Quantitative Aptitude PYQs

Given:

AB = 16 cm, AC = 12 cm and BD = 4 cm 

And AD is the angle bisector of ∠A

Concept Used:

Angle bisector theorem 

If AD is the angle bisector of ∠A, then 

\(\frac{AB}{AC} = \frac{BD}{CD} \)

Calculation:

Let the length of CD be x cm 

According to the angle bisector theorem 

\(\frac{AB}{AC} = \frac{BD}{CD} \)

⇒ \(\frac{16}{12} = \frac{4}{x} \)

⇒ 16x = 48

⇒ x = 3 cm

∴ The length of CD is 3 cm.

Given:

Side of equilateral triangle = x unit 

The radius of the circle (OA or OB or OC) = y unit 

Concept Used:

For an equilateral triangle 

All the centers i.e circumcentre, orthocentre, centroid, incentre lie on the same point

Centroid divides median in the ratio of 2 :

1. Calculation:

According to the question

ABC is an equilateral triangle 

So, The center of the circle is also the centroid of ΔABC

Since The Height or median of the equilateral triangle (AD) = √3x/2

And we know that centroid divides the median into 2 : 1 ratio

i.e OA : OD = 2 : 1

⇒ OA : AD = 2 : 3

⇒ y : √3x/2 = 2 : 3

⇒ 2y/√3x = 2/3

⇒ 3y = √3x

⇒ √3y = x

∴ The correct relation between x and y is √3y = x.

Given:

D be the midpoint on AC

BD = 6.5 cm

Concept used:

The midpoint of the hypotenuse of a right triangle is the circumcenter of the triangle.

All the vertices of a triangle are at an equal distance from the circumcenter. 

Calculation:

According to the concept,

In ΔABC

BD = AD = DC = 6.5 cm

So, AC = 6.5 × 2

⇒ AC = 13 cm

As ΔABC is a right angle triangle so, AB2 + BC2 = AC²

⇒ AB2 + BC2 = 132

⇒ AB2 + BC2 = 169 cm²

∴ AB2 + BC2 is equal to 169 square cm.

Given:

Height of cylindrical tank = 1m

Diameter of tank = 2m

Radius of spherical droplet = 1 cm

Formula used:

The volume of the cylinder  = π (radius)² × height

The volume of the sphere  = (4/3) × π × (radius)³

Calculation:

The volume of the cylinder = π (1)² × 1 cubic meter

Since 1 meter = 100 cm

⇒ The volume of the cylinder = π(100)² × 100   

The volume of spherical droplet = (4/3)π × 1³

⇒ The volume of spherical droplet = 4π/3 

Therefore, 4π/3 cm³ volume decreases per second.

Hence,

Time required to empty tank = Volume of cylinder/water discharge

⇒  [π(100)2 × 100]/(4π/3)

⇒ 75 × 10⁴ sec

⇒  75 × 104/3600 hr    [∵ 1 hr = 3600 sec]

⇒  208.33 ≈ 210 hr

The tank will be emptied in approximately 210 hour.

Given:

The ratio of the cuboid length, breadth, and height is 27 : 8 : 1, respectively.

Total Surface area of cuboid = p

Total surface area of cube = q

Formula Used: 

Volume of  cuboid = length × breadth × height

Volume of cube = (side)³

Total surface area of cuboid = 2(lb + bh + hl)

Total surface area of cube = 6(Side)²

Calculation:

Let the cuboid length, breadth, and height be 27k, 8k, k respectively.

The total surface area of cuboid = 2(27k × 8k + 8k × k + k × 27k)

⇒ p = 2k²(216 + 8 + 27)

⇒ p = 2k²(251)

⇒ p = 502k²            ------(1)

The cuboid is melted and recast into a cube of side 'a'.

