SSC CGL (Tier II) Practice Test - 10 - SSC CGL MCQ

# SSC CGL (Tier II) Practice Test - 10 - 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 - 10

SSC CGL (Tier II) Practice Test - 10 for SSC CGL 2024 is part of SSC CGL Tier II Mock Test Series 2024 preparation. The SSC CGL (Tier II) Practice Test - 10 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL (Tier II) Practice Test - 10 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 - 10 below.
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SSC CGL (Tier II) Practice Test - 10 - Question 1

### How many two-digit prime numbers are there between 10 and 100 which remain prime numbers when the order of their digits is reversed?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 1
Prime numbers from 10 to 100 that remain prime even after their digits are reversed are 11, 13, 17, 31, 37, 71, 73, 79 and 97.

They are 9 in total.

SSC CGL (Tier II) Practice Test - 10 - Question 2

### A taxi goes from City A to City B at an average speed of 84 km/hr. In the return journey, due to traffic the average speed of the taxi falls by 24 km/hr. Find the average speed of the taxi (in km/hr) for the total journey.

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 2
Speed of taxi from City A to City B = 84 km/hr

Speed of taxi from City B to City A = 84 - 24 = 60 km/hr

We know that if the distance is same, then the average speed is the harmonic mean of the given speeds.

So,

Required average speed

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SSC CGL (Tier II) Practice Test - 10 - Question 3

### Five friends purchased some books from a book shop. Four of them purchased books worth \$30 each. If the fifth one purchased books worth \$40 more than the average of all of them together, what was the total cost of the books purchased by all?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 3
Let the average of purchase of books for all 5 friends be \$X.

Therefore, total purchases made = \$5X

Now, (30 × 4) + (X + 40) × 1 = 5X (Given)

Therefore, 160 = 4X

⇒ X = 40

Thus, total cost of the books purchased = \$200.

SSC CGL (Tier II) Practice Test - 10 - Question 4

A grain trader has 100 bags of rice. He sold some bags at 10% profit and rest at 20% profit. His overall profit on selling these 100 bags was 14%. How many bags did he sell at 20% profit?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 4
Let the number of bags sold at the 10% profit be x.

Number of bags sold at 20% = 100 - x

According to the question:

1.10x + 1.20(100 - x) = 114

1.10x + 120 - 1.20x = 114

120 - 114 = 1.20 - 1.10x

6 = 0.1x

x = 60

Number of bags sold at 20% profit = 100 - 60 = 40

SSC CGL (Tier II) Practice Test - 10 - Question 5

Two dice are thrown at the same time and the product of numbers appearing on them is noted. The probability that the product is less than 9 will be

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 5
Total number of outcomes = 36

Favourable events = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (6, 1) i.e. 16 outcomes

Number of favourable events = 16

Required probability = 16/36 = 4/9

Hence, answer option 2 is correct.

SSC CGL (Tier II) Practice Test - 10 - Question 6

The ratio of the bank balances of three brothers A, B and C is 10 : 12 : 5, respectively. B transfers Rs. 60,000 from his account to C's. The new ratio of the bank balances becomes 10 : 9 : 8. What is the bank balance of A (in Rs.)?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 6
Suppose A, B and C have Rs. 10x, Rs. 12x and Rs. 5x, respectively.

When B made the transfer, B is left with 12x - 60,000; and C is left with 5x + 60,000.

So,

A : B : C = 10x : 12x - 60000 : 5x + 60,000 = 10 : 9 : 8

90x = 120x - 6,00,000

30x = 6,00,000

x = 20,000

So, bank balance of A = 10x = 10 × 20,000 = Rs. 2,00,000

SSC CGL (Tier II) Practice Test - 10 - Question 7

If 3x + 6y + 9z = 20/3 , 6x + 9y + 3z = 17/3 and 18x + 27y - z = 11/9 , then what is the value of 75x + 113y?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 7
The given equations are

3x + 6y + 9z = 20/3

Or 9x + 18y + 27z = 20 ... (i)

6x + 9y + 3z = 17/3

Or 18x + 27y + 9z = 17 ... (ii)

And 18x + 27y - z = 113/9

Or 162x + 243y - 9z = 113 ... (iii)

Adding equations (ii) and (iii), we get

180x + 270y = 130

Or 18x + 27y = 13 ... (iv)

Multiplying equation (ii) with 3, we get

54x + 81y + 27z = 51 ... (v)

Now, subtracting equation (i) from equation (v), we get

45x + 63y = 31 ... (vi)

On solving equations (iv) and (vi), we get

x = 2/9 and y = ⅓

Thus, 75x + 113y = 75(2/9) + 113(1/3) = 163/3

Hence, answer option (1) is correct.

