Practice Test: Quantitative Aptitude - 7 - SSC CGL MCQ

We know that the formula of the volume of a hollow sphere is
4π(R³ – r³)
Here, R = external radius and r = internal radius
A hollow sphere of internal and external radius 3 cm and 5 cm respectively
So the volume of the hollow sphere = [4×π×(5³ – 3³)]/3 cc
Now, after melting this sphere, we will get a right circular cone, which’s diameter is 8 cm
So, radius of that cone = 8/2 cm = 4 cm
We know that the formula of the volume of a right circular cone is πr²h/3
Here, r is the radius of the cone and h is the height of the cone
From the question,
 we can make the equation,
π × 4² × h/3 =  [4×π×(5³ – 3³)]/3
⇒ 4h = 98
⇒ h = 24.5
So, the height of the cone is 24.5 cm

If A and B can together do a piece of work in 28 days which means in 1 day, A and B will finish 1/28^th of the work.

If A, B and C can together finish the work in 14 days which means in 1 day, A, B and C will finish 1/14^th of the work.

Putting the first value in second

then they will collide i.e. cars will reach at the same time.

Assume,
1) Mixture 1: mixture of 2/5 part water & rest sugar syrup.
2) Mixture 2: mixture of 3/5 part water & rest sugar syrup.
3) Mixture 3: mixture of mixture 1 & mixture 2.
Amount of water and sugar syrup in different mixtures are represented in tabulated form for better understanding.

∴ The ratio of water & sugar syrup in the new mixture = 6:5

We have area of an equilateral triangle = (√3/4) × side × side
⇒ 121√3 = (√3/4) × side × side
⇒ side × side = 121 × 4
⇒ side = 22
Perimeter = 3 × side
= 3 × 22
= 66 cm
We now have a circle with perimeter 66 cm.
⇒ 2 × (22/7) × radius = 66 cm
⇒ (44/7) × radius = 66 cm
⇒ Radius = 7 × 1.5
= 10.5 cm

Total number of students = 50
20 students had an average score of 70
∴ Total score of these 20 students
= Average score × Number of students
= 70 × 20 = 1400
Remaining number of students = 50 – 20 = 30
Remainingstudents had an average score of 50
∴ Total score of these 30 students
= Average score × Number of students
= 50 × 30 = 1500
∴ Sum of scoresof these 50 students = 1400 + 1500 = 2900
∴ Average score of these 50 students
= Sum total of these 50 students/50 = 2900/50 = 58

Let the total number of workers in the factory be X. Then,
Total number of workers above 30 years of age

Also it is given that out of these 75% are male. Thus
Total number of male workers above 30 years of age

It is given that total number of male workers above 30 years
= 1350
Hence,
⇒ 0.45X = 1350
⇒ X = 3000
∴ Total workers in the factory are 3000.

We know that the difference between the simple interest (SI₂) & compound interest (CI₂) for 2 years [compounded annually] on a sum of Rs. P at a rate R is,

Now the given information, P = Rs. 2900, CI₂ – SI₂ = Rs. 14.21
Putting the values we get,

The annual rate of interest is 7%

Let, the third proportional is ‘x’

Formulas:

Area of a triangle 
Area of a trapezium = 
Triangle ABC and A’B’C are similar by SAS

By property of similar triangles,
∠B = ∠A’B’C and 2A’B’ = AB
∴ A’B’ is parallel to AB.
Thus quadrilateral AA’ B’B is a trapezium.
In quadrilateral AA’ B’B,
Height = BB’
⇒ Height = BC/2
Sum of parallel sides = (A’B’ + AB) = 3AB/2
Area of quadrilateral AA’ B’B  
⇒ Area of quadrilateral AA’ B’B 
Area of triangle ABC  
Ratio area of quadrilateral AA’ B’B to area of triangle ABC 
⇒ Ratio of area of quadrilateral AA’ B’B to area of triangle ABC = 3 : 4

Let the multiplying factor be ‘x’.
Materials = 3x, Labor = 4x and Overheads = 2x
Given that, materials cost = Rs.33.60
As per sum,
⇒ 33.60 = 3x
⇒ x = 11.20
⇒ Labour = 4x = 4(11.20) = 44.80
⇒ Overheads = 2x = 2(11.20) = 22.40
∴ Cost of article = 33.60 + 44.80 + 22.40 = 100.80.

