Practice Test: Quantitative Aptitude - 10 - SSC CGL MCQ

A cistern 5 m long and 8 m wide, contains water up to depth of 2 m 45 cm.
Here, cistern is in cuboid form
Length of cistern = 5 m
Breadth of cistern = 8 m
Height of cistern upto which its wet = 2 m 45 cm = 2.45 m
Total area of wet surface = {2(length + breadth) × height} + (length × breadth)
= {2(5 + 8) × 2.45} + (5 × 8)
= 63.7 + 40
=103.7 m²

Appling Pythagoras theorem on triangle ABC
AC² = AB² + BC²
⇒ y² = (2x)² + BC²
⇒ BC² = y² – 4x²
Now applying Pythagoras theorem on triangle DBC
DC² = DB² + BC²
⇒ z² = x² + BC²
⇒ BC² = z² – x²
Now, on comparing the value of BC² we get
y² – 4x² = z² – x²
∴ z² = y² – 3x²

∴ z = √(y² – 3x²)
Now in Triangle DBC,
sin θ = DB/DC
∴ sin θ = x/z
∴ sin θ = x/√(y² – 3x²)

Area of a circle = π(radius)²
A small circle of radius 3 cm is drawn within a larger circle of radius 5 cm.
Area of large circle = π(5)²
= (22/7) × (5)²
= (22 × 5 ×5)/7
= 550/7
Area of small circle = π (3)²
= (22 × 3 × 3)/7
= 198/7
Area of annular zone 
= 352/7
The ratio of the area of the annular zone to the area of the larger circle = area of annular zone: area of large circle 
= 16: 25

Given expression is,

Rationalising, we get

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

= 4 × √2 × √3
= 4√6

The given expression is:


[Using the formula (a + b)³ = a³ + b³ + 3ab(a + b)]

Trick:
(a + 2) + 1/(a + 2) = 2
If x + 1/x = k then the value of x³ + 1/x³ will always be equal to (k³ – 3k)
Here k = 2

∴ Answer is 2

Given that during the year 1997 total perfumed production was 5000 units.
From the pie chart we can see, the percentage production of jasmine perfume in the year 1997 = (80 – 50) = 30% of total production.
∴ The production of jasmine perfume in the year 1997
= 30% of 5000
= 1500
∴ The production of jasmine perfume in the year 1997 = 1500

From the pie chart we can see, the percentage production of rose perfume in the year 1995 = (45 – 0) = 45% of total production.
Similarly, from the pie chart we can see, the percentage production of rose perfume in the year 1997 = (30 – 0) = 30% of total production.
The ratio of percentage production of rose perfume during 1995 to that during the year 1997 = 45: 30 = 3 : 2

The percentage of production of sandal perfume during the year 1995 = (65 – 45) = 20%
The percentage of production of sandal perfume during the year 1997 = (50 – 30) = 20%
∴The percentage of production of sandal perfume during the year 1995 over that during 1997 = 0%

From the diagram,
AB = height of the cliff = 120 m
Angle of depression = ∠ CAE = 60°
BC = Distance of the ship to the bottom of the cliff
∴ ∠ CAB = 90° - 60° = 30°
From ΔABC
tan ∠CAB = perpendicular/Base
⇒ tan 30° = BC/AB
⇒ 1/√3 = BC/120
⇒ BC = 120/√3 m
⇒ BC = 120√3/3 m = 40√3 m

Given that,
x = m sec α cos β
Squaring both side we can write,
x² = m² sec²α × cos² β

Similarly we can write,

Now we have to calculate the value of 
Putting the values:

Sinθ = cos(90° - θ) and sin²θ + cos²θ = 1 and Sec²θ = 1 + tan²θ
Sin²32° + sin²58° + cot²58°
= sin²32° + cos²32° + cot²58°
= 1 + cot²58°
= cosec²58°

= cot²58°/cos²58°
= cot²58°/sin²32°
= cosec²32°cot²58°

First circle subtend angle = 60°
2nd circle subtend angle = 75°
For the first circle:
⇒ angle1 = 60 = 60 × π/180 = π/3 radians
For the second circle:
⇒ angle2 = 75 = 75 × π/180 = 5 π/12 radians
We have the formula 

radius1 : radius2 = 5 : 4

We know that the sum of angles of a triangle is 180°.
∴ ∠A + ∠B + ∠C = 180°
Given: ∠A + ∠B = 65°
Replacing this value we get,
⇒ 65 + ∠C = 180
⇒∠C = 115°
Also, ∠B + ∠C = 140° (given)
⇒∠B + 115° = 140°
∴ ∠B = 25°

Property of circumcentre
In any ΔXYZ, if O be circumcentre(where perpendicular bisectors of all three sides meet) of ΔXYZ.
Then, ∠XOZ = 2∠Y
∠ZOY = 2∠X
∠YOX = 2∠Z

Now, in ΔABC
S is the circumcentre of ΔABC and ∠A = 50°
∴ ∠BSC = 2∠A
= 2 × 50°
= 100°
Now in ΔBSC,
∠BCS = ∠CSB (as, BS and CS are radius of the circumcentre and angles opposite to the equal sides are equal)
∵ Sum of all angles of a triangle = 180°
∴ ∠BSC + ∠BCS + ∠CSB = 180°
⇒ 100° + 2∠BCS = 180°
⇒ 2∠BCS = 80°
⇒∠BCS = 40°

