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SSC CGL (Tier II) Practice Test - 1 - SSC CGL MCQ


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150 Questions MCQ Test SSC CGL Tier II Mock Test Series 2025 - SSC CGL (Tier II) Practice Test - 1

SSC CGL (Tier II) Practice Test - 1 for SSC CGL 2024 is part of SSC CGL Tier II Mock Test Series 2025 preparation. The SSC CGL (Tier II) Practice Test - 1 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL (Tier II) Practice Test - 1 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for SSC CGL (Tier II) Practice Test - 1 below.
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SSC CGL (Tier II) Practice Test - 1 - Question 1

Suppose AB is the diameter of the circle as shown, BC is tangent to the circle, ∠BAC = 30° and CD = √3. What is the distance between A and B?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 1
It is well-known that CB2 = CD x CA.

Also, as ΔACB is a 30° – 60° – 90° triangle, we have CA = 2CB.

Thus, CB2 = 2CD.CB, which implies CB = 2CD = 2√3

Again, using that ΔACB is a 30° – 60° – 90° triangle, we obtain AB = √3CB = 6.

SSC CGL (Tier II) Practice Test - 1 - Question 2

Which of the following statement(s) is/are TRUE?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 2

Hence, only statement I is correct.

SSC CGL (Tier II) Practice Test - 1 - Question 3

In a set of three numbers, the average of first two numbers is 21, the average of the last two numbers is 24, and the average of the first and the last numbers is 15. What is the average of the three numbers?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 3
According to the question

Let the numbers be a, b, and c.

a + b = 42 ... (1)

b + c = 48 ... (2)

a + c = 30 ... (3)

Adding equations (1), (2) and (3) we get:

2a + 2b + 2c = 120

a + b + c = 60

Average = (a + b + c)/3 = 60/3 = 20

SSC CGL (Tier II) Practice Test - 1 - Question 4

Three dice are rolled together. What is the probability of getting the same number on the three?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 4

Total number of possible cases = 63 = 216

Favourable number of cases = 6

Probability = 6/216

= 1/36

SSC CGL (Tier II) Practice Test - 1 - Question 5

A, B and C can build a wall in 12 days together. C is four times as productive as B and A alone can build the wall in 48 days. In how many days can A and B working together build the wall?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 5

A, B and C can build a wall together in 12 days. Let A, B and C alone complete the work in A days, B days C days, respectively.

So, one day's work of A, B and C is:

--- (i)

C is 4 times as productive as B.

C's 1 day's work = 4(B's 1 day's work)

So, C's one day's work is:

Also, one day's work of A is:

Putting these values in (i),

B = 16 × 5 = 80

So,

So, A and B's 1 day's work =

Thus, A and B together can finish in 30 days.

SSC CGL (Tier II) Practice Test - 1 - Question 6

In a trapezium PQRS, PQ is parallel to RS, PQ = 20 cm, RS = 3 cm, ∠PQR = 30° and ∠QPS = 60°. What is the length of the line joining the midpoints of PQ and RS?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 6

PS and QR are extended to meet at T.

PTR = STR = 90° (Angle sum property of Δs STR and PTQ)

Note:

TSR = TPQ = 60° (Corresponding angle axiom)

TRS = TQP = 30°

cm [Using the property of median to hypotenuse of right-angled triangle]

= 10 cm

UV = TV – TU = 10 – 1.5 = 8.5 cm

SSC CGL (Tier II) Practice Test - 1 - Question 7

If the angles A, B and C of a triangle are in the ratio 2 : 3 : 7, then the sides a, b and c will be in the ratio

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 7

Let the angles be 2x, 3x and 7x, respectively.

2x + 3x + 7x = 180°

⇒ x = 15°

Angles are 30°, 45° and 105°.

Sides are in the ratio of sines of angles (by sine formulae), i.e. a : b : c = sin A : sin B : sin C.

So, a : b : c = sin 30° : sin 45° : sin 105° =

Multiply throughout by 2√2, we get

SSC CGL (Tier II) Practice Test - 1 - Question 8

A bank gives Rs. 25,000 on saving a certain principal in 2 years at 8% rate of interest. How much will the bank give (in Rs.) on the same principal in 4 years at the same rate of interest compounded annually?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 8

Given: A = Rs. 25,000, r = 8%, t = 2 years

We know that the formula for compound interest is:

... (i)

According to the question, suppose Rs. x is the amount after 4 years. Then,

x = Rs. 29,160

SSC CGL (Tier II) Practice Test - 1 - Question 9

Directions: Study the information carefully to answer the question that follows.

