SSC CGL Tier 1 and Tier 2 (Tier II) Practice Test - 9 Free Online Test

= (1 + 4 + 7 + 10 + ...) up to 20 terms

Now, 1 + 4 + 7 + 10 + ... is an AP with a = 1, n = 20 and d = 3, where the symbols have their usual notations.

Sn = (n/2)(2a + (n - 1)d)

Thus, sum = (20/2)(2 x 1 + 19 x 3) = 590

Now,

= (2/3 + 1/12 + 1/20 + ....)

= (3/4 + 1/20 + ...)

= (4/5 + 1/30 + ...)

It is to be noted that if n terms of the above series are added, the result is n/(n + 1).

Thus, the sum of 20 such terms is 20/21.

Thus, final sum = 590 + 20/21 = 12410/21

Hence, answer option 1 is correct.

Let AB be x cm.

In triangle ABN, let NB be the base.

Hence, AN = x/2 and NE or side of the square = x cm

And cos 30° = BN/AB

In triangle BFE, using Pythagoras theorem,

BE = BN + NE

By solving this, we get

Area of the square = (side)²

= (24√3 - 36)² [(a - b)² = a² + b² - 2ab]

= 1728 + 1296 - 2 × 24√3 × 36

= 3024 - 1728√3

Initial expense of the firm = 2720 lakh

Initial savings of the firm = 4360 - 2720 = 1640 lakh

After the changed situation,

Income = 4360 + 45 = 4405 lakh

Expense = 2720 + 50 = 2770 lakh

Savings = 4405 - 2770 = 1635 lakh

Therefore,

Required % = x 100 = 0.304878% = 0.31% Loss

Hence, this is the required result.

Let the height of the building AB be x metres.

Base = (4 + x) m ... (i)

Now,

Base = √3 ... (ii)

From (i) and (ii),

After rationalising, we get

Interior angle = x + 120°

n(120° + x) = (n – 2) 180°

120°n + 360° = 180°n – 360°

360° + 360° = 60°n

720° = 60°n

n = 12

Investment made by Bhaskar = 3600 × 8 + 3000 × 4 = 40,800

Investment made by Keshav = X × 8 = 8X

Calculating the ratio of their investments as the profit sharing ratio will also be the same.

Arjun : Bhaskar : Keshav = 28,800 : 40,800 : 8X = 3600 : 5100 : X

Sum of ratio = 3600 + 5100 + X = 8700 + X

It is given that

22,500X = 8,000X + 69,600,000

22,500X - 8,000X = 69,600,000

14,500 X = 69,600,000

X = 48,00

This is the correct answer.

⇒ 2cosθ - 4cos²θ = 0

⇒ 2cosθ(1 - 2cosθ) = 0

i.e. cosθ = 0 or 1 - 2cosθ = 0

θ = 90° or cosθ = 2/4 = 1/2 ...(1)

As for θ = 90°, denominator becomes zero, which is not possible.

Also, cos60° = 1/2 ...(2)

From (1) and (2), we have

cosθ = cos60° or θ = 60°

5 + (2² - 1) = 8

8 + (2³ - 1) = 8 + 7 = 15

15 + (2⁴ - 1) = 15 + 15 = 30

30 + (2⁵ - 1) = 30 + 31 = 61 (But it is mentioned as 60)

61 + (2⁶ - 1) = 61 + 63 = 124

Hence, "60" is the correct answer.

Let the total work be W.

Now, 12 x 9 x ω = W

Or ω = w/12 x 9

Now, work done in 3 days = 3 x 12 x ω = w/3

Work left to be done = 2/3 x w

Number of men now = 12 - 2 + 6 = 12 + 4 = 16

Let 'd' days be taken to complete the remaining work.

Now, d x 16 x w = 2/3w

Or d =

7² = 49

7³ = 343

7⁴ = 2401

7⁵ = 16807

The units digit repeats after every 4th term.

71 4 gives remainder 3.

7³ = 343,

Thus, units digit in 7⁷¹ = 3

6¹ = 6

6² = 36

6³ = 216

6⁴ = 1296

The number always ends with 6.

Units digit in 6⁵⁹ = 6

3¹ = 3

3² = 9

3³ = 27

3⁴ = 81

3⁵ = 243

The units digit repeats after every 4th term.

65/4 gives remainder 1.

3¹ = 3

Units digit in 3⁶⁵ = 3

Now, 3 × 6 × 3 = 54

Here, the units digit is 4.

Thus, units digit in 7⁷¹ × 6⁵⁹ × 3⁶⁵ = 4

in 1 hour, the second machine will copy 1500/12 = 125 workbooks.

Thus, working together, the two machines will copy workbooks in 1 hour.

So, in order to copy 1500 workbooks, time required will be = 4.8 hours.

