SSC CGL (Tier II) Practice Test - 4 - SSC CGL MCQ

# SSC CGL (Tier II) Practice Test - 4 - 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 - 4

SSC CGL (Tier II) Practice Test - 4 for SSC CGL 2024 is part of SSC CGL Tier II Mock Test Series 2024 preparation. The SSC CGL (Tier II) Practice Test - 4 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL (Tier II) Practice Test - 4 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 - 4 below.
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SSC CGL (Tier II) Practice Test - 4 - Question 1

### The area of an equilateral triangle PQR is 9 times the area of another equilateral triangle XYZ. If perimeter of ΔPQR is 18 cm, what is the length of each side (in cm) of ΔXYZ?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 1
Perimeter of ΔPQR = 18 cm

Side of ΔPQR = 6 cm

Area of equilateral ΔPQR = 9 x area of equilateral ΔXYZ

√3/4 x (6)2 = 9 x √3/4 x a2 (where a is the side of ΔXYZ)

36 = 9a2

a = 2

SSC CGL (Tier II) Practice Test - 4 - Question 2

### If f(x) = (x - 2)(x2 + Px + 4) and (x - 3) is a factor of f(x), then what is the value of P?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 2
Since x - 3 is a factor of f(x), this implies that f(3) = 0

f(x) = (x - 2)(x2 + Px + 4),

f(3) = (3 - 2)(9 + 3P + 4)

0 = 1(13 + 3P)

3P = -13

P = -13/3

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SSC CGL (Tier II) Practice Test - 4 - Question 3

### A 64 litre mixture contains water and milk in the ratio 3 : 5. How much water must be added to this mixture to make the ratio 1 : 1?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 3
We have, Water + Milk = Mixture

Original ⇒ 24 + 40 = 64 litres

According to the question

24 + ? = 40

? = 40 - 24

? = 16 litres

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

Directions: Study the given information and answer the following question.

The pie chart given below shows the expenditure on various items and savings of a family during the year 2009.

Q. If the total income of the family was Rs. 1,50,000, then the money spent on food was

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 4
Expenditure on food = Rs. = = Rs. 34,500
SSC CGL (Tier II) Practice Test - 4 - Question 5

If 2x = 3y and 2y = 3z, then x : z is equivalent to

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 5
2x = 3y and 2y = 3z

⇒ x : y = 3 : 2 and y : z = 3 : 2

Thus, x : z = = 9 : 4

SSC CGL (Tier II) Practice Test - 4 - Question 6

PA and PB are two tangents drawn to two circles of radius 3 cm and 5 cm, respectively. PA touches the smaller and larger circles at points X and Y, respectively. PB touches the smaller and large circles at points U and V, respectively. The centres of the smaller and larger circles are O and N, respectively. If ON = 12 cm, then what is the value (in cm) of PY?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 6

Angle OXP = Angle NYP (Right angles)

Angle P = Angle P (Common)

Triangles XPO and YPN are similar. (By AA similarity)

So,

3PO + 36 = 5PO

2PO = 36

PO = 18 cm

In triangle PYN, using Pythagoras theorem,

PN2 = PY2 + YN2

302 = PY2 + 52

900 - 25 = PY2

PY2 = 875

PY = 5√35 cm

SSC CGL (Tier II) Practice Test - 4 - Question 7

Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the drawn ticket contains a number, which is a multiple of 3 or 7?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 7
Favourable cases = 3, 6, 9, 12, 15, 18, 7, 14

Probability = 8/20 = 2/5

SSC CGL (Tier II) Practice Test - 4 - Question 8

sin 1° sin 2° sin 3° … sin 210° is equal to

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 8
The given product contains the factor sin 180° = 0. Thus, the value of the product becomes 0.
SSC CGL (Tier II) Practice Test - 4 - Question 9

Directions: Study the given information and answer the following question.

The pie chart given below shows the expenditure on various items and savings of a family during the year 2009.

