Quantitative Aptitude PYQs

 

Z is the midpoint of range of x for every data

sum of all F × Z = 875 + 35P

mean of above data is 25.2 & sum of all F = 8 + 12 + 10 + P + 9 = 39 + P

mean \(= \frac{{875 + 35P}}{{39 + P}} = 25.2\)

⇒ 9.8 P = 107.8

⇒ P = 11

First arrange the data in least to greatest

 

n = 42

so, \(\frac{{n + 1}}{2} = \frac{{42 + 1}}{2} = 21.5\)

so cumulative frequency just greater than 21.5 is 25 & the corresponding value of x is 8

so, our median is 8

Shortcut TrickConcept:

1. When numbers are in Arithmetic progression with a common

the difference is 1 and their average is X, then adding n next terms

will cause an increase in average by n/2

2. Consecutive numbers are always in AP i.e common difference is the same

Calculation:

Initial average = x

4 new numbers are added, then;

New average = x + (4/2) = x + 2

Alternate Method

Let the numbers are 1, 2, ....50

We know that, 

Average of n natural number = (n +1)/2

⇒ Average of 50 natural number 

⇒ (50 + 1)/2 = x

⇒ x = 51/2       ---(1)

When 4 new consecutive numbers included 

⇒ Average = (54 + 1)/2 = 55/2 = 

⇒ 51/2 + 4/2 

⇒ x + 2 

∴ The required answer is x + 2.

Given:

Consider two-digit numbers which remain the same when the digits interchange their positions.

Concept used:

Total weight = Average weight × Number of boxes

Calculation:

Such two-digit numbers are

11, 22, 33, 44, 55, 66, 77, 88, 99

Their average is

\(⇒ \frac{{11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99}}{9}\)

It is an arithmetic progression

So, the sum of the numerator is \(= \frac{9}{2}\left[ {11 + 99} \right] = 55 \times 9 \)

Average \(= \frac{{55 \times 9}}{9} \)

⇒ 55

∴ The average of such two-digit numbers is 55.

Pictogram - It is represented in form of pictures.

So, bar diagram & pie diagram are the only diagrammatic representation

There are two types of data:

1. Primary data 2. Secondary data

Primary data is a term for data collected at source. This types of information is obtained directly from first hand sources by means of surveys, observation, and experimentation and not subjected to any processing or manipulation.

Secondary data is a term for data collected by someone other than the user i.e. the data is already available and analysed by someone else. Common sources of secondary data include various published or unpublished data, book, magazines, newspaper, etc.

Collection of primary data:

1.Observation method

2.Interview method

3.Questionnaire 4. Schedules

Collection of secondary data

1. Published data

2.Unpublished data

Maximum height of the diagram is 15 cm. Which is corresponding to Rs. 6000

so, for Rs. 6000 height is 15 cm

Then education = Rs. 480, height will be \(\frac{{15}}{{6000}} \times 480 = 1.2;cm\)

& miscellaneous = Rs. 1660, height will be \(\frac{{15}}{{6000}} \times 1660 = 4.15;cm\)

So, n observations are

a₁, a₂, a₃, -------- a_m -------- a_n

mean of “m” observation is “n”

\(\Rightarrow \frac{{{a_1} + {a_2} + {a_3} \ldots \ldots {a_m}}}{m} = n\)       ---- (1)

& mean of remaining observation is m so number of such observation are n - 3

Mean = \(\frac{{{a_{m + 1}} + \ldots + {a_n}}}{{n - m}} = m\)         ----(2)

by rearranging & adding equation (1) & (2)

a₁ + a₂ + …… a_{m + 1} …… + a_n = nm + m(n-m)

= 2mn - m²

So total observation is “n”

so mean is \(\frac{{2mn - {m^2}}}{n}\)

\(= 2m - \frac{{{m^2}}}{n}\)

Ogive : An ogive is a graph showing the curve of a cumulative frequency distribution and this Ogive graph generally use to find the Median of a data 

So, we can say that option 3^rd and option 4^th is incorrect

Now, whenever we find the Mean and Median of a data the Median gives the exact value (average) of the data.