Speed = Distance/Time

44 = D/T

D = 44T ... (1)

77 = D/T- 3 ... (2)

Using (1) in (2), we get:

44T = 77(T - 3)

44T = 77T - 231

231 = 77T - 44T

T = 231/33

T= 7 hours

Distance = Time × Speed = 44 × 7 = 308 km

In the above series pattern is:

160

160 + 12 = 161

161 + 32 = 170

170 + 52 = 195

195 + 72 = 244

244 + 92 = 325

325 + 112 = 446

Thus, the wrong number in the series is 324.

Now,

Or d = 0.5

Total distance travelled = 4d km = 4 x 0.5 km = 2 km

New cost price = 0.9 CP

New selling price = SP - 132

According to the given condition,

× 100 = profit %

0.2 CP - 132 = 0.09 CP

0.11 CP = 132

CP = Rs. 1200

• cot θ = Base / Perpendicular
• sin θ = Perpendicular / Hypotenuse
• cos θ = Base / Hypotenuse

In a right-angle triangle,

Hypotenuse² = Base² + Perpendicular²

Calculation:

Given that

cot x = 4/3 = Base / Perpendicular

⇒ Base = 4k, Perpendicular = 3k

We know that;

Hypotenuse² = Base² + Perpendicular²

⇒ Hypotenuse² = (4k)² + (3k)² = 25k²

⇒ Hypotenuse = 5k

By using the above formula

⇒ sin θ = 3/5

⇒ cos θ = 5/5

Hence, the required value

Speed = 40 kmph and 30 kmph (on return)

Total time = 14 hours

Formula:

Distance = Speed × Time

Calculation:

Let the distance be ‘D’ km

According to the question,

D/40 + D/30 = 14

⇒ (3D + 4D)/120 = 14

⇒ 7D/120 = 14

⇒ D = 240 km

∴ Distance between A & B = 240 km

Rajib, Barun, and Anil engaged in work together

Rajib would invest Rs 4875 for 8 months, Barun Rs 8400 for 5 months, and Anil Rs 15000 for 2 months.

Rajib wants to be the working member for which, he was to receive 5% of the profits

The total profit earned was Rs 7400

Concept used:

Their share of profit is distributed as their investment ratio.

Calculation:

For working member, Rajib received = 5 % of the 7400

⇒ Rs 370

Remaining amount , 7400 - 370 = Rs 7030

The ratio of their investment =

(4875 × 8) : (8400 × 5) : (15000 × 2)

⇒ 39000 : 42000 : 30000

⇒ 13 : 14 : 10

∴ Anil's share = 7030 × (10 / 37)

⇒ 190 × 10

⇒ Rs 1900

So the share of Anil will be Rs 1900

In the figure given below, ∠CBE = ∠AFC, AF = 12 cm, CB = 6 cm, BE = 12 cm and CF = 6 cm,

AC = 2 cm,

Calculation:

According to the question, the required image is:

In ΔAFC and ΔCBE,

∠CBE = ∠AFC

AF = BE

CF = CB

Therefore, from SAS criteria ΔAFC ≅ ΔCBE.

Since ΔAFC ≅ ΔCBE,

Therefore, AC = CE = 2 cm

Now,

⇒ AE = AC + CE

⇒ AE = 2 + 2

⇒ AE = 4 cm

The required ratio = CE : AE = 2 : 4 = 1 : 2

Therefore, '1 : 2' is the required answer.

The total cost of fencing the right-angled triangular park = Rs. 102,600 per m

The length of the smallest side = 20 m

The length of the second longest side = 20 + 28 = 48 m

Formula used:

Pythagoras theorem:

Hypotenuse² = Perpendicular² + Base²

Calculation:

According to the question,

Since the park is right-angled triangle,

Therefore,

The length of the largest side = h

⇒ h² = 20² + 48²

⇒ h² = 2704

⇒ h = 52 m

The required perimeter of the park = 48 + 20 + 52 = 120 m

The required rate at which fencing is done = 102600/120 = Rs. 855 per m

Therefore, 'Rs. 855' is the required answer.

A student of a school has to score 25% of marks to get throughout.

He gets 50 marks and fails by 25 marks.

Calculation:

Let the maximum marks be x.

Now,

According to the question

⇒ 25% of x = 50 + 25

⇒ 25x/100 = 75

⇒ x = (3 × 100)

⇒ x = 300

∴ The maximum marks set for the examination is 300.

EF = 30 cm

Radius = 17 cm

Concept:

Pythagorean theorem

h² = p² + b²

Where h, p & b be the hypotenuse, perpendicular & base of triangle

Calculation:

Triangle OEG is a right-angle triangle.

EG = FG = OG/2

The distance of chord from center

∴ The length of the perpendicular OG is 8 cm.