SSC CGL (Tier II) Practice Test - 10 - Question 8

A water tap fills a tub in 'p' hours and a sink at the bottom empties it in 'q' hours. If p < q and both tap and sink are open, the tank is filled in 'r' hours; then

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 8
We have, 1/p work is done by tap in unit time.

work is done by sink in unit time (to empty the tank).

1/r is total work done by pipe and sink together.

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

The square root of

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

[∵ (a – b)(a2 + ab + b2) = a3 – b3]

1/0.25 = 100/25

= 4

So, square root of 4 is 2.

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

In a triangle PQR, PX bisects QR. PX is the angle bisector of angle P. If PQ = 12 cm and QX = 3 cm, then what is the area (in cm2) of triangle PQR?

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

In triangles PQX and PRX,

∠1 = ∠2

PX = PX (Common)

QX = XR (X is the mid-point)

So, triangles PQX and PRX are similar.

Hence, PR = 12 cm

Now, let PQ = a = 12 cm, PR = b = 12 cm and QR = c = 6 cm

By Heron's formula:

Area of a triangle

SSC CGL (Tier II) Practice Test - 10 - Question 11

The arithmetic mean of the following numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6 and 7, 7, 7, 7, 7, 7, 7 is

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 11
Total number of elements = 28

Sum of all numbers = 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 + 7 × 7 = 1 + 4 + 9 + 16 + 25 + 36 + 49 = 140

Arithmetic mean = 140/28 = 5

SSC CGL (Tier II) Practice Test - 10 - Question 12

Two trees are standing along the opposite sides of a road. The distance between the two trees is 400 metres. There is a point on the road between the trees. The angles of depression of the point from the top of the trees are 45° and 60°. If the height of the tree which makes 45° angle is 200 metres, then what will be the height (in metres) of the other tree?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 12

AB = 200 metres

Angle AEB = 45°

CD = second tree = h metres

Angle CED = 60°

E = point on the road

In triangle ABE,

tan 45° = AB/BE

⇒ 1 = 200/BE metres

BE = 200 metres

DE = 200 metres

In triangle CDE,

tan 60° = CD/DE

h = 200√3

Hence, answer option (b) is correct.

SSC CGL (Tier II) Practice Test - 10 - Question 13

Directions: Study the following information to answer the question.

The table given below shows the ratio of cars and bikes manufactured by five different companies. The table also shows the ratios of three different types of cars C1, C2 and C3 and three different types of bikes B1, B2 and B3 manufactured by these five different companies. The total numbers of cars and bikes together manufactured by D, E, F, G and H are 3,00,000, 2,80,000, 3,20,000, 4,00,000 and 4,80,000, respectively.

Total number of bikes manufactured by company D is what percentage of the total number of cars of type C1 manufacture by company G?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 13
Total number of bikes manufactured by company D = 2/3 of 3,00,000 = 2,00,000

Number of cars manufactured by company G = 4,00,000 × 3/4 = 3,00,000

Total number of cars of type C1 manufactured by company G = 2/6 of 3,00,000 = 1,00,000

SSC CGL (Tier II) Practice Test - 10 - Question 14

Directions: Study the following information to answer the question.

The table given below shows the ratio of cars and bikes manufactured by five different companies. The table also shows the ratios of three different types of cars C1, C2 and C3 and three different types of bikes B1, B2 and B3 manufactured by these five different companies. The total numbers of cars and bikes together manufactured by D, E, F, G and H are 3,00,000, 2,80,000, 3,20,000, 4,00,000 and 4,80,000, respectively.