Therfore, we can say that the Graphical location of a data shows the value of a Median.

We know that, Frequency polygon is a polygon which represents the distribution of frequency with respect to the class marks of the respective classes.

while drawing frequency polygon, we put

Class Mark towards X- axis

and Frequency distribution towards Y-axis

Now, let one the coordinates of polygon be (a, b) 

⇒ a = x- coordinate (Class mark of a class) and b = y- coordinate (frequency of that class)

and also while drawing frequency polygon we get Base line of the polygon on X- axis

And we know that area of polygon ≠ total frequency 

∴ 4^th option is correct option

Length is 3 times its breadth

⇒ L = 3B

Now length & breadth are increased by 30% & 10%

New length = L + 0.3 L = 1.3 L

New breadth = B + 0.1B = 1.1 B

Perimeter of previous rectangle was = 2 [L + B]

= 2 × [3B + B]

= 8B

Perimeter of new rectangle is = 2 [1.3L + 1.1B]

= 2[(1.3) × 3B + 1.1B] = 10 B

Percentage increase in perimeter

\(= \frac{{10B - 8B}}{{8B}} \times 100 = \frac{2}{8} \times 100 = 25%\)

lengths of triangle a, b & c

\(S = \frac{{a + b + c}}{2}\)

from Heron's formula

\(Area = {\rm{\Delta }} = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}\)        ----(1)

Now sides are halved

\(\frac{a}{2},\frac{b}{2}& \frac{c}{2}\)

So \(s' = \frac{{\frac{a}{2} + \frac{b}{2} + \frac{c}{2}}}{2} = \frac{{a + b + c}}{4} = \frac{s}{2}\)

\({\rm{\Delta '}} = \sqrt {\frac{{s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}}{{16}}}\)

\({\rm{\Delta '}} = \frac{1}{4} \times \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}\)

\({\rm{\Delta '}} = \frac{{\rm{\Delta }}}{4}\)

So, percentage change in area is

\(\frac{{{\rm{\Delta }} - \frac{{\rm{\Delta }}}{4}}}{{\rm{\Delta }}} = \frac{3}{4} = 75% \)

Volume of sphere is \(\frac{4}{3}\pi {r^3}\) and surface area of sphere is 4πr²

Now volume is enclosed by 700%

⇒ new volume \(= \frac{4}{3}\pi {r^3} + \frac{{700}}{{100}} \times \frac{4}{3}\pi {r^3}\)

\(= 8 \times \frac{4}{3}\pi {r^3}\)

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

It means radius is doubled

So, now surface area is = 4 π (2r)²

= 4 × 4πr²

percentage change is surface area

\(= \frac{{4 \times 4\pi {r^2} - 4\pi {r^2}}}{{4\pi {r^2}}} \times 100 = 300% \)

Length if two parallel chords are 12 & 16 cm Radius is 10 cm

AB is diameter & CD is chord, O is center 90 circle

OC² = OE² + CE²

OE² = OC² - CE²

[CE is either 12/2 = 6 cm or 16/2 = 8 cm]

for CE = 6 cm

OE² = 10² - 6²

= 100 - 36

OE² = 64

⇒ OE = 8

For CE = 8 cm

OE² = 10² - 8²

= 100 - 64

= 36

⇒ OE’ = 6

Now both chords are either is same side or opposite sides

so, distance between chords are 8 - 6 = 2 cm or 8 + 6 = 14 cm

So, there are two circular arcs

Arc 1 with center c is BDC

Arc 2 with center A is ABD

Now join B & D

So, area of \({\rm{\Delta }}ABD = \frac{1}{2} \times base \times height\)

\(= \frac{1}{2} \times a \times a\)

\(= \frac{{{a^2}}}{2}\)

Area of arc \(ABD = \frac{1}{4} \times \pi {\left( {radius} \right)^2}\)

\(= \frac{1}{4} \times \pi {a^2}\)