The difference between the expenditure on marketing and Transportation is Rs 2400

Calculation:

The difference between the expenditure on marketing and Transportation (32 - 24) % = 8%

Total expenditure on transport, Electric Bill, and others together = (24 + 8 + 20) %

⇒ 52 %

As per the question:

Value of 8% = Rs 2400

⇒ 52% = (2400 / 8) × 52

⇒ 52 × 300 = Rs 15,600

The expenditure on transport, Electric Bill, and others together is Rs 15,600

We have,

2b = a + c ⇒ a - b = b - c

Now, x^{b - c} y^{c - a} z^{b - c}

⇒ (xz)^{b - c} y^{c - a}

⇒ y²(^{b - c}) y^{c - a} [ ∵ y² = zx]

⇒ y^{2b - 2c + c - a} = y^{2b - (a + c)}

⇒ y^{2b - 2b} = y⁰ = 1

Perimeters of Δ ABC and Δ DEF are 36 cm and 54 cm respectively

Both triangles are similar.

Concept used:If two triangles are similar then the ratio of the corresponding sides is equal to the ratio of perimeter of the triangle.

Calculation:

Here, ΔABC ~ Δ DEF

If two triangles are similar then the ratio of the corresponding sides is equal to the ratio of perimeter of the triangle

Perimeter of ΔABC/Perimeter of ΔDEF = AB/DE = BC/EF = AC/DF

Perimeter of ΔABC/Perimeter of ΔDEF = AB/DE

⇒ 36/54 = 12/DE

⇒ DE = (54 × 12)/36

⇒ DE = 18 cm

∴ The value DE is 18 cm.

The compound interest earned for the 2nd year = Rs. 240,

R = 20% p.a.

Formula used:

(1.) C.I_n+1 = (1 + R%) × C.I_n

Where,

C.I_n = Compound interest earned for n^th year

C.I_n+1 = Compound interest earned for (n + 1)th year

R = rate of interest applied

(2.) S.I = PRT/100

Where,

P = Principal

R = Rate of interest applied

T = Time invested

Concept used:

Simple interest and Compound interest are the same for the first year.

Calculation:

According to the question,

⇒ C.I₂ = (1 + 20%) × C.I₁

⇒ Rs. 240 = (1 + 20%) × C.I₁

⇒ Rs. 240 = 1.2 × C.I1

⇒ C.I₁ = Rs. 200

Now,

Let P be the principal invested.

Since simple interest and compound interest are the same for the first year,

Therefore,

⇒ P = Rs. 1000

Therefore, 'Rs. 1000' is the required answer.

Total mobile phones in the box are 35.

The average weight of 12 mobile phones is 50 grams.

The average weight of 11 mobile phones is 40 grams.

The average weight of the remaining mobile phones is 45 grams.

Concept:

Often "average" refers to the arithmetic mean, the sum of the numbers divided by how many numbers are being averaged.

Formula used:

Total weight = Average weight × total number of items

Average = Total sum of all numbers / Total numbers of items

Calculation:

Total mobile phones in the box are 35.

The average weight of 12 mobile phones = 50 grams.

Then, the total weight of 12 mobile phones

⇒ 12 × 50 = 600 grams

The average weight of the next 11 mobile phones is 40 grams.

Then, the total weight of 11 mobile phones

⇒ 11 × 40 = 440 grams

Remaining number of mobile phones

⇒ 35 – 12 – 11 = 12

The average weight of the remaining mobile phones is 45 grams.

Then, the total weight of the remaining mobile phones

⇒ 12 × 45 = 540 gram

Now the total weight of 35 mobile phones

⇒ 600 + 440 + 540 = 1580 grams

Now to find the average weight of 35 mobile phones

Average = Total weight / Total numbers of an item

⇒ 1580 / 35 = 45.12 grams

∴ the average weight of all mobile phones is 45 grams.

Rupam's age is 5/12 times Manju's age.

The HCF of the age of Rupam and Manju is 4.

Concept used:

The HCF of two or more than two numbers is the greatest number that divides each of them exactly.

Calculation:

Let Manju's age is 12x years

∴ Rupam's age = (5/12) × 12x years = 5x years

Now,

HCF of (12x and 5x) is 4

The greatest common factor of (12x and 5x) is x

∴ x = 4

Manju's age = 4 × 12 = 48 years

Rupam's age = 4 × 5 = 20 years

∴ The difference between Manju's age and Rupam's age = 48 -20 = 28 years

Total volume of milk in 150 liter solution

⇒ Total volume of water in the solution = 150 - 45 = 105 l

Let the quantity of milk added be "a"

According to the question-

⇒ a + 45 = 75

∴ a = 30 l

Total quantity of milk in the final solution = 45 + 30 = 75 l

The quantity of milk that should be added is 75 l.

Time taken by X to complete a work = 45 days

Time taken by Y to complete a work = 40 days

Time taken by Y to complete the remaining work = 23 days

Formula used:

Work = Efficiency × Time

Calculation:

LCM of 45 and 40 is 360

Efficiency of X = (360/45) = 8 units/day

Efficiency of Y = (360/40) = 9 units/day

Work done by Y in 23 days = (23 × 9) = 207 units

Work done by (X + Y) in = (360 − 207) = 153 units

Time is taken by X and Y to complete the 153 units

⇒ [(153/(8 + 9)] days

⇒ (153/17) = 9 days

∴ The required value of n is 9.