Number of students that appeared in MBA entrance exam from 2008-2011 = 8,00,000

Percentage breakup of number of students that appeared for MBA entrance exam in 2008:

(Note: 0-1 year experienced means the person is having work experience greater than 0 years and up to 1 year.)

Q. The students who appeared for MBA entrance exam in 2008 having more than 2 years of experience, were what percent less than the students who appeared for MBA exam in 2011?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 9
The students that appeared for MBA entrance exam in 2008 having more than 2 years of experience = 0.04 × 0.32 × 8,00,000 = 10,240

Total number of students that appeared for MBA exam in 2011 = 0.28 × 8,00,000 = 2,24,000

Percent less = (2,24,000 - 10,240) × 100/2,24,000 = 95.43%

SSC CGL (Tier II) Practice Test - 1 - Question 10

Two dice are thrown. Find the probability that the product of two numbers received is even.

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 10

Probability = 1 - Both off numbers / Total cases

Total cases = 62

Both odd numbers

=

Probability =

= 3/4

SSC CGL (Tier II) Practice Test - 1 - Question 11

If 2x + 1/3x = 6, then 3x + 1/2x is equal to

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 11

2x + 1/3x = 6

Multiplying both sides by 3/2, we have

3x + 1/2x = 6 x 3/2 = 9

SSC CGL (Tier II) Practice Test - 1 - Question 12

If (1/21) + (1/22) + (1/23) ... + (1/210) = 1/k, then what is the value of k?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 12

The given series is a geometric series.

Here, a is the first term, r is the common ratio and n is the number of terms.

a = 1/2, r = 1/2

n = 10

Sum of terms:

SSC CGL (Tier II) Practice Test - 1 - Question 13

Directions: Study the information carefully to answer the question that follows.

Number of students that appeared in MBA entrance exam from 2008-2011 = 8,00,000

Percentage breakup of number of students that appeared for MBA entrance exam in 2008:

(Note: 0-1 year experienced means the person is having work experience greater than 0 years and up to 1 year.)

Q. If the MBA entrance test fees increased by 15% from 2008 to 2009, what was the percent change in the total fees paid by the students in 2009 to that in 2008?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 13

Let the MBA entrance test fees in 2008 be Rs. 100.

MBA entrance test fees in 2009 = Rs. 115

Percent change =

= 13.75%

SSC CGL (Tier II) Practice Test - 1 - Question 14

Directions: Study the information carefully to answer the question that follows.

Number of students that appeared in MBA entrance exam from 2008-2011 = 8,00,000

Percentage breakup of number of students that appeared for MBA entrance exam in 2008:

(Note: 0-1 year experienced means the person is having work experience greater than 0 years and up to 1 year.)

Q. If, out of the total number of students that appeared for MBA entrance exam during 2009-2011, exactly 41% were 0-1 year experienced. Find the total number of students that appeared for MBA entrance exam during 2008-2011, who were having 0-1 year experience.

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 14
Total number of students who appeared for MBA entrance exam during 2008-2011 with 0-1 year experience = 0.21 × 0.32 × 8,00,000 + 0.41 × (0.24 + 0.16 + 0.28) × 8,00,000 = 0.21 × 0.32 × 8,00,000 + 0.41 × (0.68 × 8,00,000)

= 53,760 + 2,23,040 = 2,76,800

Thus, answer option c is correct.

SSC CGL (Tier II) Practice Test - 1 - Question 15

A certain sum of money amounts to $1600 in 4 years and further to $1800 in 6 years at simple interest. How much will it amount to in 10 years?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 15
We have, P + S.I. for 4 years = $1600 …(i)

Also, P + S.I. for 6 years = $1800 …(ii)

Thus, (ii) – (i) ⇒ S.I. for 2 years = $200

Thus, S.I. for 1 year = $100

Hence, we get

P + S.I. for 10 years = 1800 + (4 × 100) = $2200

SSC CGL (Tier II) Practice Test - 1 - Question 16

If and a sinθ - b cosθ = 0 then find the value of a2 + b2. (If sinθ, cosθ ≠ 0)

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 16

Let θ = 45°

Now,

a sinθ - b cosθ = 0

⇒ a sin θ = b cos θ

⇒ a = b (sin45° = b cos45°)

Then,

⇒ a sin3θ + b cos3θ = 8sinθ.cosθ

(a = b)

⇒ a = 4√2 = b

⇒ a2 + b2 = (4√2)2 + (4√2 )2 = 64

The value of a2 + b2 = 64.

