Practice Test: Quantitative Aptitude - 3 - SSC CGL MCQ

Before comparing 1^st we will try to bring the given numbers to their simplest form.
(0.5)³ = 0.125
√0.49 = √ (0.7)² = 0.7

0.22
Now, comparing the 4 numbers we can easily say, 1^stone is the smallest.
∴ The smallest number is (0.5)³ = 0.125

Before we solve this completely, we simplify the LHS using BODMAS rule, and a few rules from indices.
First, we express every quantity in terms of powers of 5.
Hence,
5√5 = 5^3/2
Hence, we have
5^3/2 × 5⁵ ÷ 5^-3/2 = 5^a+2
Now here, we use BODMAS rule. It is described as follows:
B – Brackets
O – Of
D – Division
M – Multiplication
A – Addition
S – Subtraction
The above is the order in which the operations are to be solved. We observe that there are no brackets or “Of” in the given expression. Hence, we go ahead with division.
5⁵ ÷ 5^-3/2 = 5^5+3/2 = 5^13/2
Now, we have multiplication
⇒ 5^3/2 × 5^13/2 = 5^16/2 = 5⁸
Equating the above with RHS,
5⁸ = 5^a+2
⇒ 8 = a + 2
⇒ a = 6 

Let the initial number of students be ‘n’.
Given, number of students in a class is increased by 20% and the number now becomes 66
n + 20% of n = 66
⇒ 1.2n = 66
⇒ n = 55

Suppose the weight of the jewel is W and its price is P then according to the question
P ∝ W²,
⇒ P = kW²    where K is the constant.
The ratio between the weights of A and B is 3 : 5. So let’s assume the weight of A is 3x and weight of B is 5x.
So, Price of A = k(3x)² = 9kx²
And price of B = k(5x)² = 25kx²
Now, weight of C = weight of A + weight of B = 3x + 5x = 8x
So, price of C = k(8x)² = 64kx²
Combined price of A and B = 9kx² + 25kx² = 34kx²
So the required ratio = 64kx² : 34kx² = 32 : 17.

Quantity of spirit in 4 litres solution with 40% concentration = (40/100) × 4 = 1.6 litres
Quantity of spirit in 6 litres solution with 50% concentration = (50/100) × 6 = 3.0 litres
Thus, amount of spirit in resulting mixture = 1.6 + 3.0 = 4.6 litres
Total amount of resulting mixture = 4 + 6 = 10 litres
∴ Strength of spirit in resulting mixture = (4.6/10) × 100 = 46%

Solution: Let the larger number be x and the smaller number be y.
It is given that LCM is thrice of the larger number,
Therefore, LCM = 3x      ----(i)
It is also given that difference between the smaller number and the HCF is 14,
Therefore, y – HCF = 14
HCM = y – 14      ----(ii)
We know that product of two numbers = LCM × HCF
Therefore, x × y = 3x × (y – 14)      (from (i) and (ii))
y = 3(y – 14)
y = 3y – 42
2y = 42
y = 21
Hence, the smaller number is 21.

Let the ages of A and B be 11x and 13x respectively. Then,
According to the question, after 7 years the ratio of their ages will be 20 : 23.
⇒ (11x + 7)/(13x + 7) = 20/23
⇒ 260x + 140 = 253x + 161
⇒ 260x – 253x = 161 – 140
⇒ 7x = 21
⇒ x = 3
∴ The difference between their ages = 13x – 11x = 2x = 2 × 3 = 6 years
Hence, the required difference between the ages of A and B is 6 years.

On Squaring we get,

On Cubing we get,
x⁶ = 27 × 9x
⇒ x⁵ = 243
⇒ x = 3

Let the CP of 1 m cloth = Rs x
Loss% = 10%
SP of 1 m cloth = Rs 9
Therefore,
Loss = x – 9
10 = ((x – 9)/x × 100%
x – 9 = x/10
x = 10
now CP of 1 m cloth = Rs 10
let the selling price be Rs y
profit% = 15% = ((y – 10)/10 × 100%)
3 = 2(y – 10)
2y – 20 = 3
y = 23/2
y = 11.5
SP of 1 m cloth = Rs. 11.50

Given, 3,000 is divided among A, B and C, so that A receives 1/3 as much as B and C together receive and B receives 2/3 as much as A and C together receive

⇒ 3A = B + C……….(1)

⇒ 3B = 2A + 2C…………(2)
⇒ Putting the value of C from (1) into (2), we get,
⇒ 3B = 2 A + 2 (3A – B)
⇒ 8A = 5B …………….(3)
⇒ Similarly, we can solve and will get relation between B and C.
⇒ 7B = 8C …………….. (4)
⇒ Multiplying (3) with 7 and (4) with 5, we get,
⇒ 56A = 35B = 40C


⇒ A : B : C = 5 : 8: 7

We know that,
Where, A = Final amount, P = Principal amount, R = % rate of interest, T = time period in years
Here, P = Rs. 1600, T = 2 years, A = 1730.56

Thus, rate of R% per annum is 4%

Speed = distance/time
Let the distance be x km. Then,
(Time taken to walk x km) + (Time taken to ride x km) = 37 min.
⇒ (Time taken to walk 2x km) + (Time taken to ride 2x km) = 74 min.
Given, time taken to walk 2x km = 55 min.
Time taken to ride 2x km = (74 - 55) min = 19 min

Let the distance travelled by first train till the point they meet be X km. Since the second train travelled 125 Km more than the first one, thus
Distance travelled by second train when they meet = X + 125
We know that, Time = Distance/Speed
We know that both trains travelled for the same time till they meet. Hence,