Given,
a = √2 + 1
b = √2 – 1
Given expression is,

= 2/2
= 1

We know that, Summation and Subtraction between two odd integers gives an even integer.
Given, a and b are two odd positive integers.
∴ a + b = 2k₁ ……(1)
And a – b = 2k₂ ........(2)
Where, k₁ and k₂are two integers.
(1) + (2) ⇒ 2a = 2(k₁ + k₂) ⇒ a = k₁ + k₂
(1) - (2) ⇒ 2b = 2(k₁ - k₂) ⇒ b = k₁ - k₂
∴ a² + b²
= (k₁ + k₂)² + (k₁ - k₂)²
= k₁² + k₂² + 2k₁k₂ + k₁² + k₂² - 2k₁k₂
= 2(k₁² + k₂²) ………..(3)
Given expression is,
a⁴ – b⁴
= (a² – b²)(a² + b²)
= (a + b) (a - b)(a² + b²)
Putting values from (1), (2) and (3)
= 2k₁ × 2k₂ × 2(k₁² + k₂²)
= 8× k₁ × k₂ × (k₁² + k₂²)
∴ a⁴ – b⁴ = 8× k₁ × k₂ × (k₁² + k₂²) is always divisible by 8.

If (x - 2) and (x + 3) are the factors of the equation, then that means that
⇒ x – 2 = 0
 ⇒ x = 2
Similarly,
⇒ x + 3 = 0
⇒ x = -3
Putting the value of one of the x in the equation such that the value is zero.
⇒ x² + k₁x + k₂
Putting the value of ‘x’ as 2, we get
⇒ (2)² + k₁(2) + k₂ = 0
⇒ 4 +2k₁+ k₂ = 0
Putting the value of ‘x’ as -3, we get
⇒ (-3)² + k₁(-3) + k₂ = 0
⇒ 9 -3k₁+ k₂ = 0
Subtracting both the equations:
⇒ 4 +2 k₁+ k₂ - (9 -3 k₁+ k₂) = 0
⇒ 4 +2 k₁+ k₂ - 9 + 3 k₁- k₂ = 0
⇒ 5 k₁ - 5 = 0
⇒ 5 k₁ = 5
⇒ k₁ = 1
Putting the value of k₁ = 1,
⇒ 4 +2 k₁+ k₂ = 0
⇒ 4 +2 (1)+ k₂ = 0
⇒ 4 +2 + k₂ = 0
⇒ 6 + k₂ = 0
⇒ k₂ = -6
Alternate method:-
(x – 2) and (x + 3) are factors of x² + k₁x + k₂
∴ (x – 2)(x + 3) = x² + k₁x + k₂
⇒ x² + x – 6 = x² + k₁x + k₂
Comparing, we get, k₁ = 1 and k₂ = - 6

In the given question let us add (ab + bc + ca) to all the denominators, such that the equation becomes:

Taking L.C.M of the above equation

⇒ 0

Given, heights of two poles are 180 m and 60 m.
We know that the distance between the two poles acts as base and will be same so,
For the first pole,
tan 60° = height/base
⇒ √3 = 180m / b
⇒ b = 60√3
tan θ = h / 60√3
⇒ tan θ = 1/√3
⇒ θ = 30°

Formula-
sin²θ + cos² θ = 1
Given,
3 sin θ + 5 cos θ = 5 …….eq(1)
Let, 5 sin θ – 3 cos θ = x …..eq(2)
On squaring and adding both equations
(3 sin θ + 5 cos θ)²+ (5 sin θ – 3 cos θ)² = 25 +x²
⇒ 9sin² θ + 25 cos² θ +30sinθ cosθ + 25 sin² θ + 9cos²θ – 30sinθ cosθ = 25 +x²
⇒ 9sin² θ + 9cos² θ+ 25 cos² θ +25 sin² θ = 25 +x²
⇒ 9(sin² θ +cos² θ) + 25(sin² θ +cos² θ) =25 +x²
⇒ 9 + 25 = 25 +x²
⇒ x²= 9
⇒ x = ±3

Given, selling price (SP) = 1510, 
Marked price – 49/2% of marked price = 1510
⇒ Marked price = 1510/0.755 = Rs. 2000
Marked price is the selling price without discount.
Given, there is a loss of 10%.

