SSC CGL (Tier II) Practice Test - 6 - SSC CGL MCQ

So, the numbers are 104, 117, 130, ... 390.

Thus, the number of terms in the series is 23.

Sum of this AP

(2 × 104 + 22 × 13) = 5681

Radius of the base of the cone = Radius of the base of the cylinder = r

Let the height of the cone be h₁ and height of the cylinder be h₂.

r = 1 cm (Given in the figure)

h₂ = 4 cm (Given in the figure)

h₁ + h₂ = 7 cm (Given in the figure)

Hence, h₁ = 3 cm

Volume of the solid = Volume of the cone + Volume of the cylinder

Substituting the variables by their numeric values in the equation, we get

Required average = 1420/20 = 71

This means- 16 : 20 = 20 : x

16x = 400

x = 25

If first dice gives 1, the second dice can be any number from 2 to 6.

Similarly, the favorable events are (1, 2) … (1, 6), (2, 3)...(2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6).

So total number of favorable events = 15

Required Probability = 15/36 = 5/12

Let the principal money be P and time be T years.

According to the question,

(1.15)^T > 2

Now, substituting values of T

T = 2,

(1.15)² = 1.3225

T = 3,

(1.15)³ = 1.5208

T = 5,

(1.15)⁵ = 2.011

For T = 5,

(1.15)^T > 2

So, time = 5 years

3^x > 9^y

3^x > 3^2y

x > 2^y

x > y

Also,

2^y > 4^z

2^y > 2^2z

y > 2^z

y > z

Thus, x > y > z

Hence answer option 1 is correct.

1² + 2 = 3

2² + 1 = 5

3² + 0 = 9

4² - 1 = 15

5² - 2 = 23

6² - 3 = 33

∴ Required square root = √225 = 15

34 = 3 + x + y + 3 + 3 + 3 + 3 + 3 + 4

34 = 22 + x + y

x + y = 12 cm

When they are moving in the same direction (x > y), then (x – y) x 5 = 100

⇒x – y = 20 …..(i)

When they are moving in opposite directions, then (x + y) = 100 ….(ii)

Adding these, we get

2x = 120

⇒ x = 120/2 = 60 km/hr and y = 40 km/hr

Expenditure in 1986 is 75% of the income i.e.

= Rs. 1,50,000

Therefore, expenditure on education

Given:

The ratio between the length and breadth of a rectangular field is 5 ∶ 4.

The breadth is 20 metres less than the length.

Concept used:

Perimeter of a rectangle = 2(Length + Breadth)

Calculation:

Let the length and breadth of the field be 5q and 4q meter respectively.

According to the question,

5q - 4q = 20

⇒ q = 20

Length = 5 × 20 = 100m

Breadth = 4 × 20 = 80m

Now, the perimeter of the field = 2(100 + 80)

⇒ 360m

∴ The perimeter of the field is 360m.

Given:

5 cot θ + 12 = 0, where π / 2 < θ < π .

Concept used:

θ lies in the 2^nd quadrant, So the value of tan θ, cot θ, and cos θ is negative and

the value of sin θ is only positive hear.

Calculation:

5 cot θ + 12 = 0

⇒ 5 cot θ = - 12

⇒ cot θ = - 12 / 5

Value of tanθ = - 5 / 12

In triangle ABC,

AB = 5 , BC = 12 so value of AC = 13

So sinθ = 5 / 13 and cos θ = - (12 / 13)

So, the value of 36 tan θ + 26 sin θ - 39 cos θ

⇒ 36 (- 5 / 12) + 26 (5 / 13) - 39 {- (12 / 13)}

⇒ - 15 + 10 + 36 = 31

Given:

Pipes A and B can fill a tank in 12 hrs and 16 hrs respectively.

The pipe C can empty it in 20 hrs.

Pipes A and C first opened for 5 hrs

then Pipe A closed and B opened.

