Practice Test: Quantitative Aptitude - 6 - SSC CGL MCQ

From the pie chart 8% of total employees go to office by metro rail.
Total employees = 1400
∴ Number of employees going to office by Metro rail = 8% of 1400 = 112
∴ 112 employees go to office by metro rail.

From the pie chart 21% of total employees go to office by car.
Total employees = 1400
Number of employees going to office by car = 21% of 1400 = 294
∴ 294 employees go to office by car.

From above question 294 employees go to office by car.
From the pie chart 15% of total employees go to office by two wheelers.
Total employees = 1400
Number of employees going to office by two wheelers = 15% of 1400 = 210
∴ 210 employees go to office by two wheelers.
∴ The ratio of two wheelers and cars being used as modes of transport is 

From above question 294 employees go to office by car.
From the pie chart 36% of total employees go to office by train.
Total employees = 1400
Number of employees going to office by train = 36% of 1400 = 504
∴ 504 employees go to office by train.
∴ Employees availing train - employees availing car = (504 – 294) = 210

We know that angle subtended by any two points from the circumference of the circle on the center is twice the angle subtended by those two points on any other part of the circle, so,
Points Q and S subtend an angle of 65° on the circumference so they subtend 65° × 2 = 130°.
Now, ∠ SOQ(outer) = 360° - 130° = 230°
Now similar to the previous part, ∠ SOQ(outer) is twice of angle B, so ∠ B = 230°/2 = 115°

Centroid of a triangle = [(x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3] where (x₁ , y₁), (x₂, y₂) and (x₃, y₃) are co-ordinates of vertices of triangle
Given, two of its vertices are at (2, 3) and (4, -2) and the centroid is at (1, 1)
Let the coordinate of third vertices be (x, y)
∴ 1 = (2 + 4 + x)/3
⇒ x = - 3
And,
1 = (3 - 2 + y)/3
⇒ y = 2
So third coordinate is (-3, 2)

Now we see that Δ ADB is isosceles as CD = CB and the sum of ∠ CBD and ∠ CDB is 70° (exterior angle property of a triangle). So,
∠CBD = ∠CDB    (isosceles triangle)
∠CBD + ∠CDB = 70°
⇒ 2 × ∠CDB = 70°
⇒ ∠ CDB = 35°

We have,
x + (1/x) = 6
on squaring both side
⇒x² + 1/x² = 36 – 2 = 34

According to the figure,
DE is the electric pole (height = 50m), AB is the building.
∠ADC = 30° & ∠AEB = 60°
We have to find out the length of AB.
As, DC ⊥ AB & EB ⊥ AB so, DC || EB
⇒ DE = BC
Also, DE || AB so, DC = EB
Now, AD × cos30° = DC & AE × cos 60° = BE
⇒ AD × cos30° = AE × cos 60°
⇒ √3 ×AD = AE
From the figure, AB = AC + BC
⇒ AB = AC + DE [As, DE = BC]
⇒ AE × sin 60° = AD × sin 30° + 50
⇒ √3 ×AD × sin 60° = AD × sin 30° + 50
⇒ AD = 50
Now, AB = AD × sin 30° + 50
⇒ AB = 25 + 50 = 75 meter.
The height of the building = 75 meter

Formula:
(a + b)² = a² + b² + 2ab

Squaring both sides we get,

We know that cotθ = tan(90° - θ)
Cot63° = tan(90° - 63°)
Cot63° = tan27°
And cot81° = tan9°
cot9° cot27° cot63° cot 81°
= cot9° cot27° tan27° tan9°
We know that tanθcotθ = 1
= 1

Formulas :-

Given the equation   
Adding both the equation:

Since we know a³ + b³ + 3ab (a + b) = (a + b)³
⇒ (x + 1/x)³ = (2a)³
⇒ x + 1/x = 2a
Similarly, subtracting both the equation:

Since we know a³ - b³ - 3ab (a - b) = (a - b)³
⇒ (x - 1/x)³ = (-2b)³
⇒ x - 1/x = -2b
Adding both the resultant equation,
⇒ x + 1/x + x – 1/x = 2a – 2b
⇒ 2x = 2a – 2b
⇒ x = a – b
Similarly, subtracting both the resultant equation,
⇒ x + 1/x - x + 1/x = 2a + 2b
⇒ 2/x = 2a + 2b
⇒ 1/x = a + b
Multiplying both the final equation:
⇒ (a + b)(a - b) = x × 1/x = 1
⇒ a² – b² = 1

ABC is an equilateral triangle
Given,
AB = BC = CA = 2 cm
∠A = ∠B = ∠C = 60°
Height of equilateral triangle = (√3/2) × (side)
= (√3/2) × 2 = √3 cm
∴ Area of the triangle = (1/2) × √3 × 2 = √3 cm²
Area inside the triangle inscribed by three arcs = 3 × area of one arc
= 3 × (θ/360°) × πr²
= 3 × π × 1² × (60/360) = π/2
∴ Required area = (√3 - π/2) cm²

