendicular drawn from the center bisects the chord MN.

i.e., MQ = QN = 8 cm

Calculation:

Using Pythagoras theorem,

⇒ OM2 = OQ2 + MQ2

⇒ OQ2 = OM2 - MQ2

⇒ OQ2 = 102 - 82

⇒  OQ2 = 100 - 36 

⇒ OQ2 = 36

⇒ OQ = 6 cm

Let ∠MOQ = ∠ MOP = θ 

In Δ OMQ, 

tanθ = \({MQ \over OQ}\) 

⇒ tanθ = 8/6 

⇒ tanθ = 4/3

In ΔPMO, 

tanθ = \({PM \over OM}\)

⇒ 4/3 = PM /10

⇒ PM = 40/3

∴ The length of PM is 40/3 cm

Given:

OM = 10 cm, MN = 16 cm

Formula Used:

Area of Triangle = \({1\over 2}\) × Base × Height

Concept Used:

A perpendicular drawn from the center bisects the chord MN.

i.e., MQ = QN = 8 cm

Calculation:

Using Pythagoras theorem,

⇒ OM² = OQ² + MQ²

⇒ OQ² = OM² - MQ²

⇒ OQ² = 10² - 8²

⇒  OQ² = 100 - 36 

⇒ OQ² = 36

⇒ OQ = 6 cm

Now, 

Area of Δ OMN = \({1\over 2}\) × MN × OQ

⇒ Area = \({1\over 2}\) × 16 × 6 

⇒ Area = 48 cm²

∴ The area of ΔOMN is 48 cm²