What is the difference between the total number of cars of type C3 manufactured by companies E and G together and the number of bikes of type B1 manufactured by company H?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 14
Number of cars manufactured by company E = 2,80,000 × 3/4 = 2,10,000

Number of cars of type C3 manufactured by company E = 1/3 of 2,10,000 = 70,000

Number of cars manufactured by company G = 4,00,000 × 3/4 = 3,00,000

Number of cars of type C3 manufactured by company G = 1/6 of 3,00,000 = 50,000

Together = 70,000 + 50,000 = 1,20,000

Number of bikes manufactured by company H = 4,80,000 × 2/3 = 3,20,000

Number of bikes of type B1 manufactured by company H = 2/8 of 3,20,000 = 80,000

Required difference = 1,20,000 - 80,000 = 40,000

SSC CGL (Tier II) Practice Test - 10 - Question 15

Directions: Study the following information to answer the question.

The table given below shows the ratio of cars and bikes manufactured by five different companies. The table also shows the ratios of three different types of cars C1, C2 and C3 and three different types of bikes B1, B2 and B3 manufactured by these five different companies. The total numbers of cars and bikes together manufactured by D, E, F, G and H are 3,00,000, 2,80,000, 3,20,000, 4,00,000 and 4,80,000, respectively.

A = Total number of cars manufactured by all the companies

K = Difference between the number of C3 type cars manufactured by company H and the number of B3 type bike manufactured by company E

What is the value of A : K?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 15
A = Total number of cars manufactured by all the companies.

Total cars by D = (3,00,000 × 1/3) = 1,00,000

E = (2,80,000 × 3/4) = 2,10,000

F = (3,20,000 × 1/2) = 1,60,000

G = (4,00,000 × 3/4) = 3,00,000

H = (4,80,000 × 1/3) = 1,60,000

A = 1,00,000 + 210,000 + 160,000 + 3,00,000 + 1,60,000 = 9,30,000

K = Difference between the number of C3 type cars manufactured by company H and the number of B3 type bike manufactured by company E

Total C3 cars by H = 1/4 of 1,60,000 = 40,000

Total bikes manufactured by company E = 2,80,000 × 1/4 = 70,000

Number of B3 type bikes manufactured by company E = 2/7 of 70,000 = 20,000

Required difference (K) = 40,000 - 20 ,000 = 20,000

So, A : K = 9,30,000 : 20,000

= 93 : 2

SSC CGL (Tier II) Practice Test - 10 - Question 16

Find the value of sin4 30° + cos4 30° - sin 25° cos 65° - sin 65° cos25°.

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 16

Given:

sin4 30° + cos4 30° - sin 25° cos 65° - sin 65° cos 25°

Concept used:

sin(90° - θ) = cosθ

cos(90° - θ) = sinθ

sin2θ + cos2θ = 1

Calculation:

sin4 30° + cos4 30° - sin 25° cos 65° - sin 65° cos 25°

⇒ (1/2)4 + (√3/2)4 - sin 25° sin 25° - cos 25° cos 25°

⇒ 1/16 + 9/16 - sin2 25° - cos225°

⇒ 1/16 + 9/16 - 1(sin2 25° + cos225°)

⇒ 10/16 - 1

⇒ (10 - 16)/16

⇒ - 6/16

⇒ - 3/8

∴ Required answer is - 3/8

SSC CGL (Tier II) Practice Test - 10 - Question 17

A is driving a car along the highway at a speed of 90 km/hr. B, driving a car at a constant speed on the same highway and in the same direction, crossed it. After 12 seconds, B has crossed A, and the distance between their cars is 100 meters. The speed (in km/h) of B's car is:

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 17

Given:

A is driving a car along the highway at a speed of 90 km/hr.

B, driving a car at a constant speed in the same direction.

After 12 seconds, B has crossed A, and the distance between their cars is 100 meters.

Formula used:

Convertion of kmph to mps = 1 × 5/18

Speed = distance / time

Calculations:

(x - 90) × 5/18 = 100/12 (where, x is the constant speed of B)

⇒ (x - 90) × 1/6 = 5

⇒ (x - 90) = 30

⇒ x = 90 + 30 = 120

∴ The answer is 120 kmph

SSC CGL (Tier II) Practice Test - 10 - Question 18

A thief is chased by police and caught. The thief was 40 m ahead of the police and the speeds of the thief and the police are 10 m/s and 15 m/s, respectively. Find the total distance travelled by police.

