Quantitative Aptitude PYQs

Given:

A circle of center “O” is inscribed in a triangle ABC.

Circle touches the sides AB and AC at M and N respectively.

Measurement of angle A = 70°

Concept Used:

1.The radius from the center of the circle to the point of tangency is perpendicular to the tangent line. 2. Sum of the angles of a quadrilateral is always 360°.

Calculation:

According to the concept used,

∠AMO = ∠ANO = 90°.

Again, it is given that ∠MAN = 70°.

Again, from the concept used,

We can say for quadrilateral AMON,

⇒ ∠AMO + ∠MON + ∠ONA + ∠NAM = 360°

⇒ 90° + ∠MON + 90° + 70° = 360°

⇒ ∠MON = 110°

∴ The value of ∠MON is 110°.

Given:

The sum of the squares of sides of a right-angled triangle is 8,450 square units.

Concept Used:

In right-angle ΔABC,

If ∠B = 90° 

Then by using pythagorus theorem,

(Hypotenuse)² = (Pependicular)² + (Base)²

AC² = AB² + BC²

Calculation:

Let the sides of the right-angled triangle be “a” and “b” units respectively.

If “h” is the length of the hypotenuse,

Then according to the concept used,

⇒ a² + b² = h²      ----(1)

Now, it is given that the sum of the squares of sides of a right-angled triangle is 8,450 square units.

a² + b² + h² = 8450      ----(2)

Now from equation (1) and (2),

2h² = 8450

⇒ h² = 8450/2

⇒ h² = 4225

⇒ h = 65.

∴ The length of hypotenuse is 65.

Given:

A triangle and a parallelogram have equal areas and equal bases.

The altitude of the triangle is k times the altitude of the parallelogram.

Formula used:

Area of parallelogram = base × height

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

Calculation:

Let the length of the base of a triangle and a parallelogram be “b”.

According to the question,

Base of triangle = Base of parallelogram = b.

If “a” is the altitude of the parallelogram,

Then, altitude of the triangle = Ka.

Again, It is given that area of a parallelogram is equal to the area of the triangle,

⇒ ab = 1/2 × ka × b

⇒ k = 2

∴ The value of k is 2.

Given:

Ratio of area of two squares = m² ∶ n⁴

Formula used:

Area of square = (side)²

Perimeter of square = 4 × (side)

Calculation:

As we know that

Area of square = (side)²

Area of Square₁ = m² = (side₁)²

⇒ m = side₁

And, Area of Square₂ = n⁴ = (side₂)²

⇒ (n²)² = (side₂)²

⇒ n² = side₂

Now,

The ratio of Perimeters = Perimeter of Square₁ ∶ Perimeter of Square₂

⇒ (4 × side₁) ∶ (4 × side₂)

⇒ m ∶ n²

∴ The ratio of their perimeters is m ∶ n².

Given:

AD is the median of the triangle ABC

Concept used:

A line segment joining the mid-point of a side to the opposite vertex is called Median

The medians of a triangle are concurrent and a median bisects the area of the triangle

Calculation:

In ΔABC, AD is the median of ΔABC 

Area of ΔABD = Area of ΔADC      ----(1)

In ΔPBC, PD is also the median of ΔPBC

Area of ΔPBD = Area of ΔPCD      ----(2)

Subtracting equation (2) from (1), we have

Area of ΔABD - Area of ΔPBD = Area of ΔADC - Area of ΔPCD

⇒ Area of ΔPAB = Area of ΔPAC   

∴ The Area of ΔPAB is equal to the Area of ΔPAC.

Calculation:

Area of segment = Area of sector - area of triangle 

⇒ \(\rm \frac {\pi r^2 θ}{360^0} - r^2 sin {θ \over2}⋅ cos {θ \over 2}\)

⇒ \(\rm {1\over 2} r^2 ({{\pi θ}\over {180^0}} - 2⋅ sin {θ\over 2}⋅ cos {θ\over 2})\)               (∴ 2 sin θ cosθ = sin 2θ )

⇒ \(\rm {1\over 2} r^2 ({{\pi θ}\over {180^0}} - sin \theta )\)

Given :

ABC is a triangle right angled at C in which P be any point on AC and Q be any point on BC.