⇒ 100X = 75X + 9375
⇒ X = 375 Km
∴ X + 125 = 500 Km
Hence the distance between the points P and Q = Distance travelled by train A & B
= 375 + 500
= 875 Km

Let the total work be W units.
Amount of work done by Aman in one day = Am units
Amount of work done by Akbar in one day = Ak units
W = (Am + Ak) 18
Now this piece of work has been started, but both worked for only 4 days, after which Aman took over,
⇒ W = (Am + Ak) 4 + (Am) 42
⇒ (Am + Ak) 18 = (Am + Ak) 4 + (Am)42
⇒ (Am + Ak) 14 = (Am) 42
⇒ Am + Ak = 3 Am
⇒ Ak = 2 Am
Substituting the value of Ak in the very first equation, we have
W = (Am + 2 Am)18
⇒ W = 54 Am
Hence, Aman alone would have taken 54 days to complete the work.

Given data-
Sides of a right-angled triangle forming right angle are in the ratio 5 : 12
Let sides of triangle forming right angle be 5x & 12x respectively, where x is constant of proportionality.
Area of the triangle = ½ × base × height = ½ × 5x × 12x = 30x²
30x² = 270
⇒ x² = 9
⇒ x = 3
∴ sides of triangle forming right angle are-
5x = 5 × 3 = 15 cm
& 12x = 12 × 3 = 36 cm
Using Pythagoras theorem
hypotenuse² = 15² + 36²
⇒ hypotenuse² = 225 + 1296
⇒ hypotenuse² = 1521
∴ Hypotenuse = √1521 = 39 cm

Volume of a cylinder = πr² × h
Hence the ratio of their volumes
= (πr₁² × h₁)/(πr₂² × h₂)
= (9² × 3)/(4² × 7)
= 243/112

Perpendiculars drawn from the centre on the chord bisects them.
Given, AB and CD are two parallel chords on the opposite sides of the centre of the circle.
AB = 18 cm, CD = 24 cm and the radius of the circle is 15 cm
From the figure,
OM and OL are perpendiculars, OC and OA are radius
∴ CM = CD/2 = 12 cm and AL = AB/2 = 9 cm
By Pythagoras theorem:
OC² = OM² + MC²
⇒ OM² = 15² – 12² = 81
⇒ OM = 9 cm
Similarly,
OA² = OL² + AL²
⇒ OL² = 15² – 9² = 144
⇒ OL = 12 cm
Distance between the two chords = OL + OM
⇒ Distance between the two chord = 12 + 9 cm = 21 cm

We know that,
Property- If MNO is a secant to a circle intersecting the circle at M and N and OT is a tangent segment, then
OM × ON = OT²
From the diagram,
PAB is a secant to a circle with center O₁ intersecting the circle at A and B and PT is a tangent segment,
∴ PA × PB = PT²  ……. (1)
Again, PAB is a secant to a circle with center O₂ intersecting the circle at A and B and PQ is a tangent segment,
∴ PA × PB = PQ²  ……. (2)
From (1) and (2)
PT² = PQ²
⇒ PT = PQ

As we know, in an equilateral triangle all sides are equal.
∴ XY = YZ = ZX
Let XY = 2a units
∴ YZ = ZX = 2a units
And XS is perpendicular bisector of side YZ
∴ YS = SZ = a units
Thus, XY: YS = 2: 1

As stated in above figure ABCD is a cyclic trapezium with Line AD and Line BC parallel to each other.
As per the properties of cyclic quadrilateral,
∠ABC + ∠ADC = 180°
72°+ ∠ADC = 180°
∠ADC = 108°
Also, Properties of parallel lines
∠ADC + ∠BCD = 180°
∠BCD = 180° - ∠ADC = 180° - 108° = 72°
∴ ∠ABC = ∠BCD
⇒ ∠BCD= 72°

∴ sinθ = kx and cosθ = ky
sin²θ + cos²θ = 1
⇒ k²x² + k²y² = 1

sinθ + cos θ = kx + ky

Consider AB as height of building and CD be as height of tower
AB=?
From trigonometric formulas,
Tan θ = perpendicular/base
∴ in ΔBCD
tan 60° = CD/BD
⇒ √3 = CD/BD
⇒ 125/√ 3 = BD
Also, in ΔABD
tan 30° = AB/BD
⇒ 125/√ (3 × 3) = AB
⇒ AB = 125/3 = 41.66m

2cos²20° + 3cos²70° + sin²70°
= 2cos²20° + 2cos²70° + cos²70° + sin²70°
= 2cos²20° + 2cos² (90 – 20)° + 1
= 2cos²20° + 2sin²20° + 1
= 2 + 1 = 3

Let's solve this step by step:

1. Five years ago:

   • The average age of the family of 3 members was 19 years.
   • Therefore, the sum of their ages 5 years ago was 19×3=57.

2. Today:

   • Each of the original three members has aged by 5 years. So, the sum of their ages today is 57+3×5=57+15=72 years.

3. Including the baby:

   • The average age of the family is still 19 years, and now there are 4 members.
   • So, the total sum of the ages of the 4 members today is 19×4=76 years.

4. Age of the baby today:

   • The sum of the ages of the original three members today is 72 years.
   • The total sum of the ages of the four members is 76 years.
   • Therefore, the age of the baby today is 76−72=4 years.

5. Age of the baby after 8 years:

   • The baby will be 4+8=12 years old.

The correct answer is (b) 12 years.

Using Sum of Arithmetic Progression
Total odd numbers between 11 and 61, n = 24
Sum = n/2[a + l] = 24/2[13 + 59] (a = first term, l = last term)
= 24/2 × 72
= 864