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 18

Given:

Distance between thief and police = 40 m

Speed of thief = 10 m/s

Speed of police = 15 m/s

Formula used:

Speed = Distance/Time

Calculations:

Let Police will catch the thief after thief has run further x m.

Distance travelled by the police = (40 + x) m

Time taken by police to travel (40 + x) m,

⇒ (40 + x)/15 sec

Time taken by thief to travel x m,

⇒ x/10 sec

As, time taken by both them will be same,

(40 + x)/15 = x/10

⇒ 400 + 10x = 15x

⇒ 5x = 400

⇒ x = 80 m

Total distance travelled by police = 40 + 80

⇒ 120 m

∴ Total distance travelled by police is 120 m

SSC CGL (Tier II) Practice Test - 10 - Question 19

A can do a piece of work in 82 days, B can do in 123 days and C can do the same work alone in 164 days. If on the first day A worked alone and on the second day B and C worked together and on the third day A and C worked together. If they repeat the cycle, then in how many days total work can be completed?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 19

Given :

Time taken by A to complete a work = 82 days

Time taken by B to complete a work = 123 days

Time taken by C to complete a work = 164 days

Concept used :

Total work = efficiency x Time taken

Calculation :

Let the total taken be 492 units

The efficiency of A = 492 / 82 = 6 units

The efficiency of B = 492 / 123 = 4 units

The efficiency of C = 492 / 164 = 3 units

Total work of A on the first day = 6 units

Total work of B and C on the second day = 7 units

Total work of A and C on the third day = 9 units

Total work in three days = 22 units

Total work in 67 days = 22 x 22 + 6 = 490 units

Work remaining for 68th day = 2 units

B and C together work on the 68th day

So, The total work is completed in 67(2/7)

SSC CGL (Tier II) Practice Test - 10 - Question 20

A milkman has two types of milk of equal quantity. He sells one at Rs.61/ltr. and other at Rs.57/ltr. After selling ½ of both the milk i.e. 33ltrs of each type of milk he experienced loss of 60%. Now he mix both type of milk and sell the mixture at Rs.P/ltr. Find the value of P if he wishes to cover all his losses.

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 20

Given:

Sold 33 ltrs out of 66 ltrs.

Calculation:

Total SP = 33(61 + 57) = 3894

Till now loss incurred is 60%,

⇒ SP/CP = 2/5

ATQ, 2/5 = 3894/CP

⇒ Total CP = 9735

Sum of CP of 1 ltr of 1st and 2nd milk is

⇒ 33(x(CP of unit price of 1st) + y(CP of unit of 2nd)) = 9735

⇒ x + y = 9735/33 = 295.

Now loss incurred in Rs. is = 9735 – 3894 = 5841

So now he wishes recover 5841 during selling of remaining 33 ltrs of milk.

⇒ Recover rate per ltr = 5841/66 = 88.5

The value of P = Recovery rate/ltr + Sum of CP of both milk/unit

⇒ P = 295 + 88.5 = 383.5

∴ The value of P is Rs.383.5

SSC CGL (Tier II) Practice Test - 10 - Question 21

The expenditures (in thousands) of two companies (A and B) on various heads in a given year are provided in the following bar graph.

Q. The expenditure made by both companies together on salary was approximately what percentage of their expenses on Infrastructure?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 21

Calculation:

The expenditure made by both companies together on salary = (5117 + 4752)

⇒ 9869

The expenses on Infrastructure made by both companies together = (7310 + 5598)

⇒ 12908

The required percentage = (9869/12908 × 100)

⇒ 76.45% ~ 76%

∴ The required percentage is 76%

SSC CGL (Tier II) Practice Test - 10 - Question 22

Two circles of radius 13 cm and 15 cm intersect each other at points A and B. If the length of the common chord is 24 cm, then what is the distance between their centres?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 22

Given:

Radii of circles are 13 cm and 15 cm

Length of common chord = 24 cm

We have to find the length of the PQ

Concept Used:

Perpendicular from the center on the chord bisects the chord

Calculation:

Let M be the midpoint of AB

Now, In ΔAPM and ΔBPM

AP = BP [ Radius of 1st circle]