Concept used :

In a right angled triangle

Base² + height² = Hypotenuse²

Calculations :

As per the right angled triangle 

For triangle AQC 

AQ² = QC² + AC²      ....(1) 

For triangle BPC 

BP² = BC² + CP²       ....(2) 

Now adding both the equations 

AQ² + BP² = QC² + AC² + BC² + CP² 

⇒ QC² + CP² + AC² + BC² 

⇒ PQ² + AB² 

For condition 2 

Ab = 2PQ only when P and Q are the midpoints of AC and BC respectively 

Here it is given that P and Q are general points, the position is not specified 

So, 

Only only statement 1 is the correct choice.

∴ Option 1 will be the correct choice.

Given:

The uncovered area of square = 42 cm²

Formula used:

Area of Square = (side)²

Area of circle = πr²

where, r = radius

Calculation:

Let the side of the square be 'a' cm and 'r' be the radius of each circle.

According to tot the question

Side of square = r + r

⇒ a = 2r

Now,

Uncovered area of square = Area of Square - [4 × {(1/4) × area of a circle)}]

⇒ 42 cm² = (side)² - (πr²)

⇒ 42 = a² - [(22/7) × (a/2)²]

⇒ 42 = a² - (11a²/14)

⇒ 42 = (14a² - 11a²)/14

⇒ (42 × 14)/3 = a²

⇒ 14 × 14 = a²

⇒ 14 cm = a

So, Radius of each coin, r = a/2

⇒ 14/2

⇒ 7 cm

∴ The radius of each coin is 7 cm.

Given:

Radii of base(r) = x

Radii of top(R) = 2x

Height of bucket(h) = 3x

Formula used:

Let R and r be the radii of the top and base (R > r) respectively and the height of the frustum of the cone be h, then

Volume of a frustum of right circular cone = [(πh/3)(R² + r² + Rr] cu units 

Calculation:

As we know that

Volume of a frustum of right circular cone = [(πh/3)(R2 + r2 + Rr] 

⇒ [{(22/7) × (3x/3)}{(2x)² + (x)² + (2x × x)}]

⇒ [(22x/7)(4x² + x² + 2x²)]

⇒ [(22x/7)(7x²)]

⇒ [(22x)(x²)]

⇒ 22x³ cu. units

∴ The capacity of bucket is 22x³  cu. units

Given :

p, q, r, s, and t represent length, breadth, height, surface area and volume of a cuboid respectively.

Formula used :

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

Volume of cuboid = lbh 

(Where l, b, and h are length, breadth and height of cuboid respectively) 

Calculations :

The surface area of cuboid (s) = 2(pq + qr + rs)

Volume(t) = pqr 

(1/p) + (1/q) + (1/r)

⇒ (pq + qr + qr)/pqr         

⇒ (s/2)/t

⇒ s/(2t) 

∴ Option 3 is the correct answer.

Given:

Total number of candidates who appeared in examination = 15

Marks of candidates who passed in the examination = 9, 6, 7, 8, 8, 9, 6, 5, 4 and 7

Concept used:

If the number of observations is n. Then, arrange the observation in ascending or descending order.

If n is an odd number, then

Median = Value of [(n + 1)/2]^th observation

Calculation:

Let the marks of those 5 candidates who failed in examination be a₁, a₂, a₃, a₄ and a₅

Marks of candidates who passed in the examination = 9, 6, 7, 8, 8, 9, 6, 5, 4 and 7

Now,

Arranging the given marks of all 15 candidates in ascending order, we have

Marks of all 15 candidates = a1, a2, a3, a₄, a₅, 4, 5, 6, 6, 7, 7, 8, 8, 9, 9

As we know that

Median = Value of [(n + 1)/2]th observation

⇒ [(15 + 1)/2]^th observation

⇒ 8^th observation

⇒ 6

∴ The median of marks of all the 15 candidates who appeared in the examination is 6.