Concept used:

Efficiency = Total work / Total time taken

Calculation:

Let the total volume of the tank is LCM of 12, 16, and 20 = 240

The efficiency of Pipe A = 240 / 12

⇒ 20 units

The efficiency of Pipe B = 240 / 16

⇒ 15 units

The efficiency of Pipe C = 240 / 20

⇒ 12 units

Pipes A and C first opened for 5 hrs

So, In 5 hrs A pipe filled = 20 × 5 units

⇒ 100 units

In 5 hrs C pipe emty = 12 × 5

⇒ 60 units

So total filled = 100 - 60

⇒ 40 units

Remaining Amount = 240 - 40

⇒ 200 units.

Now pipe A will be closed and Pipe B will be opened

So, In 1 hr Both Pipe B and C filled 15 - 12 = 3 units

So a total of 200 units will be filled in (200 / 3) = 66.66 hrs

The total tank will be filled in (66.66 + 5) = 71.66 hrs

71.66 hours the tank will be filled

Given:

The amount invested by Chand = Rs. 8560

The time of investment for Chand = 24 months

The amount invested by Sooraj = Rs. 6420

The time of investment for Sooraj = 12 months

The total profit earned = Rs. 1815

Concept:

Profit ∝ Amount of investment

Profit ∝ Time of investment

Therefore,

The ratio of profit earned by two people is equal,

To the ratio of the product of the Amount of investment and the Time of investment.

Calculation:

Let S and Chand be the profit earned by Sooraj and Chand

According to the question and as per the above concept,

The profit earned by Sooraj = 3/11 × 1815 = Rs. 495

The profit earned by Chand = 8/11 × 1815 = Rs. 1320

The required difference = Rs. 1320 - Rs. 495 = Rs. 825

Therefore, 'Rs. 825' is the required answer.

Given:

Calculation:

According to the question,

Add 3 on both sides of the equation.

Since x + y + z ≠ 0,

Therefore,

Therefore, is the required answer.

Calculation:

The total number of computers sold by dealer A on Monday, Tuesday, and Friday = 75 + 80 + 78 = 233

Thus, statement I is incorrect.

The total number of computers sold by dealer B on Wednesday, Thursday, and Saturday = 60 + 80 + 75 = 215

Thus, statement II is correct.

∴ Only statement II is correct.

Given :

Marked price : Cost price = 7 ∶ 3 and Cost price : Selling price = 3 ∶ 5

Marked price : Cost price = 7 ∶ 2 and Cost price : Selling price = 1 ∶ 2

Marked price : Cost price = 3 ∶ 1 and Cost price : Selling price = 5 ∶ 6

Marked price : Cost price = 8 ∶ 3 and Cost price : Selling price = 2 ∶ 3

Formula used:

Discount % =

Where, S.P = Selling Price

M.P = Marked Price, C.P = Cost Price

Calculation:

From the option,

Step 1: Option 1

Marked price : Cost price = 7 ∶ 3 -----(1)

Cost price : Selling price = 3 ∶ 5 ------(2)

Marked Price : Cost Price : Selling Price = 7 : 3 : 5

Let the value of Marked Price , Cost Price and Selling Price be 7x , 3x and 5x respectively

Discount % =

= 200/7

= 28.57 %

Step 2: Option 2

Marked price : Cost price = 7 ∶ 2 -----(3)

Cost price : Selling price = 1 ∶ 2------(4)

multiply(4) by 2 to make the Cost Price equal

Marked Price : Cost Price : Selling Price = 7 : 2 : 4

Let the value of Marked Price , Cost Price and Selling Price be 7x , 2x and 4x respectively

Discount % =

= 300 / 7

= 42.85 %(approx.)

