Quantitative Aptitude PYQs

Calculation:

⇒ For country A = (38 - 35)/35 × 100 = 8.5%

⇒ For country B = (47 - 45)/45 × 100 = 4.4%

⇒ For country C = (93 - 88)/88 × 100 = 5.6%

⇒ For country D = (79 - 75)/75 × 100 = 5.3%

⇒ For country E = (60.9 - 58)/58 × 100 = 5%

∴ From above, in A, C, D and E countries production of cars increased by more than or equal to 5% in 2016 over 2015.

Calculation:

⇒ 50% marks of total marks = 80 × 50/100 = 40

⇒ Total number of students who have obtained less than or equal to 40 marks

= 5 + 8 + 10 + 13 = 36

⇒ Total students = 90 

⇒ Required percentage = 36/90 × 100 = 40%

∴ 40% is the correct option.

Shortcut Trick  

 -1, 2, 2, 3, 6, 8, 9

The middle number is 3

Alternate Method 

Concept: 

Median:

Case 1: If the number of observations (n) is even

\({\rm{Median;}} = {\rm{;}}\frac{{{\rm{value;of;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{;observation;}} + {\rm{;;value;of;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{;}} + 1} \right)}^{{\rm{th}}}}{\rm{;observation}}}}{2}\)

Case 2: If the number of observations (n) is odd

\({\rm{Median;}} = {\rm{value;of;}}{\left( {\frac{{{\rm{n}} + 1}}{2}} \right)^{{\rm{th}}}}{\rm{;observation}}\)

Calculation:

Data arranged in ascending order = -1, 2, 2, 3, 6, 8, 9.

∴ Median = The number in the middle = 3

Important Points

If the number of observations is odd, the number in the middle of the list is the median.

Given:

First ten composite number = 4, 6, 8, 9, 10, 12, 14, 15, 16, 18.

Formula:

Arithmetic mean = total sum of sample/ number of sample

Calculation:

Arithmetic mean of first ten composite numbers

⇒ (4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18)/10

⇒ 112/10 = 11.2

∴ Arithmetic mean of the first ten composite numbers is 11.2.

Important Points

Composite numbers can be defined as the whole numbers that have more than two factors.

Calculation:

⇒ Total new marks of five students = (21 + 5) + (27 + 5) + (19 + 5) + (26 + 5) + (32 + 5) = 150

∴ Required revised average marks = 150/5 = 30

Given:

Mean of m observations = p

Mean of n observations = q

Mean of (m + n) observations = c 

Concept:

Mean: The mean is defined as the sum of the observations

divided by the total number of observations. 

Calculation:

According to the question,

⇒ Total sum of m observations = pm

⇒ Total sum of n observations = qn

⇒ Total sum of m and n observations = pm + qn

⇒ c = (pm + qn)/(m + n)

As, p ≤ q,

∴ From above, p ≤ c ≤ q is option is correct.

Calculation:

⇒ Total production of type1 product = 160 + 200 + 135 + 240 + 180 + 210 = 1125

⇒ Total production of type2 product = 80 + 150 + 35 + 95 + 110 + 150 = 620

⇒ Total production of type3 product = 70 + 85 + 44 + 120 + 85 + 100 = 504

⇒ Total production of type4 product = 90 + 160 + 95 + 80 + 95 + 92 = 612

⇒ Total production of type5 product = 75 + 100 + 85 + 120 + 115 + 110 = 605

∴ From above, type3 product is least produced over 2000 - 2005.

Calculation:

⇒ Total tablets produced in 2000 = 160 + 80 + 70 + 90 + 75 = 475

⇒ Total tablets produced in 2001 = 200 + 150 + 85 + 160 + 100 = 695

⇒ Total tablets produced in 2002 = 135 + 35 + 44 + 95 + 85 = 394

⇒ Total tablets produced in 2003 = 240 + 95 + 120 + 80 + 120 = 655

⇒ Total tablets produced in 2004 = 180 + 110 + 85 + 95 + 115 = 585

⇒ Total tablets produced in 2005 = 210 + 150 + 100 + 92 + 110 = 662

For (2003, 2005), difference = 662 - 655 = 7

For (2001, 2005), difference = 695 - 662 = 33

For (2003, 2004), difference = 655 - 585 = 70

For (2000, 2002), difference = 475 - 394 = 81

∴ For (2003, 2005), the difference in total number of tablets produced between them is minimum.

Shortcut TrickHere it is asked about the ratio and we know that in ratio we reduce all the values to their standard form.

⇒ Decrease in production in 2004 compared to 2001 = 20

⇒ Decrease in production in 2000 compared to 2001 = 40

⇒ Ratio = 20 : 40 = 1 : 2 

∴ The correct answer is 1 : 2.

Calculation:

⇒ Total tablets produced in 2000 = 160 + 80 + 70 + 90 + 75 = 475

⇒ Total tablets produced in 2001 = 200 + 150 + 85 + 160 + 100 = 695

⇒ Total tablets produced in 2004 = 180 + 110 + 85 + 95 + 115 = 585

⇒ Percentage drop in production in 2004 compared to 2001 = (695 - 585)/695 × 100 = 110/695 × 100

⇒ Percentage drop in production in 2000 compared to 2001 = (695 - 475)/695 × 100 = 220/695 × 100

∴ Required ratio = (110/695 × 100) : (220/695 × 100) = 1 : 2

Calculation:

For 2000, 160 = 70 + 90

For 2001, 200 < 85 + 160

For 2002, 135 < 44 + 95

For 2003, 240 > 120 + 80

For 2004, 180 = 85 + 95

For 2005, 210 > 100 + 92

∴ From options, in 2003, the production of Type 1 is more than the sum of the production of Type 3 and Type 4.