⇒ Volume of cuboid = Volume of a cube

⇒ l × b × h = a × a × a

⇒ 27k × 8k × k = a³

⇒ 216k³ = a³

⇒ a = 6k                   ------(2)

Now the total surface area of a cube

q = 6(6k)²

⇒ q = 216k²             -----(3)

Therefore, the required ratio

\(\frac{p}{q} = \frac{502k^2}{216k^2}\)

⇒ \(\frac{p}{q} = \frac{251}{108}\)

∴ The required value of \(\frac{p}{q}\) is \(\frac{251}{108}\)

Concept:

Criteria for Similarity of Triangles:

Angle–Angle–Angle: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

Pythagoras theorem: In a right angle triangle

Hypotenuse² = Base² + Perpendicular²

Calculation:

In an Δ ABC & Δ ABD

∠A = ∠A = common

∠ABC = ∠BDA  = 90° 

Hence, Δ ABC is similar to Δ ABD. Therefore, both triangles have the same ratio of corresponding sides

⇒ \(\frac{AD}{AB}=\frac{AB}{AC}\)

⇒ AB² = AD × AC

⇒ AB2 = 4 × 9 = 36

⇒ AB = 6 unit

In and Δ ABC, using Pythagoras theorem

⇒ AC² = AB² + BC²

⇒ 81 = 36 + BC² 

⇒ BC = √45 = 3√5 

⇒  AB + BC = 6 + 3√5 

⇒  AB + BC = 6 + 3 × 2.23

∴ AB + BC ≈ 12.6

Additional Information

SSS (Side–Side–Side): If in two triangles, the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

SAS (Side–Angle–Side):  If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Shortcut Trick

According to the Internal Angle Bisector Theorem, the ratio of sides is equal to the ratio of the segments of the base.

AB / AC = BD / DC

12 / AC = 10 / 5 ⇒ 12 / AC = 2 / 1

⇒ AC = 6 cm

Perimeter = AB + (BD + DC) + AC = 12 + 15 + 6 = 33 cm.

∴ The correct answer is 33 cm.

Alternate Method

Given:

Length of side AB = 12 cm

Length of segment BD = 10 cm

Length of segment DC = 5 cm

AD is the angle bisector of ∠BAC.

Formula Used:

In ΔABC, if AD is the internal bisector of ∠BAC, then AB / AC = BD / DC.

Calculations:

⇒ First, calculate the length of side BC:

⇒ BC = BD + DC

⇒ BC = 10 + 5 = 15 cm

⇒ Applying the Angle Bisector Theorem:

⇒ AB / AC = BD / DC

⇒ 12 / AC = 10 / 5

⇒ 12 / AC = 2

⇒ AC = 12 / 2 = 6 cm

⇒ Now, calculate the perimeter of ΔABC:

⇒ Perimeter = AB + BC + AC

⇒ Perimeter = 12 + 15 + 6

⇒ Perimeter = 33 cm

∴ The correct answer is 33 cm.

Additional Information

External Angle Bisector Theorem

The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle: AB / AC = BD / CD.

Length of Angle Bisector (AD)

The length of the internal bisector AD can be calculated using the formula: AD² = AB × AC − BD × DC.

Incenter Property

The point of intersection of all three internal angle bisectors is called the incenter, which is equidistant from all sides of the triangle.

Shortcut Trick

Check if the triangle is right-angled: 4² + 7.5² = 16 + 56.25 = 72.25 = 8.5².

For a right-angled triangle, the inradius (r) is calculated as: r = (Base + Perpendicular − Hypotenuse) ÷ 2.

r = (4 + 7.5 − 8.5) ÷ 2

r = 3 ÷ 2 = 1.5 cm.

∴ The correct answer is 1.5 cm.

Alternate Method

Given: Sides of the triangle are a = 4 cm, b = 7.5 cm, and c = 8.5 cm.

Formula Used: Inradius (r) = Area (Δ) ÷ Semi-perimeter (s)

⇒ Semi-perimeter (s) = (4 + 7.5 + 8.5) ÷ 2 = 20 ÷ 2 = 10 cm

&

Given:

Height of rain water = 5 cm

Area of ground = 2 hectares = 2 × 10⁸ cm²

Concept used:

Volume = Area of base × Height

1 hectare = 100000000 cm² or 10⁸ cm²

1 cm = 1/100 m or 1/10² m

Calculation:

Volume of water that falls on the ground = 2 × 10⁸ × 5

⇒ 10 × 10⁸ cm³

⇒ 109 cm3

Now we need to convert it into m³

⇒ 109 × (1/10)⁶ m³        [∵ {(1/10)²}³ = (1/10)⁶]

⇒ 10³ m3

⇒ 1000 cubic meter

∴ The volume of water that falls on 2 hectares of ground is 1000 cubic meters.