Additional Information

For m is even then 

Median = [(n/2)th term + {(n/2) + 1}th term]/2

For m is odd Then

Median = (n + 1)/2}th term

Given:

The yield is 180g, 191g, 175g, 111g, 154g, 141g and 176g

Calculations:

First of all, Arrange it in the ascending order

⇒ 111g, 141g, 154g, 175g, 176g, 180g and 191g

Then, The median is the central number which is 175g.

∴ The median yield is 175g.

Concept used:

Mode: A mode is defined as the value that has a higher frequency in a given set of values.

Calculation:

Manufacture would like to know the size of shoes which has maximum demand. Here mode is the most appropriate measure because it is repeated the highest number of times in the series. So, we used Mode to determine the average of the shoe sold in the shop.

∴ The required answer is Mode.

Concept:

Histogram: A histogram is a display of statistical information that

uses rectangles to show the frequency of data items in successive

numerical intervals of equal size.

Explanation:

This is the image of the histogram. 

The height of the rectangle in the histogram represents the frequency

Shortcut TrickMean age of family = (2 + 5 + 12 + 18 + 38 + 40 + 60)/7

⇒ Mean = 175/7 = 25

Total age of family after 5 years = 175 + (7 × 5) = 210

According to the question,

(210 + x) / 8 = 25 + 1.5

⇒ 210 + x = 26.5 × 8

⇒ x = 212 - 210 = 2 years

Given:

The ages of 7 family members are 2, 5, 12, 18, 38, 40 and 60 years respectively

A new member is added after 5 years

Then, the average age increases by 1.5 years

Formula used:

Mean = Sum of observations/Number of observations

Calculations:

Let the age of new member be x years

According to the question, we have

Mean age of 7 family members is

⇒ Mean = (2 + 5 + 12 + 18 + 38 + 40 + 60)/7

⇒ Mean = 175/7

⇒ Mean = 25

After 5 years, The sum of their ages will be

⇒ 175 + (7 × 5)

⇒ 210

After 5 years, A new member is added and mean increases by 1.5

⇒ New mean = 25 + 1.5

⇒ New mean = 26.5

Total sum of 8 family members are

⇒ Total age = 26.5 × 8

⇒ Total age = 212

So, Age of new member is

⇒ x = 212 - 210

⇒ x = 2 years

∴ The age of new member is 2 years.

Given:

The mean weight of 100 students is 46 kg

The mean weight of boys is 50 kg

The mean weight of girls is 40 kg

Formula used:

Mean = Sum of total observations/Number of observations

Calculations:

Let the number of boys and girls be x and (100 - x)

According to the question, we have

Mean = {50x + 40(100 - x)}/100

⇒ 46 = {50x + 40(100 - x)}/100

⇒ 4600 = 50x + 4000 - 40x

⇒ 4600 - 4000 = 10x

⇒ x = 600/10

⇒ x = 60

So, The number of boys is 60

Hence, The number of girls is 

Number of girls = 100 - x 

⇒ Number of girls = 100 - 60

⇒ Number of girls = 40

The number of boys exceeds that the number of girls by

⇒ Difference = 60 - 40

⇒ Difference = 20

∴ The number of boys exceeds that the number of girls by 20.

Shortcut Trick 

⇒ 10 units = 100 [given]

⇒ 1 unit = 10

⇒ Boys = 6 × 10 = 60

⇒ Girls = 4 × 10 = 40

⇒ Difference in boys and girls  = 60 - 40 = 20

∴ The correct answer is option (2).

Given :

A set of values 25, 65, 73, 75, 83, 76, 17, 15, 7, 14 

Formula used :

Mean = Sum of all values/Number of values 

Calculations :

Mean = (25 + 65 + 73 + 75 + 83 + 76 + 17 + 15 + 7 + 14)/10                   (Number of terms are 10)

⇒ 450/10 

⇒ 45 

Deviation From mean = (45 -25), (45 - 65), (45 - 73), (45 - 75), (45 - 83), (45 - 76), (45 - 17), (45 - 15), (45 - 7) and (45 - 14) 

⇒ 20, -20, -28, -30, -38, -31, 28, 30, 38 and 31 

Sum of deviations = 20 - 20 - 28 - 30 - 38 - 31 + 28 + 30 + 38 + 31 

⇒ 0 

∴ Option 2 will be the correct answer.