Q. If the total income of the family for the year 2009 was Rs. 1,50,000, then the difference between the expenditures on housing and on transport was

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 9
Percentage difference between expenditures on housing and on transport = (15 - 5)% = 10%

Difference between expenditures on housing and on transport = 1,50,000 × 10/100 = Rs. 15,000

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

A man can row against the current three-fourth of a kilometre in 15 minutes and returns the same distance in 10 minutes. The ratio of his speed to that of the current is

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 10
Speed of boat in still water = x kmph

Speed of current = y kmph

From equations (i) and (ii),

SSC CGL (Tier II) Practice Test - 4 - Question 11

The ratio between the ages of A and his father is 1 : 7. The difference between their ages is 30 years. What would be the ratio of their ages 10 years later?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 11
Let the ages of A and his father be x and 7x respectively.

So, 7x - x = 30

x = 5 years

Present age of A = 5 years

Present age of A's father = 7 × 5 = 35 years

So, 10 years later, the ratio would be as follows.

(5 + 10) : (35 + 10)

= 15 : 45 = 1 : 3

SSC CGL (Tier II) Practice Test - 4 - Question 12

is equal to

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 12

SSC CGL (Tier II) Practice Test - 4 - Question 13

A car dealer sold two cars and received \$560 for each car. One of these transactions amounted to a 40% profit for the dealer, whereas the other amounted to a 20% loss. What is the dealer's net profit on the two transactions?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 13
Total SP = \$560 + 560 = \$1120

Total CP = + = \$400 + 700 = \$1100

Profit = \$1120 - 1100 = \$20

SSC CGL (Tier II) Practice Test - 4 - Question 14

Directions: Study the given information and answer the following question.

The pie chart given below shows the expenditure on various items and savings of a family during the year 2009.

Q. The percentage of income which was spent on clothing, education of children and transport together was

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 14
Required percentage = 10 + 12 + 5 = 27%
SSC CGL (Tier II) Practice Test - 4 - Question 15

A husband and a wife appear in an interview for the same post. The probability of husband's selection is 1/7 and that of wife's selection is 1/5. What is the probability that none of them will be selected?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 15

Required probability =

SSC CGL (Tier II) Practice Test - 4 - Question 16

In a triangle ABC, ∠BAC = 90o and AD is perpendicular to BC. If AD = 6 cm and BD = 4 cm, then the length of BC is

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 16

The given data can be represented diagrammatically as follows.

By Pythagoras theorem,

BC2 = AB2 + AC2

⇒ (BD + DC)2 = (AD2 + DB2) + (AD2 + DC2)

⇒ (4 + DC)2 = 62 + 42 + 62 + DC2 (∵ (a + b)2 = a2 + 2ab + b2)

⇒ 16 + 8DC + DC2 = 36 + 16 + 36 + DC2

⇒ 8DC = 72

⇒ DC = 72/8 = 9 cm

BC = BD + DC = 4 + 9 = 13 cm

∴ BC = 13 cm

SSC CGL (Tier II) Practice Test - 4 - Question 17

Refer the below data table and answer the following Question.

Q. What was the Revenue of the company if its Expenditure was Rs. 475 crore in the year when its % profit was the least?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 17
According to the given information,

The company’s profit was least in 2015 and that was 10%

⇒ 10% = Revenue/Expenditure - 1

⇒ Revenue/475 = 1 + 0.1

∴ Revenue = 475 × 1.1 = Rs. 522.5 crore

Hence (d) is correct.

SSC CGL (Tier II) Practice Test - 4 - Question 18

In a square, the length of diagonal is 16 cm. The total surface area of a hemisphere is twice the area of square. Find the volume of hemisphere?(in cubic cm) (Use π = 3.14)

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 18

Formula Used:

Total area of a hemisphere = 3π r2

Volume of a hemisphere = 2/3π r3

Calculation:

Side length of square = Diagonal length / √2 = 16/√2

Area of square = 16/√2 + 16/√2 = 128 square cm

The total surface area of a hemisphere is twice the area of square.

Let radius of hemisphere be R cm.