Given:

Radius of the wheel = 35

Covered distance = 11 km = 1100000 cm

Formula used:

The perimeter of the wheel = 2πr

where r is the radius of the wheel

(2πr) × N = D

2πr = Perimeter of wheel

N = Number of revolutions

D = Covered distance

Calculation:

According to the formula,

2πr × n = 1100000

⇒ 2 × (22/7) × 35 × n = 1100000

⇒ 220n = 1100000

⇒ n = 1100000/220

⇒ n = 5000

∴ The value of n is 5000

Given:-

Total number of people below the age of 40 years is 10.5 lakhs

Formula Used:

Percentage = (Value/Total Value) × 100

Calculation:

Total percentage of people below the age 40:- 

⇒ 14.00% + 29.75% + 26.25% =  70%

⇒ 70% of population = 10.5 lakhs

⇒ 1% of population = \(\frac{10.5}{70} \) = 0.15 lakhs

⇒ 1% of population = 15000

⇒ Total population (i.e. 100%) = 15000 × 100

∴ Total population = 15 lakhs

Given:-

Total number of people below the age of 40 = 10.5 lakhs

Formula Used:

Percentage = (Value/Total Value) × 100

Calculation:

Total taxpayers of the age group below 30 years = 14%

% of persons below the age of 40 = 14 + 29.75 + 26.25 = 70%

According to the question, 

70% of total people = 10.5

⇒ 1% = \(\frac{10.5}{70} × 100\)

⇒ 1% =15 thousand

So, the total taxpayers in the age group below 30 years in 14%

⇒ 14 × 15000 = 2,10,000

Since ratio of taxpayers and non-payers in this group is 1 : 2

⇒ Number of taxpayers =  \(2.1\times\frac{1}{3} = 0.7\)

⇒ Total number of taxpayers in that age group is 0.7 (in lakhs).

Concept:

% increse = (Final value - initial value)/initial value] × 100 

Calculation:

 

∴ 2 times the increase in expenditure in a year exceeded by more than 15% as compared to the previous year.

Concept:

% increse = (Final value - initial value)/initial value] × 100 

Calculation:

 

Clearly, we can see, during 2014,

% increase = 25.9 - (-32.4) 

⇒ % increase = 58.3 %

During 2014, % increase is maximum as compared to the previous year.

Given:

The total budget allocated is ₹300.4 lakhs

The budget allotted to A, B, C, D, and E are respectively 40%, 18%, 9%, 25%, and 8%

Formula Used:

Percentage = (Value/Total Value) × 100

Calculation:

The total allocated amount of A and C 

⇒ 120.16 + 27.036 = 147.196

The total allocated amount of B, D, and E

⇒ 54.072 + 75.1 + 24.032 = 153.204

Subtract total amount of A and C from total amount of B, D and E

⇒ 153.204 - 147.196 = ₹ 6.008

∴ The amount allocated to A and C together is  ₹ 6.008 less than the amount allocated to B, D and E together.

Given:

The total budget allocation is ₹300.4 lakhs

The total budget is increased by three times.

