Practice Test: Quantitative Aptitude - 8 - SSC CGL MCQ

From the diagram,
Height of the kite = AB = 80 m
Angle of inclination of the string with horizontal = ∠ ACB = 30°
Length of the string = AC
From ΔABC,
sin ∠ACB = Perpendicular/Hypotenuse
⇒ sin 30° = AB/AC
⇒ 1/2 = 80/AC
⇒ AC = 160 m

tan θ  + cot θ = 2
∵ cot θ = 1/tan θ

Using the identity: a² – 2ab + b² = (a – b)²

∴ cot θ  = 1/tan θ  = 1/1 = 1
Given expression is,
tanⁿ θ  + cotⁿ θ
= 1ⁿ + 1ⁿ
= 1+ 1 = 2

We have to find the value of expression xy + yz + zx + 2xyz
Multiplying and dividing by xyz we have

Substituting values of x, y and z

= 1

We are given the following equation (x² + x + 1) (x² – x + 1)
Simplifying it, we get
= x⁴ - x³ + x² + x³ - x² + x + x² – x + 1
= x⁴ + x² + 1
So, coefficient of x² is 1

Number of students with marks 0 – 10 (first interval) = 4
Number of students with marks 40 – 50 (last interval) = 5
Required ratio = 4 : 5

From the above graph, we can observe that
Number of students with marks 0 – 10 = 4
Number of students with marks 10 – 20 = 6
Number of students with marks 20 – 30 = 10
Number of students with marks 30 – 40 = 8
Number of students with marks 40 – 50 = 5
Hence, minimum numbers of students are in the interval 0 – 10 marks

From the above graph, we can observe that
Number of students with marks 0 – 10 = 4
Number of students with marks 10 – 20 = 6
Number of students with marks 20 – 30 = 10
Number of students with marks 30 – 40 = 8
Number of students with marks 40 – 50 = 5
Hence, maximum numbers of students are in the interval 20 – 30 marks

From the above graph, we can observe that
Number of students with marks 0 – 10 = 4
Number of students with marks 10 – 20 = 6
Number of students with marks 20 – 30 = 10
Number of students with marks 30 – 40 = 8
Number of students with marks 40 – 50 = 5
Total students = 4 + 6 + 10 + 8 + 5

= 33

Given,
Equations are
X – y = 0 ……(1)
x + y – 4 = 0
x + y = 4 ……(2)
By solving these two equations we will get the point of intersection
Now,
(1) + (2)
x – y + x + y = 0 + 4
⇒ 2x = 4
⇒ x = 2
Putting this value in (1)
2 – y = 0
⇒ y = 2
∴ Point of intersection is (2, 2)

Given, AB, BC and the median AE in ΔABC is similar to the sides PQ, QR and median PO in ΔPQR.

 AE and PO are medians, EB = EC and OQ = OR
∵ AE and PO are medians, EB = EC and OQ = OR
∴ Triangles ABE and PQO are similar by SSS as 
By property of similar triangles, ∠ABC = ∠PQR
∴ Triangles ABC and PQR are similar by SAS as, 

If AD is the median of the triangle ABC and G be the centroid then we know that
AG:GD = 2:1
Now adding the numerator in the denominator we get,
⇒ AG : (AG + GD) = 2: (2 +1)
⇒ AG : AD = 2:3 

Perimeter of a triangle = Sum of all sides
Given, length of the sides of a triangles be represented by x + 3, 2x – 3 and 3x – 5.
Perimeter = 31
⇒ x + 3 + 2x – 3 + 3x – 5 = 31
⇒ 6x = 36
⇒ x = 6
Sides are 9, 9 and 13
Thus, Longest side = 13

We know that area of arc of a circle = ½ × r² × θ
Where θ is the angle traced by the arc in radians.
Since the length of the rope is same for all three cows thus the radius of the arc traced by them is same i.e. 7m (given).
Also total angle traced by all three cows together is 180° as it is a triangular field.
Hence, total are grazed by 3 cows together

= 77 m²
Now, Area of triangular field by using heron’s formula
where s is the semi perimeter of the triangle
Given sides of triangle are 26m, 28m and 30m.
Hence, semi perimeter, s = (26 + 28 + 30)/2
= 42 m
∴ Area of triangular field

= 336 m²
Hence, the area of ungrazed field
= Area of triangular field – total area grazed by 3 cows
= 336 – 77
= 259 m²

Pythagoras theorem:-
Hypotenuse² = perpendicular² + base²
The figure represents the side view of a frustum.
Perpendicular = 24cm
Base = 20 – 13 = 7cm
Hypotenuse = Slant height.

Let the number of days in which Ajay finishes the work be ‘d’ days.
Raman takes twice time, thus number of days in which Raman finishes the work = 2d days.
In 1 day, Ajay finishes 1/d part of the work while Ajay finishes 1/2d part of the work
Working together, in one day, they finish part of work = 1/d + 1/2d = 3/2d
∴ Number of days in which they finish the work = 2d/3
Given, they take 12 days to finish the work.
∴ 2d/3 = 12
⇒ d = 18 days
∴ Number of days Raman will take = 2d = 36 days

Given,
Total distance = 288 km
Speed = 48 kms/hr
Time taken to travel 288 km = Distance/Speed= 288/48 = 6 hour
Now we need to find out the number of stops in the 288 km travel. Given that train stops after every 18 km.
∴ Number of stops = 288/18 = 16
It means train stops 15 times before 288 km and 1 time just after 288 km.
∴ We need to take only 15 stops into consideration for the 288 km travel.
∴ Total stopping time in the 288 km travel = 15 × 18 = 270 minutes
∴ Total time needed to reach the destination
= 6 hours + 270 minutes
= 6 hours + 4 hours + 30 minutes
= 10 hours 30 minutes