⇒ 3 × 3.14 × R2 = 2 × 128

⇒ 9.42 × R2 = 256

⇒ R = √256/9.42 = √27.17 = 5.21 cm

Volume of hemisphere = 2/3 × 3.14 × (5.21)3 = 296.04 cubic cm

∴ Volume of hemisphere is 296.04 cubic cm

SSC CGL (Tier II) Practice Test - 4 - Question 19

If ABC be an equilateral triangle and AD perpendicular to BC Then

AB2 + BC2 + CA2 = ?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 19

Let, the side of equilateral triangle is = a

Then AB = BC = CA = a

BD = DC = a/2

AB2 + AC2 = 2AD2 + BD2 + CD2

AB2 + AC2 + BC2

= 2AD2 + BD2 + CD2 + BC2

Put the value of side in R.H.S.

SSC CGL (Tier II) Practice Test - 4 - Question 20

Refer the below data table and answer the following Question.

Five points are to be deducted from the student average of marks scored because of poor attendance. What will be the students net average marks scored?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 20

⇒ Total marks obtained by student = 80 + 40 + 30 + 80 + 30 = 260

⇒ Average marks = 260/5 = 52

⇒ Average marks left after deduction = 52 – 5 = 47

SSC CGL (Tier II) Practice Test - 4 - Question 21

A student appeared for a sectional test containing 10 questions. A reviewer checks the answers and finds that the student answered three questions wrong. What is the probability that the sixth reviewed question is the last wrong question?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 21

The sixth question is the last wrong question

⇒ 2 errors are in the first 5 questions

Required Probability = 2 errors out of 5 × (6th wrong Q)/ 3 errors out of 10

5C2 / 10C3

∴ 1/12

SSC CGL (Tier II) Practice Test - 4 - Question 22

Aman goes upstream in a motorboat and returns back to the same point in 50 minutes. If speed of the current is 5 km/hr and total distance travelled by him is 20 km, then what is the speed of the motorboat in still water?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 22

Let speed of the boat in still water be x km/hr

Speed of current = 5 km/hr

⇒ Speed upstream = Speed of boat - Speed of current

⇒ u = x - 5

Speed downstream = v = x + 5

Distance travelled upstream and downstream = 20 km (∵ 10 km each for upstream and downstream)

⇒ Time taken to go upstream = 10/(x - 5) = t1

⇒ Time taken to go downstream = 10/(x + 5) = t2

Total time = 50 min = 50/60 hours = 5/6 hours

⇒ t1 + t2 = 5/6

⇒ (x - 25) (x + 1) = 0

∴ x = 25 ∵ x can’t be negative.

∴ Speed of boat in still water = 25 km/hr

SSC CGL (Tier II) Practice Test - 4 - Question 23

If u : v : w = 1 : 5 : 13, then what will be the value of (3u + 2v + 4w) / (2w - u - 4v)?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 23

Let u, v and w be x, 5x and 13x.

(3u + 2v + 4w) / (2w - u - 4v)

= (3x + 10x + 52x)/ (26x - x - 20x)

= 65x/5x

= 13

∴ Ans = 13

SSC CGL (Tier II) Practice Test - 4 - Question 24

The marked price of an article is Rs. 950. A shopkeeper gives a discount of 10% and still makes a profit of 80%. What is the cost price (in Rs.) of the article?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 24

Marked price of the article = 950

Discount = 10%

Discounted price = Marked price × (90/100)

Discounted price = 950 × 90/100 = 855

Still he makes 80% profit on the discounted price

So, cost price will be = 855 × 100/180 = 475

SSC CGL (Tier II) Practice Test - 4 - Question 25

If s2 – 33s + 244 = 0, then, the value of is__.

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 25

s2 – 33s + 244 = 0

Formula Used:

(x + y)2 = x2 + 2xy + y2

Calculation:

Suppose,

Taking square both sides, we have –

SSC CGL (Tier II) Practice Test - 4 - Question 26

A truck after covering 1 / 6th of the total distance decreases its speed by 2 / 3 of its usual speed and reaches destination 1 hr 40 minutes late. Find the usual time taken (in hours) by the truck to cover the complete distance?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 26

Let the total distance be 6x and the usual speed is v.