⇒ 3 × 300.4 

⇒ ₹901.2

The total increased budget is ₹901.2 lakhs

Formula Used:

Percentage = (Value/Total Value) × 100

Calculation:

Increment in total budget allocation =  (901.2 - 300.4) = 600.8 lakhs

The amount spent on A = 40% of the total budget

⇒ 40% × 600.8

⇒ \(\frac{40}{100}× 600.8\)

⇒ 240.32

∴  TThe total amount on A is ₹240.32 lakhs

Given : 

Total candidates appeared in an examination = 500

30 candidates failed in English only

75 failed in Hindi only

50 failed in mathematics only and 

15 failed in both English and Hindi

17 failed in both Hindi and Mathematics 

17 failed in both Mathematics and English 

5 failed in all three tests

Concept used :

Venn diagram concept

Calculations :

Candidates failed in at least 2 subjects = candidates failed in 2 subjects + candidates failed in 3 subjects

Candidates failed in only English and mathematics = 12

Candidates failed in only English and Hindi = 10

Candidates failed in only Hindi and mathematics = 12

Candidates failed in Hindi, English, and mathematics = 5

Candidates failed in at least 2 subjects = 12 + 10 + 12 + 5 = 39

Percentage of candidates who failed in at least two subjects

⇒ (candidates failed in at least 2 subjects/total candidates) × 100

⇒ (39/500) × 100 = 39/5

⇒ 7.8 %

∴ The percentage of candidates who failed in at least two subjects is 7.8%.

Given : 

Total candidates who appeared in an examination = 500

30 candidates failed in English only

75 failed in Hindi only

50 failed in mathematics only

15 failed in both English and Hindi

17 failed in both Hindi and Mathematics 

17 failed in both Mathematics and English 

5 failed in all three tests

Concept used :

Venn diagram concept

Calculations :

Candidates failed in only english = 30

Candidates failed in only hindi = 75

Candidates failed in only mathematics = 50

Candidates failed in only one subject = 30 + 75 + 50

⇒ 155

The percentage of candidates who failed in only one subject

⇒ (Candidates failed in only one subject/total candidates) × 100

⇒ (155/500) × 100

⇒ 31%

∴ The percentage of candidates who failed in only one subject is 31%.

Given : 

 

Total candidates appeared in an examination = 500

30 candidates failed in English only

75 failed in Hindi only

50 failed in mathematics only and 15 failed in both English and Hindi

17 failed in both Hindi and Mathematics 

17 failed in both Mathematics and English 

5 failed in all three tests

Concept used :

Venn diagram concept

Calculations :

Candidates failed in atleast one subject

⇒ Candidates failed in one subject + candidates failed on two subject + candidates failed in three subject

Candidates failed in english = 30

Candidates failed in Hindi = 75

Candidates failed in mathematics = 50

Candidates failed in english and hindi only = 10

Candidates failed in english and mathematics only = 12

Candidates failed in Hindi and mathematics only = 12

Candidates failed in Hindi, english and mathematics = 5

Candidates failed in atleast one subject = 30 + 75 + 50 + 10 + 12 + 12 + 5

⇒ 194

Percentage of candidates failed in atleast one subject

⇒ (Candidates failed in atleast one subject/Total candidates) × 100

⇒ (194/500) × 100

⇒ 38.8 %

∴ Percentage of candidates failed in atleast one subject is 38.8%

Given :

Total candidates = 500

Failed in English only = 30

Failed in Hindi only = 75

Failed in Mathematics only = 50

Failed in English and Hindi = 15

Failed in Hindi and Mathematics = 17

Failed in Mathematics and English = 17

Failed in all three = 5

Concept used:

Candidates who failed in at least one subject are counted by summing those who failed in only one subject, those who failed in exactly two subjects, and those who failed in all three. However, remember that those who failed in exactly two subjects and all three subjects get counted extra times in the "only" categories, so we adjust for this by adding back those numbers.

Calculations :

So, the calculation goes as follows:

Total failed = (30 + 75 + 50) + (15 + 17 + 17) - 2 x 5

The subtraction by 2 x 5 is essential because the 5 candidates who failed in all three were counted three times in the single failures and once more in the double failures, requiring us to subtract them twice to adjust for the overcounting.

Calculating the number:

Total failed = 155 + 49 - 10 = 194

Therefore, the number of candidates passing in  subjects is the total number of students minus the number who failed any subject:

Passed in all subjects = Total candidates - Total failed = 500 - 194 = 306

Passed in two or more subjects =  306 + candidates failed in only one subject = 306 + 30 + 75 + 50 = 461

So, the correct answer is option 1) 461.