PM = PM [Common]

AM = AM [M is the mid point]

So, ΔAPM ≅ ΔBPM

So, ∠AMP = ∠BMP = 90°

Similarly In ΔAQM ≅ ΔBQM

So, ∠AMQ = ∠BMQ = 90°

Now, We can say that ΔAPM and ΔAQM are right-angle triangles and also PQ is a straight line [Since ∠AMP + ∠AMQ = 180°]

According to the concept used

AM = BM = AB/2 = 24/2 = 12 cm

Now, In ΔAPM

PM2 = AP2 - AM2

⇒ PM2 = 132 - 122 = 25

⇒ PM = 5

Now, In ΔAQM

QM2 = AQ2 - QM2

⇒ QM2 = 152 - 122 = 81

⇒ QM = 9

So, PQ = PM + MQ = 5 + 9 = 14 cm

∴ The distance between the centers is 14 cm.

SSC CGL (Tier II) Practice Test - 10 - Question 23

The material of a sphere of radius r is melted and recast into a hollow cylindrical shell of thickness a and outer radius b. What is its length assuming that no material is lost in recasting?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 23

Given:

Thickness = a

The radius of sphere = r

Concept used:

The volume of a sphere = (4/3)πr3

The volume of a hollow cylinder= π(R2 - r2)h

Calculation:

Let required length be h

Internal radius = (b - a)

According to the question, we have

(4/3)πr3 = π[b2 - (b - a)2]h

⇒ (4/3)πr3 = π[b2 - b2 - a2 + 2ab]h

⇒ (4/3)r3 = [2ab - a2]h

∴ The required length is

SSC CGL (Tier II) Practice Test - 10 - Question 24

A trader allows a discount of 10% on the marked price of an article and thus gains 17% on the cost price of the article. If the cost price is increased by 10%, then what percentage discount should he allow on the marked price of the article so as to get the same percentage profit as before?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 24

Given:

Discount of 10% on the marked price.

Gains 17% on the cost price.

Again, cost price is increased by 10%.

Formula used:

Profit% = SP − CP / CP × 100

SP = MP - discount% on SP

Discount% = MP − SP / MP × 100

Let CP of the article be Rs. 100

SP of the article = 100 × 117 / 100 = 117

MP of the article = 117 × 100 / 90 = 130

New CP of the article = 100 × 110 / 100 = 110

New SP of the article = 110 × 117 / 100 = 128.7

Discount percentage = 130 − 128.7 / 130 × 100

⇒ 1.3 / 130 × 100 = 1%

SSC CGL (Tier II) Practice Test - 10 - Question 25

A cube of the maximum possible size is cut from a hemisphere of radius (3√3)/2 cm. One identical cubes is placed next to this cube to form a cuboid. Find the lateral surface area (in cm2) of the cuboid?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 25

Given:

The radius of the hemisphere = (3√3)/2 cm

Concept used:

Lateral Surface area = 2(l + b) × h

If we cut a maximum possible size of a cube from a hemisphere then its side = (√2/√3) × r

Here r = radius of the hemisphere,

l = length of the cuboid,

b = breadth of the cuboid

And h = height of the cuboid

Calculation:

According to the concept,

Side of the cube = (√2/√3) × (3√3)/2

⇒ Side of the cube = 3/√2

Now, one identical cubes are placed next to this cube

So, The length of the form cuboid = (3/√2) × 2

⇒ 3√2

The breadth of the cuboid = 3/√2

The height of the cuboid = 3/√2

Now,

L.S.A = 2 (3√2 + 3/√2) × 3/√2

⇒ L.S.A = 2 [(6 + 3)/√2] × 3/√2

⇒ L.S.A = 2 [9/√2] × 3/√2

⇒ L.S.A = 2 × 27/2 = 27 cm2

∴ The lateral surface area (in cm2 ) of the cuboid 27.

SSC CGL (Tier II) Practice Test - 10 - Question 26

7/18 of the balls in a drum were red, 8/15 of the balls in the drum were blue and the rest were yellow. If there were 84 more red balls in that drum than there were yellow balls, what was the total number of balls in the drum?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 26

Given:

7/18 of the balls in the drum were red

8/15 of the balls in the drum were blue

Concept:

No. of red balls = No. of yellow balls + 84

Calculations:

Let the number of the balls be X .