SSC CGL (Tier II) Practice Test - 1 - Question 17

If the greatest number of six digit 435a1b is divisible by 45 then find the value of 101a + 99b.

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 17

Given:

six digit no. 435a1b, divisible by 45.

Concept used:

Divisibility rule for 3 states that "a number is completely divisible by 3 if the sum of its digits is divisible by 3."

Divisibility rule for 5 states that "a number is completely divisible by 5 If the last number is either 0 or 5."

Calculation:

If given number divisible by 45 = 5 × 3 × 3,

⇒ then it's also divisible of 3, 5, 9 and 15.

According to divisibility rule of 5,

⇒ the value of b is 0 or 5 but given no is greatest of six digit then

⇒ b = 5

According to divisibility rule of 3,

⇒ Sum of digit = 4 + 3 + 5 + a + 1 + 5

⇒ 18 + a

value of a = 3 or 6 or 9 but given no is greatest of six digit then

⇒ a = 9.

Now,

⇒ 101a + 99b = 101 × 9 + 99 × 5

⇒ 909 + 495 = 1404

The Value of 101a + 99b = 1404.

SSC CGL (Tier II) Practice Test - 1 - Question 18

A boat covers a distance of 18 km downstream and 14 km upstream in 9 hours and It covers a distance of 45 km downstream and 12 km upstream in 11 hours. What is the speed (in km/hr) of the current?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 18

Given:

Condition 1:

A boat covers a distance in downstream = 18 km

A boat covers a distance in upstream = 14 km

Total time taken = 9 hour

Condition 2:

A boat covers a distance in downstream = 45 km

A boat covers a distance in upstream = 12 km

Total time taken = 11 hour

Concept used:

Time = Distance/Speed

If boat stream in downstream then speed of boat = speed of boat in still water + speed of current

If boat stream in upstream then speed of boat = speed of boat in still water - speed of current

Calculation:

Let speed of boat in still water = x km/hour

and speed of current = y km/hour

⇒ If boat stream in downstream then speed of boat = x + y

⇒ If boat stream in upstream then speed of boat = x - y

According to the question,

Equation (1) multiply with 5 and equation (2) multiply with 2

Subtracting equation (4) from equation (3)

⇒ x - y = 2 ----(5)

Putting the value of (x - y) in equation (1)

⇒ x + y = 9 ----(6)

now equation (5) - equation (6)

⇒ - 2y = - 7

⇒ y = 7/2 km/hour

Speed of the current = 7/2 km/hour.

SSC CGL (Tier II) Practice Test - 1 - Question 19

Study the given pie - chart and table carefully and answer the question that follows. The percentage wise distribution of lecturers in five different subjects in a university is shown in the pie chart.

The total number of lecturers is 500.

Ratio of male to female lecturers:

Q. What is the difference in the number of female lecturers in chemistry and Mathematics?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 19

Given :

Total number of lecturers are 500

Calculations:

Lecturers in Chemistry = 20% of 500

⇒ 100

Female lecturers in Chemistry = (1/5) of 100 (Ratio of male to female is 4 : 1)

⇒ 20

Lecturers in mathematics = 30% of 500

⇒ 150

Female lecturers in mathematics = (3/10) of 150 (Ratio of male to female is 7 : 3)

⇒ 45

Difference of female lecturers of Chemistry and Mathematics = 45 - 20

⇒ 25

∴ The correct answer will be option D.

SSC CGL (Tier II) Practice Test - 1 - Question 20

Rs. 30000 is borrowed at compound interest at the rate of 8 percent per annum. If the interest is compounded quarterly, then how much amount will have to be paid after 6 months?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 20

Given:

Rs. 30000 is borrowed at compound interest at the rate of 8% p.a. and interest compounded quarterly and time = 6 months.

Formula used:

Amount = Principal ( 1 + Rate/100)Time,

Calculation:

Principal = Rs. 30000 , Rate = 8% p.a. , but compounded quarterly so actual rate = 8%/4 = 2% , Time = 6 months but compounded quarterly so actual time = 6 × 4 = 24 months = 2 years.

Amount = Principal (1 + Rate/100)Time

Amount = 30000 (1 + 2/100)2 = 30000 × 51 × 51 /2500 = 31212 Rs.