Given:

Ratio of sides of triangle is x : (x - 1) : (x - 18)

Formula:

Pythagoras theorem,

(diagonal)² = (side1)² + (side2)²

Calculation:

Let sides of right angled triangle be ax, a(x - 1), a(x - 18) respectively.

(ax)² = [a(x - 1)]² + [a(x - 18)]²

⇒ x² = x² - 2x + 1 + x² - 36x + 324

⇒ x² - 38x + 325 = 0

⇒ x = 25 or x = -13

Sides of right angled triangle = 25, 24 and 7

∴ Perimeter of right angled triangle = 25 + 24 + 7 = 56 units

Shortcut TrickPerimeter = x + x - 1 + x - 18 = 3x - 19

Now, we can see, answer (i.e. perimeter) should be divisible by 3 if

we add 19.

From option, 56 + 19 = 75

Which is divisible by 3.

∴ The perimeter of the triangle is 56.

Calculation:

AM = MN = NB

Δ AMP and Δ ABC are similar triangle as ∠AMP = ∠ABC = 90⁰ and MP ∥ BC and AM : AB = 1 : 3

⇒ MP = BC/3 = 12/3 = 4 cm

Similarly, ΔANQ and ΔABC are similar triangle as ∠ANQ = ∠ABC = 90⁰ and NQ ∥ BC and AN : AB = 2 : 3

⇒ NQ = 2BC/3 = 2 × 12/3 = 8 cm

∴ (PM + QN) = 4 + 8 = 12 cm

Calculation:

ACBD is a cyclic quadrilateral.

∠DAC = ∠ACB = ∠CBD = ∠BDA = 90^º all are angles in semicircle.

⇒ AO = DP = BP = CP = radius of a circle.

ΔAPD and ΔDPB are an isosceles triangles.

⇒ ∠PAD = ∠PDA = 60º

⇒ ∠APD = 180º - 2 × 60º = 60º

⇒ ∠APD + ∠BPD = 180º......(Linear angles)

∴ ∠BPD = 180º - 60º = 120º

Given:

ΔABC is an equilateral 

Concept used:

The angles subtended by an arc on the circumference on the same side of the arc are equal.

All the angles of the equilateral triangle are equal and the measure of each angle is 60°.

Calculation:

ΔABC is an equilateral 

∠CAB = 60° 

The angles subtended by an arc on the circumference on the same side of the arc are equal

∠CAB = ∠CDB = 60°

Calculation:

Let AM = AN = a

CM = 4 - a and BN = 6 - a

CO = 4 - a and BO = 6 - a

Then,

⇒ CO + BO = 8

⇒ 4 - a + 6 - a = 8

⇒ a = 1

Radius of three circles = (a = 1), (4 - a = 3) and (6 - a = 5)

∴ Sum of radii of three circles = 1 + 3 + 5 = 9cm

​Formula:

Tangent-secant theorem,

PT² = PA × PB

Calculation:

 

We know  that,

PT2 = PA × PB

⇒ 12² = 9 × PB

⇒ 9 × PB = 144

⇒ PB = 16 cm

∴ AB = 16 - 9 = 7cm

Shortcut Trick We know (5, 12, 13) is a Pythagorean Triplet.

Here, Hypotenuse = 13

Also, 5 + 12 +13 = 30 cm which is the perimeter of the triangle.

So, Area = (1/2) × 5 × 12 = 30 cm²

∴ The area of the triangle = 30 cm².

Calculation:

⇒ hypotenuse + base + height = 30

⇒ base + height = 17

Using pythagoras theorem,

⇒ hypotenuse² = base² + height²

⇒ base² + height² = 169

Solving,

⇒ Base × height = 60

∴ Area of right angled triangle = 1/2 × base × height = 1/2 × 60 = 30 cm²

Calculation:

Using pythagoras theorem,

⇒ AB = √(36 + 64) = 10cm

⇒ Area of Δ ABC = 1/2 × 6 × 8 = 24 cm²

⇒ Area of Δ ABC = 1/2 × p × AB

⇒ 1/2 × p × AB = 24

⇒ p = 4.8

∴ CP = p = 4.8cm

Calculation:

ABCD is a trapezium.

Draw OF || BA || CD, intersecting AD in F.

FO || AB (from figure)

⇒ DO/OB = DF/FA [basic proportionality theorem]

FO || DC (from figure)

⇒ CO/OA = DF/FA [basic proportionality theorem]

⇒ DO/OB = CO/OA

So, diagonals of a trapezium divide each other proportionally.

ΔAOB and ΔDOC are similar triangle

⇒ A(ΔAOB) : A(ΔDOC) =  AB² : DC²

∴ A(ΔAOB) : A(ΔDOC) = 9 : 4

Calculation:

Let the circle touches the sides AB, BC, CD and DA of the quadrilateral ABCD at points N, O, P and M respectively.

Since the lengths of the tangents drawn from an external point are equal. So

⇒ DP = DM (tangents on circle from point D)

⇒ CP = CO (tangents on circle from point C)

⇒ BN = BO (tangents on circle from point B)

⇒ AN = AM (tangents on circle from point A)

Adding,

⇒ DP + CP + BN + AN = DM + CO + BO + AM

⇒ CD + AB = AD + BC

∴ AD = 12 + 9 - 8 = 13cm