Quantitative Aptitude MCQ Quiz - Objective Question with Answer for Quantitative Aptitude - Download Free PDF

Given:

The line segment is divided in the ratio 3 ∶ 7.

Formula Used:

Total parts = Sum of the parts in the ratio

Calculation:

Let the line segment be divided into two parts in the ratio 3 ∶ 7.

Total parts = 3 + 7

⇒ Total parts = 10

The line segment contains 10 parts while constructing the point of division.

Given:

The tops of two poles of height 30 m and 14 m are connected by a string. If the wire makes an angle of 30° with the horizontal.

Calculation:

Let the length of the wire be h.

Height of pole 1 = 30 m

AB = 30 - 14 = 16 m

In ΔABC,

Sin30° = AB/AC

⇒ \(\dfrac{1}{2}\) = \(\dfrac{16}{h}\)

⇒ h = 32 m

∴ The length of the wire is 32 m.

Given:

Length of one diagonal of the rhombus = 40 cm

Length of the other diagonal of the rhombus = 60 cm

Formula Used:

In a rhombus, the diagonals are perpendicular bisectors of each other, and they divide the rhombus into four congruent right-angled triangles.

Let's use the Pythagorean theorem to find the length of the side of the rhombus.

Solution:

Let's denote the length of the side of the rhombus as "s" and the diagonal as d1 and d2.

According to the given information, the diagonals of the rhombus are 40 cm and 60 cm. These diagonals divide the rhombus into four congruent right-angled triangles.

Using the Pythagorean theorem, we can write the following equation for one of the right-angled triangles:

(d1/2)² + (d2/2)² = (s)²

Simplifying this equation:

(40/2)2 + (60/2)2 = (s)2

(20)2 + (30)2 = (s)2

400+ 900 = (s)2

s² = 1300

s = √1300

s = 10√13

Therefore, the length of the side of the rhombus is 10√13 cm.

Given:

Days of confinement: 6, 7, 8, 9, 10

Number of patients: 4, 6, 7, 5, 3

Formula used:

Mode is the value that appears most frequently in a data set.

Calculation:

Frequency of days of confinement:

6 → 4

7 → 6

8 → 7

9 → 5

10 → 3

The mode is the value with the highest frequency.

⇒ Mode = 8

∴ The correct answer is option (1).

Given:

Length of train = 1800 meters,

Time taken = 10 seconds,

Ratio of speeds = 8:1

Concept Used:

Speed = Distance / Time

Calculation:

Total speed = Distance / Time = 1800 m / 10 s = 180 m/s (This is the relative speed)

Speed of train = 8/9 × total speed = 8/9 × 180 = 160 m/s

In km/hr, 160 × 18/5 = 576 km/hr

Therefore the speed of the train is 576 km/hr.

Given:

x - 1/x = 3

Concept used:

a³ - b³ = (a - b)³ + 3ab(a - b)

Calculation:

x³ - 1/x³ = (x - 1/x)³ + 3 × x × 1/x × (x - 1/x)

⇒ (x - 1/x)3 + 3(x - 1/x)

⇒ (3)³ + 3 × (3)

⇒ 27 + 9 = 36

∴ The value of x³ - 1/x³ is 36.

Given:

Profit = 25 Percent

Discount = 15 Percent

Formula:

MP/CP = (100 + Profit %)/(100 – Discount %)

MP = Printed Price

CP = Cost Price

Calculation:

We know that –

MP/CP = (100 + Profit %)/(100 – Discount %)   ………. (1)

Put all given values in equation (1) then we gets

MP/CP = (100 + 25)/(100 – 15)

⇒ 125/85

⇒ 25/17

Given:

Diameter of semicircle = 14√2 cm

Radius = 14√2/2 = 7√2 cm

Total no. of chords = 6

Concept:

Since the chords are equal in length, they will subtend equal angles at the centre. Calculate the area of one sector and subtract the area of the isosceles triangle formed by a chord and radius, then multiply the result by 6 to get the desired result.