∴ Answer is 31212 Rs.

SSC CGL (Tier II) Practice Test - 1 - Question 21

A student obtained equal marks in english and history. The ratio of marks in history and maths is 3 ∶ 4. He scored an aggregate of 60 percent in the three subjects. Maximum marks for each subject is 100. How much did he score in history?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 21

Given:

Marks obtained in history = marks obtained in English.

Ratio of marks in history and maths = 3 : 4.

Total marks scored by him = 60% of maximum marks in 3 subjects.

Maximum marks for each subject = 100.

Concept used:

X% of Y is

= (X / 100) × Y

Calculation:

Let the marks obtained in history and maths be 3x and 4x respectively.

Marks obtained in English = marks obtained in history = 3x.

Combined maximum marks of all 3 subjects = 100 × 3 = 300.

Total marks scored by candidate in all 3 subjects

= 60% of 300

= (60 / 100) × 300

= 180 marks.

Thus,

3x + 3x + 4x = 180

10x = 180

x = 180 / 10

x = 18.

Score in history = 18 × 3 = 54 marks.

∴ The student scored 54 marks in history.

SSC CGL (Tier II) Practice Test - 1 - Question 22

If Sarita had walked at a speed of 15 km/h instead of 10 km/h, she would have walked 40 km more in the same time. What is the actual distance travelled by her?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 22

Given:

Actual speed = 10 km/h

Imaginary speed = 15 km/h

Formula used:

Distance = speed × time

Calculation:

Let the time be T.

According to the question,

15T – 10T = 40

⇒ 5T = 40

⇒ T = 8 hours

Actual distance covered = (10 × 8)

⇒ 80 km

∴ The actual distance traveled by her was 80 km.

SSC CGL (Tier II) Practice Test - 1 - Question 23

A man sold a car at 12 percent profit and a bike at 12 percent profit. If he had sold the car at 21 percent profit and the bike at 12 percent profit, then he would have gained Rs. 4500 more. If the total cost price of car and bike is Rs. 75000, then what is the respective cost price of bike and car?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 23

Given:

Initial profit% of car = 12%.

Initial profit% of bike = 12%.

Final profit% of car = 21%.

Final profit% of bike = 12%.

Final profit = Rs. 4500 more than initial profit.

Total cost price of bike and car = Rs. 75000

Formula used:

Let C.P. = Cost price and S.P. = Selling price then

Profit% = [(S.P. - C.P.) / (C.P.)] × 100

As the initial and final profit% is same for bike, we only consider the profit % of car for the change in net profit.

Calculation:

Let the C.P. of the car be Rs. x and bike be Rs. y.

x + y = 75000 ... (1)

As the initial and final profit% is same for bike, we only consider the profit % of car for the change in net profit.

Net profit = 4500

Final profit% - Initial profit% = 4500

21% - 12% = 4500

9% = 4500.

1% = 4500 / 9

1% = 500

100% = 50000.

This is the C.P. of the car, x = Rs. 50000.

Thus, the C.P. of bike = Rs. 75000 - Rs. 50000 = Rs. 25000.

∴ The cost price of the bike and car are Rs. 25000 and Rs. 50000 respectively.

SSC CGL (Tier II) Practice Test - 1 - Question 24

When a number is successively decreased by 35 percent, 60 percent and 10 percent, then it becomes 37440. What is that number?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 24

When a number is successively decreased by 35 percent, 60 percent and 10 percent, it becomes 37440.

Concept used:

When a number m is successively decreased by x, y, z we can use the formula

where x' is the new value of x.

Calculation:

Using above concept,

x = 37440 × 1000 / 13 × 2 × 9

x = 160000.

∴ The number is 160000.

SSC CGL (Tier II) Practice Test - 1 - Question 25

A borewell with a diameter of 20 m from the inside is dug 10 m deep. Soil taken out of it has been evenly spread all around it to a width of 30 m to form an embankment. Find the height (in m) of the embankment.

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 25

Volume of the soil dug out = πr2h

⇒ (22/7 × 10 × 10 × 10) = 3,142.85 m3.

Area of embankment = Area of embankment with well - Area of well

= (22/7 × 40 × 40) - (22/7 × 10 × 10) = 4,714.3 m2

Height of embankment = Volume/Area = 3142.85/4714.3 = 2/3 m.

