SSC CGL (Tier II) Practice Test - 9 - SSC CGL MCQ

# SSC CGL (Tier II) Practice Test - 9 - SSC CGL MCQ

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## 30 Questions MCQ Test SSC CGL Tier II Mock Test Series 2024 - SSC CGL (Tier II) Practice Test - 9

SSC CGL (Tier II) Practice Test - 9 for SSC CGL 2024 is part of SSC CGL Tier II Mock Test Series 2024 preparation. The SSC CGL (Tier II) Practice Test - 9 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL (Tier II) Practice Test - 9 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 - 9 below.
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SSC CGL (Tier II) Practice Test - 9 - Question 1

### What is the sum of ... up to 20 terms?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 1
Given expression: up to 20 terms

= (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.

SSC CGL (Tier II) Practice Test - 9 - Question 2

### An equilateral triangle of side 12 cm is drawn. What is the area (in cm2) of the largest square which can be drawn inside it?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 2

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

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SSC CGL (Tier II) Practice Test - 9 - Question 3

### Directions: Study the graph carefully and answer the following question.Q. In 2008, If the firm faces extra expense of the value Rs. 50 lakh, and earns Rs. 45 lakh more, then the approximate % change in the savings of the firm during the given years is

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 3
Initial income of the firm = 4360 lakh

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.

SSC CGL (Tier II) Practice Test - 9 - Question 4

A flag of height 4 metres is standing on the top of a building. The angle of elevation of the top of the flag from a point X is 45° and the angle of elevation of the top of building from X is 30°. Point X is on the ground level. What is the height (in metres) of the building?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 4

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

SSC CGL (Tier II) Practice Test - 9 - Question 5

Each interior angle of a regular polygon is 120° greater than each exterior angle. How many sides are there in the polygon?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 5
Exterior angle = x

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

SSC CGL (Tier II) Practice Test - 9 - Question 6

Arjun invested Rs. 2400 and Bhaskar invested Rs. 3600 in a startup. After the completion of four successful months of the startup, Keshav also invested Rs. X in this startup. After the completion of eight successful months of the startup, Bhaskar withdrew Rs. 600. When annual profit was calculated, it came out to be Rs. 22,500 and Keshav's share in this profit was Rs. 8000. What is the value of X?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 6
Investment made by Arjun = 2400 × 12 = 28,800

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

SSC CGL (Tier II) Practice Test - 9 - Question 7

Which of the following values of θ satisfies the given equation?

= 4, where θ ∈ (0, π)

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 7

⇒ 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°

SSC CGL (Tier II) Practice Test - 9 - Question 8

Directions: In the following number series, one number is wrong. Find out the wrong number.

5, 8, 15, 30, 60, 124

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 8
5

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.

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

A group of 12 men can complete a piece of work in 9 days. After 3 days, 6 men joined to replace 2 men. How many days would be required to complete the remaining work?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 9
Let the work done by one man in 1 day be ω.

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 =

SSC CGL (Tier II) Practice Test - 9 - Question 10

The digit in units place in the product 771 × 659 × 365 is

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 10
71 = 7

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

SSC CGL (Tier II) Practice Test - 9 - Question 11

A photocopy machine copies 1,500 workbooks in 8 hours, while another machine takes 12 hours to do the same job. What is the total number of hours that both the machines working together will take to complete the 1,500 workbooks?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 11
In 1 hour, the first machine will copy 1500/8 = 375/2 workbooks.

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.

SSC CGL (Tier II) Practice Test - 9 - Question 12

Ramesh travels by bus from city A to city B at an average speed of 44 km/hr. Suresh travels by taxi from city A to city B at an average speed of 77 km/hr and takes 3 hours less than the time taken by Ramesh. What is the distance (in km) between the two cities?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 12
Let the time taken by Ramesh be T hours and the distance to be travelled be D km.

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

SSC CGL (Tier II) Practice Test - 9 - Question 13

Directions: In the following number series one number is wrong. Find out the wrong number.

160, 161, 170, 195, 244, 324, 446

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 13
160, 161, 170, 195, 244, 324, 446

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.

SSC CGL (Tier II) Practice Test - 9 - Question 14

A person covered one-fourth of a distance at the speed of 2.5 kmph. He covered the remaining distance at the speed of 5 kmph. If he took half-an-hour for the entire journey, what distance did he travel?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 14
Let the total distance travelled be 4d km.