We have Mayur covering a distance of 15 km in 1 hour.
Hence time taken to complete 1 km =
As after every 1 km, he Mayur takes a rest for 6 minutes, effective time to finish 1 km = 4 + 6 = 10 min
To cover first 10 km, he needs = 10 × 10 = 100 mins
For the last 1 km, he need = 4 min (he will not take time to rest as he has reached the finish point)
Total time needed = 100 + 4 = 104 min
Or 104/60 hours
= 1 hour and 44 minutes

A fruit seller purchased 50 kg of mangoes at the rate of Rs. 25 per kilo.
Total cost price of the 50 kg of mangoes = 50 × 25 = 1250
If he incurred 20% profit overall then the selling price of 50 Kg of Mangoes 
= 1500
He sold some of the mangoes at the rate of Rs. 40 per kilo and the rest at the rate of Rs. 15 per kilo.
Suppose he sold X kg of mangoes at the rate of Rs 40 and (50-X) Kg of sold at the rate of Rs15.
We can say that,
⇒ 40 × X + 15 × (50 - X) = 1500
⇒ 40X + 750 – 15X = 1500
⇒ 25X = 750
⇒ X = 30
So he sold 30 kg of mangoes at the rate of Rs 40 and (50-30) =20 Kg of sold at the rate of Rs 15.
Hence the answer is 20 kg

Formula: Let the successive increase in percentages be a% and b%.
In that case, the total increase will be (a + b + ab/100) %
A shopkeeper gives 12% additional discount on the discounted price after giving an initial discount of 20 % on the labeled price of a radio.
So here a= - 12 and b = - 20
The total successive discount = a + b + ab/100
= (-12) + (-20) + (-12) × (-20)/100
= -12 – 20 + 2.4
= - 29.6
Hence if the Label price is 100 then after giving 29.6% discount the final sale price is (100 -29.6) = 70.4
So if the final sale price is Rs 70.4 then the label price is Rs 100
If the final sale price is Rs 1 then the label price is Rs 100/70.4
If the final sale price is Rs 3520 then the label price is Rs 
Hence the answer is Rs 5000.

We have, Average age = Sum of all ages/No. of persons
∴ 15.6 = Sum of 20 boys ages/ 20
⇒ Sum of ages of 20 boys = 20 × 15.6 = 312
After 5 new boys come, total boys = 20 + 5 = 25
Sum of ages of 5 new boys = 15.4 × 5 = 77
Sum of 25 boys ages = 312 + 77 = 389
Overall average = 389/25 = 15.56 yrs.

Given, price of a commodity rises from Rs. 6 per kg to Rs. 7.50 per kg.
Also given, the expenditure remains the same.
Thus, initial consumption × 6 = New consumption × 7.5
⇒ New consumption = 0.8 × initial consumption
Decrease in consumption = Initial consumption – Final consumption
⇒ Decrease in consumption = Initial consumption - 0.8 × initial consumption
⇒ Decrease in consumption = 0.2 × initial consumption

% decrease in consumption   
⇒ % decrease in consumption 

Let the No. of yrs for which the loan has been taken = x yrs
According to question Rate of Interest Applicable = x% pa
We know that, formula for simple interest: 

Where,
S.I. = Simple Interest = 540
P= principal = 1500
T = Time = x
R = Rate of Interest = x

⇒ 540 = 15x²
⇒ x² = 36
Taking Square root on both sides
⇒ x = 6
⇒ Percent Per Annum (PCPA) = 6%

Let X kg of nickel be added to the alloy.
Given, Initial ratio is 5:3:2.
Let the initial weight of copper, zinc and nickel in 100 kg of alloy be 5Y, 3Y and 2Y respectively.
Also, initial weight of alloy = 100 kg
∴ 5Y + 3Y + 2Y = 100 kg
⇒ Y = 10
Since X kg of nickel is added to the alloy, thus
⇒ New weight of nickel = X + 20 kg
Given in question, new ratio after this addition is 5:3:3. On comparing it with the old ratio 5:3:2 we find that the weight of nickel has increased by 10 kg i.e
New weight of nickel = 20 + 10
= 30 kg
∴ X + 20 = 30
⇒ X = 10 kg
Hence, 10 kg of nickel must be added to the alloy to get new ratio of 5:3:3.

Area of a square = side²
Area of a circle = πr²
The radius of each circle in the figure (r) = 4 cm.
We have constructed a square by joining the centers of 4 circles.
∴ Each side of the square = (4+4) cm. = 8 cm.
Now, if we consider any one of the circles it is clearly visible that the square includes ¼ th of each circle.
Now total area of all circles included inside the square
= 4 × [¼ × area of one circle]
= area of one circle
= 16 × π cm² [∵ area of circle = π × r²]
The area of the square = 8² = 64 cm²
From the figure, the area of the region bounded by the four circles = the area of the square - total area of all circles included inside the triangle
⇒ The area of the region bounded by the four circles:
= [64 - 16 × π] cm.²
= 16 × [4 – π] cm.²

Area of a square = side × side
Area of a circle = πr²,
Area of quadrant of a circle = πr²/4
Where r is radius.
Given, from each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut.
Area of the Whole square = Side² = (4 cm)² = 16 cm²
Area of circle and quadrants (radius = 1 cm) = π r² + [4 × (π r²/4)]
= π (1 cm)² + [π (1 cm)²]
= 2π
Shaded Area = [16 – 2π] cm²