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

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.

Hence, only statement I is correct.

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

Total number of possible cases = 6³ = 216

Favourable number of cases = 6

Probability = 6/216

= 1/36

• 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.

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

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

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

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%

Probability = 1 - Both off numbers / Total cases

Total cases = 6²

Both odd numbers

=

Probability =

= 3/4

2x + 1/3x = 6

Multiplying both sides by 3/2, we have

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

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:

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

MBA entrance test fees in 2009 = Rs. 115

Percent change =

= 13.75%

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

Thus, answer option c is correct.

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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%.

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.

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.

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