SSC CGL (Tier II) Practice Test - 1 - Question 26

​Mean of twenty observations is 15. If two observations 3 and 14 are replaced by 8 and 9 respectively, then the new mean will be:

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 26

Given :

Mean of 20 observations = 15

Concept used :

new mean = old mean + difference in observations / no.of observations

Calculation :

difference in observations = sum of old observations - sum of new observations added instead

Here , difference = (3+14) - (8 + 9) = 0

So, new mean = old mean + 0/20 = new mean = 15

∴ The answer is 15.

SSC CGL (Tier II) Practice Test - 1 - Question 27

Study the following table carefully and answer the question given below it:

Number of candidates appearing for an interview for a post in various Banks and percentage of candidates qualifying.

Q. The number of candidates who did not qualify in bank K was approximately what percent of the candidates who did not qualify in bank I ?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 27

Given:

The percentage of the not qualifying candidate in K is 80%

The percentage of the not qualifying candidate in I is 74%

Calculation:

⇒ Total number of candidate not qualifying in K = 980 × (80/100) = 784

⇒ Total number of candidate not qualifying in I = 2200 × (74/100) = 1628

⇒ The required percentage = (784/1628) × 100 = 48.15 ≈ 48%

∴ The required result will be 48%.

SSC CGL (Tier II) Practice Test - 1 - Question 28

The ratio of the salaries of Akhshay and Amir is 8 ∶ 9. If Akhshay’s salary is increased by 50% and Amir’s salary is reduced by 25%, then their ratio becomes 16 ∶ 9. What is the salary of Akshay?

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 28

Given:

Ratio of salary of Akshay : Amir is 8 : 9. Salary increased by 50% of Akshay and 25% decreased of Amir's salary, then ratio becomes 16 : 9.

Calculations:

Let Akshay's salary be 8x and Amir's salary be 9x

When Akshay's salary is increased by 50%,

⇒ 8x + 8x × 50%

⇒ 8x + 8x × 50 / 100

⇒ 8x + 8x / 2

⇒ 8x + 4x

∴ Salary becomes = 12x

When Amir's salary is decreased by 25%,

⇒ 9x - 9x × 25%

⇒ 9x - 9x × 25 / 100

⇒ 9x - 9x / 4

⇒ 9x - 2.25x

∴ Salary becomes = 6.75x

Now, by the problem, 12x / 6.75x = 169

But from the above calculation we can't determine the value as unknown term is cancelled out and no term is left.

∴ The answer is cannot be derermined.

SSC CGL (Tier II) Practice Test - 1 - Question 29

The cost of 2 sarees and 4 shirts is Rs 16,000 while the cost of 1 saree is equal to the cost of 6 shirts. The cost of 12 shirts is:

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 29

Given:

The cost of 2 sarees and 4 shirts = Rs. 16,000

The cost of 1 saree = the cost of 6 shirts

Calculation:

Let the cost of one saree and shirt be x and y respectively.

2x + 4y = 16,000 -----(1)

Also, x = 6y

⇒ 2(6y) + 4y = 16000

⇒ 16y = 16000

⇒ y = 1000 Rs.

Hence, the cost of 12 shirts

⇒ 12y = 12 × 1000 = 12000

∴ The cost of the 12 shirts is Rs. 12,000.

SSC CGL (Tier II) Practice Test - 1 - Question 30

If a +1/b = 1 and b +1/c = 1, then c +1/a is equal to:

Detailed Solution for SSC CGL (Tier II) Practice Test - 1 - Question 30

Calculation

a + 1/b = 1

⇒ ab + 1 = b

⇒ b – ab = 1

⇒ b(1 – a) = 1

⇒ b = 1/(1 – a)

Similarly if (b + 1/c) = 1

⇒ bc + 1 = c

⇒ c(1 – b) = 1

⇒ c = 1/(1 – b)

⇒ c + 1/a = 1/(1 – b) + 1/a

⇒ [1/(1 – 1/(1 – a)] + 1/a

⇒ [1/(1 + 1/(a - 1)] + 1/a

⇒ (a – 1)/(a – 1 + 1) + 1/a

⇒ (a – 1)/a + 1/a

⇒ (a – 1 + 1)/a

∴ c + 1/a = a/a = 1

SSC CGL (Tier II) Practice Test - 1 - Question 31

Select the term that is related to the fourth number in the same way as the first number is related to the second number and the fifth number is related to the sixth number.

17 : 293 :: ? : 488 :: 21 : 445