Given, A can fill it in 20 minutes and B can empty it in 30 minutes
Pipe A's work for one minute=1/20
Pipe B's work for one minute=1/30
∴ Total work done in two minute 
In every 2 minute, 1/60^th work is done, which means upto 114^th minute, amount of work done will be = 57 × 1/60 = 57/60
Now the remaining work i.e. 3/60 (=1/20^th) will be done by A in next 1 minute.
∴ Total time required = 114 + 1 = 115 minutes

Let, 
Total distance covered = 2x km
We know that,
Time required = Distance covered/Speed
Given,
Half of the distance i.e. x km is covered at speed 16 km/hr
∴ Time required = x/16 hours
Next half of the distance i.e. x km is covered at speed 24 km/hr
∴ Time required = x/24 hours
Given, 
Total time required = 5 hours 
∴ x/16 + x/24 = 5
⇒ (3x + 2x)/48 = 5
⇒ 5x = 48 × 5
⇒ x = 48 km
∴ Total distance covered = 2 × 48 km = 96 km

∵ Price of 12 eggs = Rs. 3.75
∵ Price of 1 egg = 3.75/12 = Rs.0.3125
∵ Price of 1600 eggs = Rs. (0.3125 × 1600) = Rs.500
∵ cost price, CP of eggs = Rs. 500
Selling price of 2 eggs = Rs. 1
∵ Selling price of 900 eggs = 
Selling price of 5 eggs = Rs. 2
∴ Selling price of remaining (1600 - 900 = 700) eggs  
∴ Total selling price, SP of eggs = Rs. 450 + 280 = Rs.730.
∵ SP > CP
∴ shopkeeper will have gain 
Gain = SP – CP = Rs. 730 - 500 = Rs. 230.

Hence, Ram will have a gain percent of 46% by selling eggs.

Given,
Ratio of milk and water is 5: 1
Amount of mixture = 24 litres
∴ Amount of milk in the mixture 
∴ Amount of water in the mixture 
If 4 litres of water is added
Amount of water in the new mixture = 4 + 4 = 8 litres
Total amount of mixture = 24 + 4 = 28 litres
And, Amount of milk in the new mixture = 20 litres
∴ Required percent
= (amount of milk/total amount in mixture) × 100%
= (20/28) × 100
= 500/7
= 

Formula:-
Sum of observations = average × number of observations
The average age of 11 players of a cricket team is increased by 2 months when two of them aged 18 years and 20 years are replaced by two new players
∴ Total age of players who are replaced= (18 +20) months
= 38 years
Age of 11 players of a cricket team is increased by (2 × 11= 22) months
Total Age of new players = 38years + 22 months
Average age of new players = (total age /no of players)
= (38years + 22 months)/2
= 19years 11 months

Let, x be the value of the machine 3 years ago
Since, the value of machine depreciates at the rate 10% every year
∴ its value after 1 year 
Its value after 2 years  
Its value after 3 years 
According to the question,
0.729x = 729
x = 729/0.729 = 1000
Alternate:-
As we know that, if the value of machine depreciates by R% every year, then the value of machine after ‘t’ years will be:-

Here, P’ = future value of machine
P = current value.
Here it is given that, current value is Rs. 729 and we will have to find out the value 3 years ago.
So, here P’ = 729 and P = ?, R = 10%

⇒ P = 1000

We know that, formula for simple interest:

Where,
S.I. = Simple Interest
P = principal
R = Rate of interest
T = Time

Let R = T = k
Hence

1600 = 25 × k × k
64 = k²
k = +8 or k = -8
Since time period cannot be negative, hence, we have
k or rate of interest = 8 %

Given, the cost of a piece of diamond varies with the square of its weight.
⇒ Price = k × weight²   (k is any constant)
Let the weight of a piece of diamond be ‘6x’.
Then, original price = k (6x)² = 36kx²
⇒ 36kx² = 5184      ……… eq(1)
The piece is cut into 3 pieces of weights ratio 1 : 2 : 3.
Now, the new price = k (x² + (2x)² + (3x)²) = 14kx²
From eq(1)
New price = (14 × 5184)/36 = 2016.
∴ The loss = 5184 – 2016 = Rs. 3168.

Follow BODMAS rule to solve this question, as per the order is given below,
Step - 1 - Parts of an equation enclosed in 'Brackets' must be solved first,
Step - 2 - Any mathematical 'Of' or 'Exponent' must be solved next,
Step - 3 - Next, the parts of the equation that contains 'Division' and 'Multiplication' are calculated,
Step - 4 - Last but not least, the parts of the equation that contains 'Addition' and 'Subtraction' should be calculated.
We know that,
(a² – b²) = (a + b)(a – b)

Let x = 4.327327327…. ………(1)
Multiply both sides by 1000
⇒ 1000x = 4327.327327… …….(2)
Subtract eq. (1) from eq. (2)
⇒ 999x = 4323
⇒ x = 4323/999

⇒ 4⁶¹ + 4⁶² + 4⁶³ + 4⁶⁴=4⁶¹ (1 + 4 + 4² + 4³)
⇒ 4⁶¹ + 4⁶² + 4⁶³ + 4⁶⁴=4⁶¹ × 85
⇒ 4⁶¹ + 4⁶² + 4⁶³ + 4⁶⁴=4⁶¹ × 5 × 17
So, the expression is divisible by 17