Given:

Five observations are given x, x + 2, x + 4, x + 6, x + 8

The mean is m

Formula used:

Mean = Sum of observations/Number of observations

Calculations:

According to the question, we have

Mean = Sum of observations/Number of observations

⇒ m = (x + x + 2 + x + 4 + x + 6 + x + 8)/5

⇒ 5m = 5x + 20

⇒ x = (5m - 20)/5

⇒ x = m - 4

Put the value of x = m - 4 in first three observation, we get

Observations are m - 4, m - 2, m

So, The mean of first three observations is 

⇒ Mean = (m - 4 + m - 2 + m)/3

⇒ Mean = (3m - 6)/3

⇒ Mean = m - 2

∴ The mean of first three observations is m - 2.

Observations are x, x + 2, x + 4, x + 6, x + 8

In a consecutive number series, the middle number is the mean

So, x + 4 = m

The first three observations are

x, x + 2, x + 4

Then, (x + 2) is the mean which two less than (x + 4)

So, Mean of first three observations is m - 2

∴ The mean of first three observations is m - 2.

Given:

The numbers are 2, 4, 6,........., 100

Formula used:

l = a + (n - 1)d

Median = [(n/2)^th term + {(n/2) + 1}^th term]/2

Where,

n is the number of terms

a is first terms 

l is last terms 

d is common difference

Calculations:

In the series, 2, 4, 6,......., 100

l = a + (n - 1)d

⇒ 100 = 2 + (n - 1)2

⇒ 49 = n - 1

⇒ n = 50

Here, n = 50, which is even

According to the question, we have

Median = [(n/2)^th term + {(n/2) + 1}^th term]/2

⇒ [(50/2)^th term + {(50/2) + 1}^th term]/2

⇒ [25^th term + (25 + 1)^th term]/2

⇒ (25^th term + 26^th term)/2

⇒ (50 + 52)/2

⇒ 102/2

⇒ 51

∴ The median of 2, 4, 6, ... 100 is 51.

 

In the consecutive number series, we can use this method

Let us consider a series

2, 4, 6, 8

Then, Median is (8 + 2)/2 = 5

Similarly, In the series 2, 4, 6, 8,........, 100

Median is (100 + 2)/2 = 51

∴ The median of 2, 4, 6, ... 100 is 51.

Additional Information

For m is even then 

Median = [(n/2)th term + {(n/2) + 1}th term]/2

For m is odd Then

Median = (n + 1)/2}th term

Given :

The harmonic mean and the geometric mean of two numbers are 10 and 12 respectively

Concept used :

(Geometric mean)² = Harmonic mean × Arithmetic mean 

Calculations :

according to the formula given above 

(12)² = 10 × Arithmetic mean 

⇒ Arithmetic mean = 144/10 

⇒ 14.4 

∴ Option 4 will be the correct answer.

Alternate Method 

Concept:

A.M. between a and b = \(a+b \over 2\)

G.M. between a and b = \(\sqrt{ab} \)

H.M. between a and b = \(2ab\over a+b \)

Calculation:

Given, G.M. = 12, H.M. = 10

\(GM = \sqrt{ab} \)

\(12^2 = ab\)

ab = 144 ........(1)

\(HM=\frac{2ab}{a+b}\)

\(10=\frac{2ab}{a+b}\)

2ab = 10a + 10b

ab = 5 (a + b)........(2)

From equation (1) and (2)

144 = 5 (a + b)

\(\frac{144}{5}=a+b\)

\(\frac{144}{10}=\frac{a+b}{2}\)

\(\frac{a+b}{2}=14.4\)

But, we know that,

A.M. between a and b = \(a+b \over 2\)

Therefore, AM = 14.4