So, area of region 1 is \(= \frac{1}{4}\pi {a^2} - \frac{{{a^2}}}{2}\)

\(= \frac{{{a^2}}}{2}\left[ {\frac{\pi }{2} - 1} \right]\)

Area of region I & II are same from geometry

⇒ Common area in these two sectors is

\(= 2 \times \frac{{{a^2}}}{2}\left[ {\frac{\pi }{2} - 1} \right] = {a^2}\left[ {\frac{\pi }{2} - 1} \right]\)

Square has sides “a” & length of regular octagon is x

Triangle PQR is an right angle triangle with other two angles are 45°

from Pythagoras theorem

PQ² + QR² = PR²

we know PQ = QR because of regular octagon inside square

2PQ² = x²

\(x = \sqrt 2 PQ\)

\(\Rightarrow PQ = \frac{x}{{\sqrt 2 }}\)

so, length of side of a square is \(x + \frac{x}{{\sqrt 2 }} + \frac{x}{{\sqrt 2 }}\) which also equal to “a”

\(\Rightarrow a = x + \frac{x}{{\sqrt 2 }} + \frac{x}{{\sqrt 2 }}\)

\(a = x\left[ {1 + \sqrt 2 } \right]\)

\(\Rightarrow x = \frac{a}{{1 + \sqrt 2 }};or;x = a\left( {\sqrt 2 - 1} \right)\)

Concept:

Pythagoras theorem: For a right-angled triangle

Hypotenuse² = Base² + Perpendicual²

Formula used:

(a + b)² = a² +  2ab + b²

(a - b)2 = a2 - 2ab + b2

Calculation:

Let the side of right-angle triangle be  (x - 1), x, x + 1

Using pythagoras theorem

(x - 1)² + x² = (x + 1)²

⇒ x² - 2x + 1 + x² = x² + 2x + 1

⇒ x² - 4x = 0

⇒ x (x - 4) = 0

⇒ x = 0, 4

But side of triangle can never be zero

⇒ x = 4

So, three consecutive number are 3, 4, 5

∴  Only 1 such set is possible

Circumference of a circle is 2πr

Then circumference of a smaller circle is 44

⇒ 2πr₁ = 44

\(\Rightarrow {r_1} = \frac{{22}}{\pi }\)

& circumferences of the bigger circle is 88 cm

⇒ 2πr₂ = 88

\(\Rightarrow {r_2} = \frac{{44}}{\pi }\)       ----(2)

Then area between two circles is, difference in the area of the larger circle to the area of the smaller circle

\(\Rightarrow \pi r_2^2 - \pi r_1^2\)

\(\Rightarrow \pi {\left( {\frac{{44}}{\pi }} \right)^2} - \pi {\left( {\frac{{22}}{\pi }} \right)^2}\)

\(= \frac{{1936}}{\pi } - \frac{{484}}{\pi }\)

\(= \frac{{1452}}{{3.14}}\)

= 462 square cm (approx.)

length & breadth of outer rectangle are 12,6

so, its area is 12 × 6 = 72 ft²

Now is there a margin of 6 inches                  {12 inches = 1 ft}

Then length of inner rectangle = 12 - 1 = 11 ft

Breadth of inner rectangle = 6 - 1 = 5 ft

so, area of inner rectangle = 11 × 5 = 55 ft²

so, area between two rectangle which is the dark red border = 72 - 55 = 17 square feet.

Given that perimeter of a right-angle triangle is k times the shortest side

⇒ AB + BC + AC = K × BC      ----(1)

Another condition given that ratio of other side to the hypotonus is 4: 5

\(\Rightarrow \frac{{AB}}{{AC}} = \frac{4}{5}\)

Or AB = 4P & AC = 5P (where P is any real number)

AC² = AB² + BC²

(5P)² = (4P)² + BC²

BC = 3P

Now, put AB, BC & AC is equation 1^st

Then

4P + 5P + 3P = K × 3P

12P = K × 3P

⇒ K = 4