Practice Test: Quantitative Aptitude - 1 - SSC CGL MCQ

From the given data,
A’s one day’s work = 1/8
B’s one day’s work = 1/12
C’s one day’s work = ¼ - (1/8 + 1/12) = 1/24
∴ A : B : C = ratio of their 1 day’s work = 1/8 : 1/12 : 1/24 = 3 : 2 : 1
∴ ratio of share of B and C = 2 : 1

From the given data
⇒ volume of cylindrical pillar/curved surface area = πr²h/2πrh = 1104/368
⇒ r = (1104 × 2/368) = 6 m
Given that curved surface area = 2πrh = 368
⇒ h = 368/(2π × 6) = 9.76 m
∴ height of the cylinder = 9.76 m

As per the given data,
Cost price of 23 kg of rice = 24 × 23 = Rs.552
Cost price of 27 kg of rice = 27 × 37 = Rs.999
∴ cost price of 50 kg of rice = Rs.552 + Rs.999 = Rs.1551
Also given that he sells at Rs.35 per kg
⇒ selling price of 50 kg of rice = 50 × 35 = 1750
∴ profit percent = (sp - cp/cp × 100)
= (1750 – 1551/1551 × 100)
= 12.83 %

From the given data
Let x be the total production
Also from the given data 30% is lost in grinding wheat, a country can export 60 lakh tons of wheat.On the other hand if 20% is lost in grinding,it can export 80 lakh tons of wheat
⇒ 80x/100 – 70x/100 = 80 lakh tons – 60 lakh tons
⇒ 10x/100 = 20 lakh tons
⇒ x = 200 lakhs tons
∴ total production of wheat in the country is 200 lakhs of tons

As per the given data
Length of the bridge = 250 m
Length of the train = 350 m
Length of the train + length of the bridge = 350 + 250 = 600m
We know that speed = (length of the train + length of bridge)/time
⇒ speed = (600)/30 = 20 m/s = 20 × 18/5 = 72 km/hr
∴ speed of the train = 72km/hr

From the given data

⇒ 8m = m² - 2m + 24
⇒ m² - 10m + 24 = 0
⇒ m² - 6m - 4m + 24 = 0
⇒ m(m - 6) - 4(m - 6) = 0
⇒ (m - 4)(m - 6) = 0
⇒ m = 4 and 6
∴ least possible value of m is 4

As per the given data
⇒ α + β = - b/a = - ( - 3)/2 = 3/2
⇒ αβ = c/a = 2/2 = 1
For the new equation roots are α² and β²
∴ sum of the roots α² + β² = (α + β)² - 2αβ = (3/2)² - 2(1) = 9/4 – 2 = ¼
Product of the roots = α²β² = (αβ)² = 1² = 1
⇒ Required equation = x² – (sum of the roots)x + product of the roots = 0
⇒ x² – (1/4)x + 1 = 0
⇒ 4x² - x + 4 = 0

It is possible for only an equilateral triangle as all the corresponding sides and all the corresponding angles are equal

As per the given data
⇒ ∠B = 70° and ∠C = 30°
∴ ∠OBC = 70/2 = 35° and ∠BCO = 30/2 = 15°
We know that sum of the angles in a triangle = 180°
⇒ ∠BOC + ∠OBC + ∠BCO = 180°
⇒ 35 + 15 + ∠BOC = 180°
⇒ ∠BOC = 180 – 50 = 130°
∴ the value of ∠BOC = 130° 

From the given data
Let the ratio of parallel sides be 5x : 9x
Also given that perpendicular distance between them h = 21 cm
we know that area of trapezium = ½ (a + b)h
⇒ 441 = ½ (5x + 9x) 21
⇒ 21 = ½ (14x)
⇒ 42 = 14x
⇒ x = 3
Substitute the value of x in 5x and 9x
⇒ 5x = 5(3) = 15 cm
⇒ 9x = 9(3) = 27 cm
∴ longer side of the parallel side = 27 cm

As per the given data,
Given that cos A = 3/5
 4/5 and tan A = SinA/CosA = 4/3
Also given that sin B = 5/13
 12/13 and tan B = SinB/CosB = 5/12
We know that

From the given data
Let the number be x
We know that dividend = divisor × quotient + remainder
⇒ x = 13p + 11…eq 1
Also given that when the same number is divided by 17, the remainder is 9
⇒ x = 17q + 9...eq 2
From eq 1 and eq 2, we get
⇒ 13p + 11 = 17q + 9
⇒ 17q – 13p = 2
⇒ q = ( 2 + 13p)/17
The least value of p for which q = (2 + 13p)/17 is a whole number is p = 26
Substitute p = 26 in x = 13p + 11
⇒ x = 13× 26 + 11
⇒ x = 349

As per the given data
Let the cost price be x and selling price be y
Given that shopkeeper sells an article at 20% loss
⇒ x – 20x/100 = y
⇒ 80x/100 = y
⇒ x = 5y/4
⇒ 4x - 5y = 0.…eq 1
Also given that if he sells it for 100 more, he would earn a profit of 10%
⇒ x + 10/100 x = y + 100
⇒ 11x = 10y + 1000
⇒ 11x - 10y = 1000.…eq 2
Multiply eq 1 with 2 , we get
⇒ 8x – 10 y = 0 …eq 3
⇒ subtract eq 3 from eq 2 , we get
⇒ 3x = 1000
⇒ x = 333.33
Therefore cost price of the article = Rs.333.33