The actual time taken by the truck ⇒ (x / v) + [5x / (v / 3)] = 16x / v

The usual time the truck would have taken ⇒ 6x / v

⇒ Difference in time = 16x / v - 6x / v = 5 / 3 [1 hr 40 minutes = 5 / 3 hour]

⇒ x / v = 1 / 6

The usual time is taken by the truck to cover the complete distance = 6x / v = 6 × 1 / 6 = 1

∴ the usual time taken by the truck to cover the complete distance is 1 hour.

SSC CGL (Tier II) Practice Test - 4 - Question 27

An ore contains 25 % of an alloy that has 90 % iron. Other than this, in the remaining 75% of the ore, there is no iron. How many kilograms of the ore are needed to obtain 60 kg of pure iron?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 27

Concept:

Unitary method:

The unitary method is a method in which you find the value of a unit and then the value of a required number of units.

Calculation:

Given:

25 % of an alloy has 90 % iron, Pure iron = 60 kg

Let there be 100 kg of ore.

∵ 25 % or 25 kg of an alloy has 90 % iron

⇒ Amount of iron = 90/100 × 25 = 22.5 kg

So, 22.5 kg of iron is contained in 100 kg alloy

⇒ 1 kg of iron is contained in 100 / 22.5 kg alloy.

∴ 60 kg of iron is contained in 100 / 22.5 × 60 = 266.66 kg alloy.

SSC CGL (Tier II) Practice Test - 4 - Question 28

In a class of 45 students, 40% are girls and the remaining are boys. The average marks of the girls are 64 and that of the boys is 60. What is the average marks of the whole class?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 28

Total number of students = 45

Number of girls in class = 45 × 40/100 = 18

Number of boys in class = 45 - 18 = 27

Average marks of girls = 64

Sum of marks of girls = 64 × 18 = 1152

Average marks of boys = 60

Sum of marks of boys = 60 × 27 = 1620

Total sum of marks of all class = 1152 + 1620 = 2772

Average marks of whole class = 2772/45 = 61.6

SSC CGL (Tier II) Practice Test - 4 - Question 29

Solutions A and B contain acid and water in the ratio 4 ∶ 3 and 9 ∶ 5, respectively. Four litres of solution A is mixed with 5 litres of solution B. What is the ratio of acid and water in the resulting mixture?

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 29

Given:

Solutions A and B contain acid and water in the ratio 4 ∶ 3 and 9 ∶ 5

Calculation:

Total acid after mixing of two solution = 4 × (4/7) + 5 × (9/14)

⇒ 16/7 + 45/14

⇒ (32 + 45)/14

⇒ 77/14

⇒ 11/2

Total water after mixing of two solution = 4 × (3/7) + 5 × (5/14)

⇒ 12/7 + 25/14

⇒ (24 + 25)/14

⇒ 49/14

⇒ 7/2

Ratio = 11/2 : 7/2

⇒ 11 : 7

∴ The ratio of acid and water in the resulting mixture is 11 : 7.

SSC CGL (Tier II) Practice Test - 4 - Question 30

In ΔABC, O is any point inside the triangle. If OA = 12 cm, OC = 15 cm, ∠AOB = ∠BOC = ∠COA and also ∠BAC = 60°. Find OB.

Detailed Solution for SSC CGL (Tier II) Practice Test - 4 - Question 30

Given:

ΔABC,

OA = 12 cm, OC = 15 cm

∠AOB = ∠BOC = ∠COA

∠BAC = 60°

Concept used:

Concept of similarity

Calculation:

If ∠AOB = ∠BOC = ∠COA., then all the angles are 120° [Complete angle = 360°]

In ΔOAC

Let ∠OAC be θ

So, ∠OAB = 60° - θ [∠OAC + ∠OAB = 60°]

∠OCA = 60° - θ [sum of angles of triangle is 180°]

In ΔAOB

∠OBA = θ [sum of angles of triangle is 180°]

From the above calculation, we can say

ΔOAB ∼ ΔOCA [AAA Similarity]

So,

OA/OC = OB/OA

⇒ 12/15 = OB/12

⇒ OB = 122 ÷ 15

⇒ OB = 144 ÷ 15

⇒ OB = 9.6

∴ The length of OB is 9.6 cm.

## SSC CGL Tier II Mock Test Series 2024

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## SSC CGL Tier II Mock Test Series 2024

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