Formula used:

Area of sector = (θ/360°) × πr²

Area of triangle = 1/2 × a × b × Sin θ

Calculation:

The angle subtended by each chord = 180°/no. of chord

⇒ 180°/6

⇒ 30°

Area of sector AOB = (30°/360°) × (22/7) × 7√2 × 7√2

⇒ (1/12) × 22 × 7 × 2

⇒ (77/3) cm²

Area of triangle AOB = 1/2 × a × b × Sin θ

⇒ 1/2 × 7√2 × 7√2 × Sin 30°

⇒ 1/2 × 7√2 × 7√2 × 1/2

⇒ 49/2 cm²

∴ Area of shaded region = 6 × (Area of sector AOB - Area of triangle AOB)

⇒ 6 × [(77/3) – (49/2)]

⇒ 6 × [(154 – 147)/6]

⇒ 7 cm²

∴ Area of shaded region is 7 cm²

Formula used

Area = length × breath

Calculation

The garden EFGH is shown in the figure. Where EF = 220 meters & EH = 70 meters.

The width of the path is 4 meters.

Now the area of the path leaving the four colored corners

= [2 × (220 × 4)] + [2 × (70 × 4)]

= (1760 + 560) square meter

= 2320 square meters

Now, the area of 4 square colored corners:

4 × (4 × 4)

{∵ Side of each square = 4 meter}

= 64 square meter

The total area of the path = the area of the path leaving the four colored corners + square colored corners

⇒ Total area of the path = 2320 + 64 = 2384 square meter

∴ Option 4 is the correct answer.

Given:

Valid votes = 75% of total votes

Winning Candidate = 70% of Valid votes

He won by a majority of 3630 votes

Losing Candidate = 30% of Valid votes

Calculation:

Let 100x be the total number of votes polled

Valid votes = 75% of total votes

= 0.75 × 100x

= 75x

Majority of the Winning Candidate is 3630

Then, Difference between Winning and Losing Candidate = (70 % - 30 %) of valid votes

= 40% of the valid votes

Valid Votes = 75x

Then,

= 0.40 × 75x

= 30x

Hence, 30x is Majority of winning candidate

30x = 3630

x = 121

Total number of votes is 100x

= 100 × 121

= 12100

Answer is 12100.

Concebt used

a.b̅ = a.bbbbbb

a.0b̅ = a.0bbbb

Calculation

0.7 = 0.700000 ̇....

\(0.\bar7 = 0.77777 \ldots\)

\(0.0\bar7 = 0.077777 \ldots\)

\(0.\overline {07} = 0.070707 \ldots\)

Given:

u : v = 4 : 7 and v : w = 9 : 7

Concept Used: In this type of question, number can be calculated by using the below formulae

Calculation:

u : v = 4 : 7 and v : w = 9 : 7

To make ratio v equal in both cases

We have to multiply the 1st ratio by 9 and 2nd ratio by 7

u : v = 9 × 4 : 9 × 7 = 36 : 63 ----(i)

v : w = 9 × 7 : 7 × 7 = 63 : 49 ----(ii)

Form (i) and (ii), we can see that the ratio v is equal in both cases

So, Equating the ratios we get,

u ∶ v ∶ w = 36 ∶ 63 ∶ 49

⇒ u ∶ w = 36 ∶ 49

When u = 72,

⇒ w = 49 × 72/36 = 98

∴ Value of w is 98

GIVEN :

If the price of petrol has increased from Rs. 40 per litre to Rs. 60 per litre

CALCULATION :

Let the consumption be 100 litres.

When price is Rs. 40 per litres, then, the expenditure = 100 × 40

⇒ Rs. 4,000.

At Rs. 60 per litre, the 60 × consumption = 4000

Consumption = 4,000/60 = 66.67 litres.

∴ Required decreased % = 100 - 66.67 = 33.33%

Solution:

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 25/2 + 37/3 + 73/6

= (75 + 74 + 73)/6

= 222/6

= 37

Shortcut Trick

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 12 + 12 + 12 + (1/2 + 1/3 + 1/6)

= 36 + 1 = 37

Solution:

We have to follow the BODMAS rule

Calculation:

\(? = \sqrt[5]{{{{\left( {243} \right)}^2}}}\)

\(⇒ ? = (243)^{\frac{2}{5}}\)

\(⇒ ? = (3 × 3 × 3 × 3 × 3)^{2⁄5}\)

\(⇒ ? = (3^5)^{2⁄5}\)

⇒ ? = 3²