As per the given data,
Average of 100 numbers = 50
⇒ (x₁ + x₂ + ….. + x₁₀₀)/100 = 50
⇒ x₁ + x₂ + ….. + x₁₀₀ = 5000
⇒ x₁ + x₂ + … + x₉₈ = 5000 - 60 - 80
⇒ x₁ + x₂ + … + x₉₈ = 4860
∴ sum of the 98 numbers = 4860
⇒ average of 98 numbers = 4860/98 = 49.6

As per the given data
Let their simple interest be I₁ and I₂
Let their rate of interest be R₁ and R₂
⇒ I₁ – I₂ = (PTR₁/100 – PTR₂/100)
⇒ I₁ – I₂ = (R₁ - R₂) (PT/100)
⇒ 65 = (R₁ - R₂) (13000 × 2/100)
⇒ R₁ - R₂ = ¼ = 0.25

As per the given data,
⇒ a – b = 3
Squaring on both sides, we get
⇒ a² + b² - 2ab = 9
⇒ 29 – 2ab = 9
⇒ 29 – 9 = 2ab
⇒ 20 = 2ab
⇒ ab = 10
We know that (a + b)² = a² + b² + 2ab
⇒ (a + b)² = 29 + 2(10) = 49
⇒ a + b = √49 = ±7

As per the given data
Let AD be the length of the median subtending from vertex A and to the side BC
⇒ BD = DC = BC/2 = 10/2 = 5 cm
By Apollonius theorem, we get
⇒ AB² + AC² = 2 (AD² + BD²)
⇒ 12² + 6² = 2 (AD² + 5²)
⇒ 90 = AD² + 25
⇒ 65 = AD²
⇒ AD = 8.06 cm
∴ length of the medium = AD = 8.06 ≈ 8 cm

As per the given data
Converting the degree into radians
⇒ 30° = π/6 rad and 60° = π/3 rad
Let r₁ and r₂ be the radii of the two circles and l be the length of the each arc
We know that l = r θ
⇒ r₁θ₁ = r₂θ₂
⇒ r₁/r₂ = θ₂/θ₁
⇒ r₁/r₂ = (π/3)/(π/6)
⇒ r₁/r₂ = 6/3 = 2/1
∴ the required ratio = 2 : 1

From the given data
Let the each side of a square be a, then its area will be a²
∴ area of circle will also be a²
Also given that the perimeter of square and equilateral triangle is same,
⇒ 4a = 3s
the each side of equilateral triangle is 4a/3
∴ area of equilateral triangle = √3/4 × (4a/3)² = 4√3 a²/9
∴ ratio between the area of circle and the area of equilateral triangle = a²/(4√3 a² × 9) = 9 : 4√3

As per the given data,
Let AB and CD be the two buildings, and AC be the river with a length of x meters

Let us consider AB be the building of height = 36 m and CD be the height of other building of h meters
⇒ ∠ACB = 45° and ∠EDB = 30°
From ΔABC, we get
⇒ AB/AC = tan 45°
⇒ 1 = AB/AC
⇒ AB = AC = 36 m
∴ DE = AC = 36 m
From ΔEDB, we get
⇒ tan 30° = BE/DE
⇒ 1/√3 = BE/DE
⇒ BE = DE/√3 = 36/√3 = 20.7 m
We know that CD = AE = AB – BE = (36 - 20.7) m = 15.3 m
∴ height of the other building = 15.3 m

As per the given data
Percentage of expenditure that is spend on tennis = 10%
Amount of expenditure that is spend on tennis = 10/100 × 5,00,000 = Rs.50,000

From the given data
Percentage of expenditure that is spend on cricket = 18%
Amount of expenditure that is spend on cricket = 18/100 × 5,00,000 = Rs. 90,000
Percentage of expenditure that is spend on hockey = 13%
Amount of expenditure that is spend on hockey = 13/100 × 5,00,000 = Rs. 65,000
∴ ratio of expenditure that is spend on cricket and hockey = 90,000 : 65,000 = 18 : 13

From the given data
Percentage of expenditure that is spend on basketball = 34%
Amount of expenditure that is spend on basketball = 34/100 × 5,00,000 = Rs. 1,70,000
Percentage of expenditure that is spend on football = 25%
Amount of expenditure that is spend on football = 25/100 × 5,00,000 = Rs. 1,25,000
Difference of expenditure that is spend on basketball and football = 170000 - 125000 = Rs. 45,000
∴ Amount of Rs. 45,000 more is spend on basketball than football

From the given data
Total expenditure spend on all the sports = Rs. 5,00,000
Percentage of expenditure that is spend on football = 25%
Amount of expenditure that is spend on football = 25/100 × 5,00,000 = Rs. 1,25,000
∴ ratio of expenditure that is spend on football to the total expenditure = 1,25,000 : 5,00,000 = 1 : 4