Now,

Or d = 0.5

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

SSC CGL (Tier II) Practice Test - 9 - Question 15

A man sells an article at a gain of 10%. If he had bought it at 10% less and sold it for Rs. 132 less, he would have still gained 10%. The cost price of the article is

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 15
SP = 1.1 CP

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

SSC CGL (Tier II) Practice Test - 9 - Question 16

If cot x = 4/3 then find the value of

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 16
Formula used:
• 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

SSC CGL (Tier II) Practice Test - 9 - Question 17

A person goes from A to B with speed 40 km/hr & return from B to A with speed 30 km/hr. Whole journey takes 14 hr, then find the distance between A & B in Km.

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 17
Given

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

SSC CGL (Tier II) Practice Test - 9 - Question 18

Rajib, Barun, and Anil engaged in work together. it was agreed that 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. Calculate the share of Anil in the profit.

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 18
Given

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

SSC CGL (Tier II) Practice Test - 9 - Question 19

In the figure given below, ∠CBE = ∠AFC, AF = 12 cm, CB = 6 cm, BE = 12 cm and CF = 6 cm. If AC = 2 cm, then find the ratio of CE : AE.

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 19
Given

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.

SSC CGL (Tier II) Practice Test - 9 - Question 20

The total cost of fencing the right-angled triangular park is Rs. 102,600. If the length of the smallest side is 20 m and the length of the second longest side is 28 m longer than the smallest side, then find the rate at which fencing is done per m. (in Rs per m)

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 20
Given

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.

SSC CGL (Tier II) Practice Test - 9 - Question 21

A student of a school has to score 25% of marks to get throughout. If he gets 50 marks and fails by 25 marks, then find the maximum marks set for the examination?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 21
Given

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.

SSC CGL (Tier II) Practice Test - 9 - Question 22

The perpendicular from the center O of the circle bisects the chord EF. If EF = 30 cm and the radius of the circle is 17 cm, then find the length of the perpendicular OG.

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 22
Given

EF = 30 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.

SSC CGL (Tier II) Practice Test - 9 - Question 23

The pie chart represents an expenditure of a family in a month. If the difference between the expenditure on marketing and Transportation is Rs 2400. Then find the expenditure on transport, Electric Bill, and others together.

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 23
Given

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

SSC CGL (Tier II) Practice Test - 9 - Question 24

If 2b = a + c and y2 = zx then find (xb - c) (yc - a) (za - b)

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 24
Calculation

We have,

2b = a + c ⇒ a - b = b - c

Now, xb - c yc - a zb - c

⇒ (xz)b - c yc - a

⇒ y2(b - c) yc - a [ ∵ y2 = zx]

⇒ y2b - 2c + c - a = y2b - (a + c)

⇒ y2b - 2b = y0 = 1

SSC CGL (Tier II) Practice Test - 9 - Question 25

The perimeters of Δ ABC and Δ DEF are 36 cm and 54 cm respectively. AB = 12 cm. If both are similar triangles, find side DE.

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 25
Given:

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.

SSC CGL (Tier II) Practice Test - 9 - Question 26

If the compound interest earned for the 2nd year is Rs. 240 and the rate of interest applied is 20% per annum, then find the principal invested.

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 26
Given

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

R = 20% p.a.

Formula used:

(1.) C.In+1 = (1 + R%) × C.In

Where,

C.In = Compound interest earned for nth year

C.In+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.

SSC CGL (Tier II) Practice Test - 9 - Question 27

There are 35 mobile phones in a box, out of which the average weight of 12 mobile phones is 50 grams and the average weight of 11 mobile phones is 40 grams and the average weight of the remaining mobile phones is 45 grams, then find the average weight of all mobile phones. (approx.)

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 27
Given:

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.

SSC CGL (Tier II) Practice Test - 9 - Question 28

Rupam's age is 5/12 times Manju's age, and the HCF of the age of Rupam and Manju is 4. Find the difference between Manju's age and Rupam's age.

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 28
Given

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

SSC CGL (Tier II) Practice Test - 9 - Question 29

A certain amount of milk is added to the 150 liters of 3 : 7 milk-water solution. Find the quantity of the milk present in the final mixture, if the ratio of the milk and water in the final solution is 5 : 7?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 29
Calculation

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.

SSC CGL (Tier II) Practice Test - 9 - Question 30

X and Y can do a piece of work in 45 days and 40 days respectively. They begin to work together, but X leaves after n days and then Y completes the remaining work in 23 days. What is n equal to ?

Detailed Solution for SSC CGL (Tier II) Practice Test - 9 - Question 30
Given

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.

## SSC CGL Tier II Mock Test Series 2024

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## SSC CGL Tier II Mock Test Series 2024

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