Practice Test: Quantitative Aptitude - 5 - SSC CGL MCQ

Given,
Ratio of time taken by A and B = 1 : 3
∴ B takes 3 days to finish a unit work while A takes 1 day to finish the same work
The difference of time is 2 days
Given,
Difference of time is 40 days
∴ Time taken by B to complete the work = [(3/2) × 40] = 60 days
∴ A takes 20 days to finish a work
⇒ Work in 1 day by A = 1/20
⇒ Work in 1 day by B = 1/60
⇒ Work in 1 day by (A+B) = [(1/20) + (1/60)] = 4/60
∴ A and B will take days to complete the work = 15 days

Let, the side of square be a
Given,
Diagonal of square (d) = 4 cm
⟹ Side of square = Diameter of inscribed circle
Using Pythagoras theorem:
⟹ d² = a² + a²
⟹ (4)² = 2a²
⟹ a = 2√2cm
∴ Radius of circle = √2cm
⟹ Area of circle = π × r²
∴ Area of circle = π × (√2)² = 2π

Original price of a TV set = Rs. 4,000
∴ after 10% discount price will be = [4000 - (¹⁰/₁₀₀) × 4000] = Rs. 3600
Given,
Increase in service contract = 5%
∴ The raised 5% service contract on new price will be = (5/100) × 3600 = Rs. 180
∴ Final selling price = 3600 + 180 = Rs. 3780

Let, the distributed amount among A, B and C are x, 3x and 5x respectively
Given,
‘A’ received amount = Rs. 500
∴ x = 500
⟹ ‘B’ received amount = Rs. (500 × 3) = Rs. 1500
⟹ ‘C’ received amount = Rs. (500 × 5) = Rs. 2500
⟹ Sum of money = Rs. (500 + 1500 + 2500) = Rs. 4500

Given,
S.P₁ = Rs.300, Profit₁ = 10% and Profit₂ = 15%
We know that,

Where, S.P = Selling Price, C.P = Cost Price

⟹ C.P = Rs. 272.72
⟹ Profit₂ = ^Profit%/₁₀₀ × C.P
⟹ Profit₂ = ¹⁵/₁₀₀ × 272.72
⟹ Profit₂ ≈ Rs. 41
⟹ SP₂ = C.P + Profit₂
⟹ SP₂ = 272.72 + 41
∴ Selling price of the bag will be = Rs. 313.72

We know that,
Percentage weight of 1st ball with respect to 2nd ball will be = ⁵/₇ × 100 = 71 ³/₇ %

Given,
Speed of bus = 40 km/hr = (100/9) m/s
Time = 5 s
We know that,
Distance = Speed × time
∴ Distance = (100/9) × 5 = 55.55 m

On solving,

Let, length of XY be x
Given,
Perimeter (P₁) of ∆XYZ = 160 cm
Perimeter (P₂) of ∆PQR = 40 cm
∵ ∆XYZ and ∆PQR are similar

Given,
Radius of circle = r
Chord length = radius of circle = r
Draw a perpendicular line of length x from center O

Using Pythagoras theorm,
OA² = Aa² + Oa²
⇒ r² = (^r/₂)² + x²
⇒ x² = 3r²/₄
⇒ x = ^√3/₂ r
Take any one triangle so formed
⟹ cos θ = ^√3/₂
⟹ θ = π/6
∴ The angle subtended by the chord at the center = 2 × π/₆ = π/3

We know that,
sin 60° = √3/2
cosec 45° = √2
cot 30° = √3
tan 45° = 1
Putting these values in the given equation,

⟹ [(3/4) – 2] + [3 + 1] = 11/4

Let, the number of males be ‘m’ and number of females be ‘n’
Salaries of male employees be (x₁, x₂, x₃,…….x_m)
Salaries of females employees be (y₁, y₂, y₃,………y_n)
Given,
Average salary of males = Rs.6000
⟹ m/n = 1
⟹ m = n  ----(iii)

Given,
x = √3-1
Put the value of x in given equation
⇒ x² + 2√3 = (√3 - 1)² + 2√3
⇒ 3 + 1 - 2√3 + 2√3 = 4

We can write,
(55)³ = (50 + 5)³
We know that,
(a + b)³ = (a)³ + (b)³ + 3 × (a × b) (a + b)
⟹ (50 + 5)³ = (50)³ + (5)³ + 3 × (50 × 5) (50 + 5)
⟹ (50 + 5)³ = (125000) + (125) + 3 × 250 × 55
∴ (50 + 5)³ = 166375

Using Pythagoras's theorem,
⇒ PR² = PQ² + QR²
⇒ PR² = 4² + 6²
⇒ PR = √52 cm = 2√13
⇒ PS = PR/2 = √13

Let, the diagonal of two squares be d₁ and d₂. And their sides and area be a₁ and a₂ & A₁ and A₂ respectively

Given,

Using Pythagoras theorm in any one triangle
⇒ d₁² = a₁² + a₁²
⇒ d₁² = 2a₁²

Similarly,

∴ The ratio of both squares

 

Let, the length of the ladder be ‘l’ which is QR

 

Use trigonometric method in given triangle PQR
⇒ cos θ = Base/Hypotenuse 
⇒ cos 30° = 5/l
⇒ l = 10/√3 m

Let, the two numbers are 15x and 15y
Given,
⇒ (15x) × (15y) = 1800
⇒ xy = 8
Value of co-primes of x and y are (1, 8)
∴ The two numbers are (1 × 15, 8 × 15) = (15, 120)

Let, the sum of the money be x. CI is compound interest and SI is Simple interest

Given,
CI – SI = 3

Let, the milk be in given mixture is 3x and water be in given mixture is 2x
According to question,
⇒ 3x + 2x = 20
⇒ x = 4
∴ the milk in given mixture is 12 litres and the water be in given mixture is 8 litres
After adding 2 litres of water in the given mixture, the quantity of water will be (8 + 2) = 10 litres
∴ Ratio of milk to water in new mixture = 12/10 = 6/5 = 6 : 5

∵ Total number of officers = 1400
∵ Percentage of officer who go to office by Bus = 20%
∴ Number of officers who go to office by Bus = 20% of 1400 = 280

∵ Total number of officers = 1400
∵ Percentage of officer who go to office by Metro Rail = 8%
∴ Number of officers who go to office by Metro Rail = 8% of 1400 = 112
मेट्रो रेल से कार्यालय जाने वाले अधिकारियों की संख्या क्या है?

∵ Total number of officers = 1400
∵ Percentage of officer using Cars = 21%
∵ Percentage of officer using trains = 36%
∴ required difference = 36% of 1400 – 21% of 1400 = 15% of 1400 = 210

∵ Total number of officers = 1400
∵ Percentage of officer using two - wheelers = 15%
∵ Percentage of officer using trains = 36%
∴ Required ratio = (15% of 1400) : (36% 0f 1400) = 15 : 36 = 5 : 12