SSC CGL (Tier II) Practice Test - 5 - SSC CGL MCQ

# SSC CGL (Tier II) Practice Test - 5 - SSC CGL MCQ

Test Description

## 30 Questions MCQ Test SSC CGL Tier II Mock Test Series 2024 - SSC CGL (Tier II) Practice Test - 5

SSC CGL (Tier II) Practice Test - 5 for SSC CGL 2024 is part of SSC CGL Tier II Mock Test Series 2024 preparation. The SSC CGL (Tier II) Practice Test - 5 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL (Tier II) Practice Test - 5 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 - 5 below.
Solutions of SSC CGL (Tier II) Practice Test - 5 questions in English are available as part of our SSC CGL Tier II Mock Test Series 2024 for SSC CGL & SSC CGL (Tier II) Practice Test - 5 solutions in Hindi for SSC CGL Tier II Mock Test Series 2024 course. Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free. Attempt SSC CGL (Tier II) Practice Test - 5 | 150 questions in 150 minutes | Mock test for SSC CGL preparation | Free important questions MCQ to study SSC CGL Tier II Mock Test Series 2024 for SSC CGL Exam | Download free PDF with solutions
SSC CGL (Tier II) Practice Test - 5 - Question 1

### If A = 0.abcabc ____, then by which of the following numbers can A be multiplied so as to get an integral value?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 1
A = 0.abcabc ... (1)

1000A = abc + 0.abc... = abc + A ... (2)

Equation (2) - Equation (1) gives:

999A = abc

A = abc/999

Now, 2997A = 3 × abc

And, 1998A = 2 × abc

Thus, A can be multiplied by both 1998 and 2997 to get an integral value.

Hence, answer option 4 is correct.

SSC CGL (Tier II) Practice Test - 5 - Question 2

### A prism has a regular hexagonal base, whose side is 12 cm. The height of the prism is 24 cm. It is cut into 4 equal parts by 2 perpendicular cuts, as shown in the figure. What is the sum of the total surface area of the four parts?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 2

Here is a 3D image of hexagonal prism, side 12 cm. If a prism is cut by two cuts MN and BE, there are 4 parts.

The surface area of one part is shown below,

Total surface area of 1 part

= 720 + 252√3

Total surface area of 4 parts

 1 Crore+ students have signed up on EduRev. Have you?
SSC CGL (Tier II) Practice Test - 5 - Question 3

### The wages of three labourers A, B and C are in the ratio 10 : 12 : 15, respectively. A's wage is increased in the ratio 5 : 6, B's wage is increased in the ratio 3 : 4 and C's wage is increased in the ratio 3 : 5. The new ratio of the wages of A, B and C respectively is

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 3
Let the wage of A be 10x.

Wage after increment

Let the wage of B be 12x.

Wage after increment

Let the wage of C be 15x.

Wage after increment

Ratio of the wages of A, B and C after increment = 12 : 16 : 25

SSC CGL (Tier II) Practice Test - 5 - Question 4

In a certain store, the profit is 250% of the cost. If the cost increases by 25%, but the selling price remains constant, then approximately what percentage of selling price is the profit?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 4
Let CP be Rs. 100.

Profit = Rs. 250

SP = Rs. 350

New CP = Rs. 125

New SP = Rs. 350

Profit = SP – CP = Rs. (350 – 125) = Rs. 225

Required percentage

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

For making a colour A, three liquids x, y and z are mixed in the ratio 3 : 5 : 7. For colour B, the ratio is 1 : 8 : 3. If 10 parts of colour A are mixed with 8 parts of colour B, what is the ratio of liquids x, y and z in the resulting mixture?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 5
Total parts in A (3 : 5 : 7) = 3 + 5 + 7 = 15

Total parts in B (1 : 8 : 3) = 1 + 8 + 3 = 12

Ratio of volumes = 15 : 12 = 5 : 4

We have to mix in the ratio 10 : 8, i.e. 5 : 4.

Final ratio x : y : z = (3 + 1) : (5 + 8) : (7 + 3) = 4 : 13 : 10

SSC CGL (Tier II) Practice Test - 5 - Question 6

Work of 2 men is equal to that of 5 women, while the work of 2 women is equal to that of 5 boys. One man alone can complete a job in 120 days. If 4 men, 10 women and 25 boys work together, in how many days would the job be completed?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 6
We have, 2 Men = 5 Women; and 2 Women = 5 Boys

Thus, 4 Men = 10 Women = 25 Boys

Thus, if 4 men, 10 women and 25 boys are put together to work, it is equivalent to 12 men (4 + 4 + 4) working for the same.

Thus, if one man can complete a job in 120 days, 12 men would take 10 days to complete the same. [Using inverse proportion]

SSC CGL (Tier II) Practice Test - 5 - Question 7

Avinash borrows Rs. 60,000 from a bank to buy a new machine. If the rate of interest is 10% per annum compounded annually, then what will be the amount payable after two years?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 7
Money borrowed (p) = Rs. 60,000

Amount to be paid after 2 years at the rate 10% p.a.