Now, No. of red balls = (7/18X)

And, No. of blue balls = (8/15X)

Therefore, No. of yellow balls

According to question,

No. of red balls = No. of yellow balls + 84

X = 270

Hence, total no. of balls = X = 270.

SSC CGL (Tier II) Practice Test - 10 - Question 27

The ratios of alcohol to water in solutions A and B are 3 ∶ 5 and 9 ∶ 7, respectively. A and B are mixed in the ratio 5 ∶ 8. In 520 ml of the resulting solution, how much water (in ml) should be mixed so as to obtain a solution in which the ratio of alcohol to water is 3 ∶ 4?

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 27

Given:

The ratios of alcohol to water in solutions A and B are 3 ∶ 5 and 9 ∶ 7, respectively.

A and B are mixed in the ratio 5 ∶ 8.

Calculation:

In 520 ml of the new solution:

Solution A = 520 × 5/13

⇒ 200 ml

Solution B = 520 × 8/13

⇒ 320 ml

Now,

Portion of alcohol in solution A = 200 ml × 3/8

⇒ 75 ml

Portion of alcohol in solution B = 320 ml × 9/16

⇒ 180 ml

Total alcohol = 75 + 180

⇒ 255 ml

Portion of water in solution A = 200 ml × 5/8

⇒ 125 ml

Portion of water in solution B = 320 ml × 7/16

⇒ 140 ml

Total water = 125 + 140

⇒ 265 ml

Now,

In the new solution amount of alcohol will be same,

So, new amount of water = 255 × 4/3

⇒ 340 ml

Extra water = 340 - 265

⇒ 75 ml

∴ Required answer is 75 ml.

SSC CGL (Tier II) Practice Test - 10 - Question 28

The mean of a distribution is 24 and the standard deviation is 6. What is the value of variance coefficient?

A. 50%

B. 25%

C. 100%

D. 75%

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 28

Given:

Mean of the distribution = 24

Standard deviation = 6

Concept used:

Coefficient of variance = standard deviation/mean × 100

Calculation:

Coefficient of variance = 6/24 × 100 = 100/4

Coefficient of variance = 25%

∴ The coefficient of variance is 25%.

SSC CGL (Tier II) Practice Test - 10 - Question 29

Choosing two different numbers from first three natural numbers, the probability of one of the numbers to be maximum of three is

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 29

Given :

First three natural numbers

Formula used :

P = (Required numbers outcomes)/(total numbers of outcomes)

Solution :

The first three natural numbers are 1, 2, 3

Choosing two different numbers are (1, 2), (2, 3), (1, 3)

⇒ Total number of outcomes = 3

One of the numbers be maximum of three are (2, 3), (1, 3)

⇒ Required numbers of outcomes = 2

⇒ P (numbers to be maximum of three) = 2/3

∴ the probability of one of the numbers to be maximum of three is 2/3 .

SSC CGL (Tier II) Practice Test - 10 - Question 30

Priyan lent a certain sum of money at a 20% p.a. rate of interest compounded quarterly. If the lender pays three equal quarterly installments of Rs. 1852.2 then find the money lent by Priyan.

Detailed Solution for SSC CGL (Tier II) Practice Test - 10 - Question 30

Given:

X = 1852.2

R = 20% p.a.

R = 5% per quarter

Formula used:

Let installment be X

P = (X/ (1 + (R/ 100)) + (X/ (1 + (R/ 100)2) + (X/ (1 + (R/ 100)3)

Calculation:

P = (1852.2/ (1 + (5/ 100)) + (1852.2/ (1 + (5/ 100)2) + (1852.2/ (1 + (5/ 100)3)

⇒ P = 1852.2 × 20/ 21 × (1 + (20/ 21) + (400/ 441))

⇒ P = 1852.2 × 20/ 21 × (1261/ 441)

⇒ P = 1852.2 × 20 × 1261/ 9261

⇒ P = 5044

∴ sum lent is Rs 5044

## SSC CGL Tier II Mock Test Series 2024

66 docs|110 tests
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## SSC CGL Tier II Mock Test Series 2024

66 docs|110 tests