In Δ OAB,
Mid- point of OA = E
Mid- point of OB = H
Therefore,
EH|| AB
Therefore,
HE = AB/2
Similarly,
HG = BC/2
FG = CD/2
EF = AD/2
EF + HG + FG + EF = (1/2)(AB + BC + CD + AD)
⇒ Perimeter of EFGH = (1/2) perimeter of ABCD
So,
The ratio of the perimeter of the quadrilateral EFGH to the perimeter of parallelogram ABCD = 1:2

Roots of equation ax² + bx + c = 0 are 
Difference between the roots = 
Given, roots differ by 7

Given,
A can do a piece of work in 18 days
∴ Part of work done by A in 1 day = 1/18
B can do a piece of work in 36days
∴ Part of work done by B in 1 day = 1/36
C can do a piece of work in 72 days
∴ Part of work done by C in 1 day = 1/72
Let,
Total number of days they work = x days
Number of days A work = 8 days
∵ B leaves7 days before the completion of the work
Number of days B work = (x – 7) days
And, Number of days C work= x days
∴ {(1/18) × 8 } + {(1/36) × (x – 7)} + {(1/72) × x} = 1
⇒ 4/9 + (x – 7)/36 + x/72 = 1
⇒ (32 + 2x -14 + x)/72 = 1
⇒ 18 + 3x = 72
⇒ 3x = 54
⇒ x = 18 days

Let speed of A be a kmph and speed of B be b kmph
Then we have
⇒ a + b = 60/6
⇒ a + b = 10
or 2a + 2b = 20------[1]
also,
⇒ 2/3a + 2b = 60/5 = 12--------------[2]
[1] – [2] leads to
⇒ (4/3)a = 8
a = 6 kmph

Given, container contains mixture of milk and water in the ratio of 4 : 7.
Let the volume of milk and water be 4a and 7a.
Now, 8 litres are removed and replaced by water.
In 1 litre of mixture has 4/11 litres of milk and 7/11 litres of water are present.
Thus in 8 litres 32/11 and 56/11 litres of milk and water are present.
∴ Volume of milk = 4a – 32/11
Volume of water = 7a – 56/11 + 8
New ratio = 4 : 9

⇒ a = 52/11
Now, initial amount of water = 7a = 364/11

We know that investment × time = profit
Assume that the total sum = P & total profit is M
∴ Sum invested by Ram = ⅓P & sum invested by Shyam = ⅔P

For Ram we can say,

For Shyam we can say,

[Here T = time for which investment of Shyam was in the business]

∴ Investment of Shyam was in the business for 3 months.

Formula for mean: 
Where, X = Mean, ∑X = sum of all items, N = number of items
Given, mean of 20 items is 55

⇒ two items are removed, so number of items = 18
⇒ removed items = 45 + 30 = 75
⇒ ∑X = 1100 – 75 = 1025

Mass of ore extracted from mine of lead ore = 8000 kg.
Percentage of metals in a mine of lead ore = 60%
Mass of metal = 60% of 8000
= (60/100) × 8000
= 4800kg.
Percentage of silver is 3/4 % of metals and the rest is lead.
Mass of silver = 3/4 % of 4800
= (3/400) × 4800
= 36kg.
Mass of lead = (4800 – 36)kg

= 4764 kg.

For CI:

Where,
A is the amount at the end of time t,
P is the principal,
t is time,
r is rate
A sum becomes Rs, 10000 after 3 years and Rs, 20000 after 6 years on compound interest.
For t = 3yrs

Diving eq 3 by eq 2 we get,

Given, ratio of monthly incomes of A and B is 7: 8.
Let the monthly incomes of A and B be 7a and 8a respectively, where a is any constant
Ratio of monthly expenditures is 3: 4
Let the monthly expenditures be 3b and 4b respectively, where b is any constant
Given, each of them saves Rs. 400
∴ 7a – 3b = 8a – 4b = 400
⇒ a = b
∴ 7a – 3a = 400
⇒ a = 100
Sum of their income = 7a + 8a = 15a
⇒ Sum of their income = Rs. 1500

The value of (1001)³ = 1001 × 1001 × 1001
= 1003003001
Or,
1001 = 1000 + 1
Cubing
(1001)³ = (1000 + 1)³
We know that, (a + b)³ = a³ + b³ + 3ab (a + b)
⇒ (1001)³ = 1000³ + 1 + 3000 × (1000 + 1)
⇒ (1001)³ = 1000000000 + 1 + 3003000
⇒ (1001)³ = 1003003001

3²ⁿ + 9n + 5 = 3 × 3^2n-1 + 9n + 3 + 2
= 3(3^2n-1 + 3n + 1) + 2
As, 3(3^2n-1 + 3n + 1) is completely divisible by 3.
Therefore,
Remainder is 2