Given, vertices of parallelogram are A(1, 2), B(4, y), C(x, 6) and D(3, 5)
We know that diagonals of a parallelogram bisect each other, so let the diagonals cross at point O.
Since O divides AC and BD in 1:1 ratio so,
Coordinate of O (w.r.t. AC) =
Coordinate of O (w.r.t. BD) 
Since both the coordinate s are same so equating them we get,
1 + x = 7
⇒ x = 6
And,
8 = 5 + y
⇒ y = 3

Pythagoras theorem:-
Hypotenuse² = perpendicular² + base²
(a - b)² = a² + b² - 2ab
Let O be the center of the circle.
Since Δ ABC is isosceles therefore line AO will bisect BC perpendicularly.
So, PB = CP
By Pythagoras theorem,
AB² = AP² + BP²
⇒ 36 = AP² + BP² …(1)
And,
BO² = BP² + OP²
⇒ 25 = BP² + (5 – AP)² [∵AP + OP = 5]
⇒ 25 = BP² + 25 + AP² – 10AP      …(2)
Substituting value of AP² + BP² in (2)
⇒ 10AP = 36
⇒ AP = 3.6 cm
⇒ BP² = 36 – 12.96
⇒ BP² = 23.04
⇒ BP = 4.8 cm
⇒ BC = 9.6 cm

Pythagoras theorem,
Hypotenuse² = perpendicular² + base²
Area of a triangle = abc/4R
Where, a, b and c are sides of the triangle and R is the radius of circumcircle.
Area of the triangle PQR = ½ × PQ × PR
Where PR = 8cm, PQ = 6cm
Area of triangle =12×8×6=24cm2=12×8×6=24cm2
Now, using Pythagoras theorem, a² + b² = c²
⇒ c² = (8)² + (6)² = 64 + 36 = 100 = (10)²
⇒ c = 10
Radius 

From the table,
Price per kg of Bajra sold by farmer D = 28
Price per kg of Bajra sold by farmer E = 30
Earning on 240 kg Bajra by D = 240 × 28 = Rs. 6720
Earning on 240 kg Bajra by E = 240 × 30 = Rs. 7200
∴ Required ratio
= 6720 : 7200 = 14 : 15

Price per kg of Bajra sold by farmer A = 22
Price per kg of Bajra sold by farmer B = 24.5
Price per kg of Bajra sold by farmer C = 21
Price per kg of Bajra sold by farmer D = 28
Price per kg of Bajra sold by farmer E = 30
∴ Average price per kg of Bajra sold by all the farmers together
= (Sum of price per kg of Bajra)/number of farmer
= (22 + 24.5 + 21 + 28 + 30)/5
= 125.5/5 = 25.1

From the table,
Price per kg of rice sold by farmer A = 30
Price per kg of corn sold by farmer A = 22.5
Price per kg of jowar sold by farmer A = 18
∴ Earning on 350 kg of rice by farmer A = 350 × 30 = Rs. 10500
Earning on 150 kg of corn by farmer A = 150 × 22.5 = Rs. 3375
Earning on 250 kg of jowar by farmer A = 250 × 18 = Rs. 4500
∴ Total earn = (10500 + 3375 + 4500) = Rs. 18375

Given,
Price per kg of paddy sold by farmer B = 25
Price per kg of rice sold by farmer B = 36
∴ Earning on 150 kg of paddy by farmer B = 150 × 25 = Rs. 3750
Earning on 150 kg of rice by farmer B = 150 × 36 = Rs. 5400
∴ Required per cent
= (3750/5400) × 100
= 3750/54 = 69.44%