Step 3: Option 3

Marked price : Cost price = 3 ∶ 1 -----(5)

Cost price : Selling price = 5 ∶ 6 ------(6 )

multiply (5) × 5 to make the Cost Price equal

Marked Price : Cost Price : Selling Price = 15 : 5 : 6

Let the value of Marked Price , Cost Price and Selling Price be 15x , 5x and 6x respectively

Discount % =

= 900/15

= 60%

Step 4: Option 4

Marked price : Cost price = 8 ∶ 3 -----(7)

Cost price : Selling price = 2 ∶ 3------(8)

multiply (8) × 3 and (7) by 2 to make the Cost Price equal

Marked Price : Cost Price : Selling Price = 16 : 6 : 9

Let the value of Marked Price , Cost Price and Selling Price be 16x , 6x and 9x respectively

Discount % =

= 700 / 16

= 43.75%

Hence, Option A has the lowest discount.

Answer is "Marked price : Cost price = 7 ∶ 3 and Cost price : Selling price = 3 ∶ 5" .

Given:

x and y are the LCM and HCF of the numbers, 30, 45, 90 and 135 respectively

Concept used:

HCF - Highest common factor

LCM - Least common multiple

Calculation:

Let's find factors of given numbers,

30 = 2 × 3 × 5

45 = 3 × 3 × 5

90 = 2 × 3 × 3 × 5

135 = 3 × 3 × 3 × 5

So, 3 and 5 are highest common factors of all the given numbers.

HCF (30, 45, 90 and 135) = 15

Y = 15

And,

30 = 2 × 3 × 5

45 = 3 × 3 × 5

90 = 2 × 3 × 3 × 5

135 = 3 × 3 × 3 × 5

So, the least common factor of all the given numbers = 3 × 3 × 3 × 2 × 5

Hence, LCM (30, 45, 90 and 135) = 270

X = 270

Value of x - 7y,

x - 7y = 270 - 7 (15)

= 270 - 105

= 165

∴ Answer is 165

Given:

Calculations:

(let k is the constant)

x = 4k, y = 5k, z = 3k

Now,

= 4

∴ The answer is 4.

Calculation:

Total expenditure on taxes for all years = 76 + 105 + 84 + 85 + 89 = 439

Total expenditure on fuel and transport for all years = 96 + 156 + 143 + 124 + 134 = 653

Ratio between the total expenditure on taxes for all the years and the total expenditure on fuel and transport for all years = 439 : 653

= 10.2(approx.) × 43 : 15.2(approx.) × 43

= 10.2 : 15.2

= 10 : 15 (approx.)

Answer is 10 : 15.

Given:

The number 678967# is exactly divisible by 72

Concept used:

If the number is divisible by 72 that means it has to be divisible by 8 and 9 as these are the two factors of 72.

Divisibility rules of 8 : when the number made by last three digits of a number is divisible by 8 then the number is also divisible by 8. Apart from this if the last 3 or more digits of a number are zero then the number is divisible by 8.

Divisibility rules of 9 : when the sum of all the digits of a number is divisible by 9 then the number is also divisible by 9.

Calculations:

The number is 678967#, so if we apply the divisibility rules of 8, the last 3 digits have to be divisible by 8.

Now, 67# has to be divisible by 8, so if we put 2 in place # it becomes 672 which is divisible by 8.

Again, applying divisibility rules of 9, by adding all the numbers = 6 + 7 + 8 + 9 + 6 + 7 + # = 43 + #

So, if we take 2 again in place of # the sum of the numbers become = 45 which is divisible by 9.

Thus, two rules satisfies the condition and the number is 2.

∴ The answer is 2.

Given∶

ΔPQR is a right angle triangle, ∠PQR = 90°, PQ = 7 cm, QR = 24 cm.

Formula Used∶

In radius of the circle = Area of triangle /Semi perimeter of triangle.

Pythagoras theorem

Semi- perimeter of a triangle = (a + b +c)/2, where a, b and c are the sides of the triangle.

Area of triangle = 1/2 × Base × Height

Calculation∶

In ΔPQR, ∠PQR = 90°, PQ = 7cm and QR = 24 cm

So, PR² = PQ² + QR² [Pythagoras theorem]

⇒ PR² = 7² + 24²

⇒ PR² = 49 + 576 = 625

⇒ PR = 25 cm

Semi perimeter of ΔPQR,

S = (PQ + QR + PR)/2

⇒ S = (7 + 24 + 25) /2

⇒ S = 28 cm

Now, Area of ΔPQR = 1/2 × PQ × QR

Area = 1/2 × 7 × 24 = 84 cm²

⇒ In radius, r = Area/semi perimeter

⇒ r = 84/28 = 3 cm

∴ The radius of the incircle is 3 cm.

Given:

PT is perpendicular to QR.

PS : ST = 3 : 1.

∠QPT = 30° and tan ∠PRQ = 4 × tan ∠SQT

Formula Used:

If θ is an angle of a right angled triangle, then tan θ = Perpendicular/Base

Calculation:

∠ PQT = 180° - (∠PTQ + ∠QPT) = 180° - (90° + 30°) = 60°

Now, tan ∠PRQ = PT/TR ----(1)

and tan ∠SQT = ST/QT ----(2)

Now multiplying (1) with (2) we get,

tan ∠PRQ/tan ∠SQT = PT/TR × QT/ST

⇒ 4 × ST/PT = QT/TR

⇒ 4 × 1/4 = QT/TR

⇒ QT = TR

It means PT bisects QR in a ratio 1 : 1

So, ∠PRQ = 60° (∵dQT/TR = PQ/PR) and thus ΔPQR is an equilateral triangle.

∴ Value of ∠PRQ is 60°

Formula used∶

Rationalization of surds, (a + b) (a – b) = a2 – b2

Calculation∶

1/(4 - √15) – 1/(√15 - √14) + 1/(√14 - √13) -1/√(13 - √12) + 1/√(12 - √11) – 1/(√11 - √10) + 1/(√10 – 3)

⇒ [1/(4 - √15) × {(4 +√15)/ (4 +√15)}] = [(4 +√15)/{42 – (√15)2}] = (4 +√15)/(16 – 15) = (4 +√15)/1 = (4 + √15)

⇒ [{1/(√15 - √14)} × {(√15 + √14)/ (√15 + √14)}] = [{(√15 + √14)}/{(√15)2 – (√14)2}] = [√15 + √14)/(15 – 14) = (√15 + √14)/1 = √15 + √14

⇒ [{1/(√14 - √13) × {(√14 + √13)/ (√14 + √13)}] = [{(√14 + √13)}/{(√14)2 – (√13)2}] = [(√14 + √13)/(14 – 13)] = [(√14 + √13)/1] = √14 + √13

⇒ [{1/(√13 -√12) × {(√13 + √12)/ (√13 + √12)}] = [{(√13 +√12)}/{(√13)2 – (√12)2}] = [(√13 + √12)/(13 – 12)] = [(√13 + √12)/1] = √13 + √12

⇒ [{1/(√12 -√11) × {(√12 + √11)/ (√12 + √11)}] = [{(√12 +√11)}/{(√12)2 – (√11)2}] = [(√12 + √11)/(12 – 11)] = [(√12 + √11)/1] = √12 + √11

⇒ [{1/(√11 -√10) × {(√11 + √10)/ (√11 + √10)}] = [{(√11 +√10)}/{(√11)2 – (√10)2}] = [(√11 + √10)/(11 – 10)] = [(√11 + √10)/1] = √11 + √10

⇒ [{1/(√10 -3) × {(√10 + 3)/ (√10 + 3)}] = [{(√10 +3)}/{(√10)2 – (3)2}] = [(√10 + 3)/(10 – 9)] = [(√10 + 3)/1] = √10 + 3

Now, put all the values in their required places in the given equation, we get

⇒ (4 + √15) – (√15 + √14) +(√14 + √13) – (√13 + √12) + (√12 + √11) – (√11 + √10) + (√10 + 3)

⇒ 4 + √15 – √15 - √14 + √14 + √13 – √13 - √12 + √12 + √11 – √11 - √10 + √10 + 3 = 7

∴ The required value is 7.

In certain coding language,

CIGARETTE is written as GICERAETT as follows,

Similarly,

Direction can be written as follows,

Hence, ‘RIDTCENOI’ is the correct answer.