SSC CGL (Tier II) Practice Test - 5 - Question 8

7 years ago, the ages (in years) of A and B were in the ratio 4 : 5 and 7 years hence, they will be in the ratio 5 : 6. The present age of B is

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 8
7 years ago, A's age = 4x years and B's age = 5x years

⇒ 25x + 70 = 24x + 84

⇒ x = 84 - 70 = 14

∴ B's present age

= 5x + 7 = 5 × 14 + 7 = 77 years

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

If a point N(a, b) divides the line segment joining the points (0, 0) and (5, 5) internally in the ratio 2 : 3, then the value of b is

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 9
Point N(a, b) divides points (0, 0) and (5, 5) in the ratio 2 : 3.

i.e. b = 2

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

In the figure, if ∠ADF = 25°, AD is a tangent to the circle and DE is parallel to AC, then what is the measure of ∠EAC?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 10
As shown in the bellow figure,

∠ADF = 25° (given)

∴ ∠DEF = ∠ADF = 25° (∵ The angle between the tangent and chord at the point of contact is equal to the angle in the alternate segment)

AC || DE (given)

∴ ∠EAC = ∠DEF = 25° (alternate angles)

SSC CGL (Tier II) Practice Test - 5 - Question 11

The centre of two concentric circles is O. The areas of the two circles are 616 cm2 and 154 cm2, respectively. A tangent is drawn through point A on the larger circle to the smaller circle. This tangent touches the smaller circle at B and intersects the larger circle at C. What is the length (in cm) of AC?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 11

Let the radius of bigger circle be R cm and radius of smaller circle be r cm

Area of the larger circle = 616 cm2

πR2 = 616

r = 14 cm

Area of the smaller circle = 154 cm2

πr2 = 154

r = 7 cm

In triangle OBA, using Pythagoras theorem

OA2 = OB2 + AB2

142 = 72 + AB2

196 - 49 = AB2

Now, AC = 2AB

SSC CGL (Tier II) Practice Test - 5 - Question 12

What is the value of cos (90° - B) sin (C - A) + sin (90° + A) cos (B + C) - sin (90° - C) cos (A + B)?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 12
Given expression: cos (90° - B) sin (C - A) + sin (90° + A) cos (B + C) - sin (90° - C) cos (A + B)

= sin B sin (C - A) + cos A cos (B + C) - cos C cos (A + B)

= sin B (sin C cos A - cos C sin A) + cos A (cos B cos C - sin B sin C) - cos C (cos A cos B - sin A sin B)

= cos A sin B sin C - sin A sin B cos C + cos A cos B cos C - cos A sin B sin C - cos A cos B cos C + sin A sin B cos C

= 0

Hence, answer option 4 is correct.

SSC CGL (Tier II) Practice Test - 5 - Question 13

The simplified value of

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 13

SSC CGL (Tier II) Practice Test - 5 - 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.

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 14
Average of the total number of cars of type C1 manufactured by the given five companies

Total number of cars manufactured by D = 1,00,000, and number of C1 type = 2/10 of 1,00,000 = 20,000

Total number of cars manufactured by E = 2,10,000, and number of C1 type = 1/3 of 2,10,000 = 70,000

Total number of cars manufactured by F = 1,60,000, and number of C1 type = 2/4 of 1,60,000 = 80,000

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

Total number of cars manufactured by H = 1,60,000, and number of C1 type = 1/4 of 160,000 = 40,000

Average = (40,000 + 100,000 + 80,000 + 70,000 + 20,000) ÷ 5 = 62,000

SSC CGL (Tier II) Practice Test - 5 - 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.

Z = Total number of B2 type bikes manufactured by all the companies

R = Total number of C1 type cars manufactured by companies F, G and D together

What is the value of Z/R?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 15
Z = Total number of B2 type bikes manufactured by all the companies

Number of bikes manufactured by company D = 3,00,000 × 2/3 = 2,00,000

Number of B2 type bikes manufactured by D = 2/5 of 2,00,000 = 80,000

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

Number of B2 type bikes manufactured by E = 3/7 of 70,000 = 30,000

Number of bikes manufactured by company F = 3,20,000 × 1/2 = 1,60,000

Number of B2 type bikes B manufactured by F = 1/4 of 1,60,000 = 40,000

Number of bikes manufactured by company G = 4,00,000 × 1/4 = 1,00,000

Number of B2 type bikes manufactured by G = 2/5 of 100,000 = 40,000

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

Number of B2 type bikes manufactured by H = 1/8 of 3,20,000 = 40,000

Total number of B2 type bikes (Z) = 80,000 + 30,000 + 40,000 + 40,000 + 40,000 = 2,30,000

R = Total number of C1 type cars manufactured by companies F, G and D together

Total number of cars manufactured by D = 1,00,000, and number of cars of C1 type 2/10 = of 1,00,000 = 20,000

Total number of cars manufactured by F = 160,000, and number of cars of C1 type = 2/4 of 160,000= 80,000

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

So, R = 1,00,000 + 80,000 + 20,000 = 2,00,000

SSC CGL (Tier II) Practice Test - 5 - Question 16

A trader decided to take loan of Rs. 500000 from XYZ Bank of India at 5% interest rate. He further invests this money and gains 20% return on the money invested such that the difference between total amount after gain and amount to be returned is Rs.75000,then for how many years he took the loan?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 16
Given:

Rate = 5% per annum

Late fee = Rs. 100000

Return on money invested = 20%

Difference between amount gained and amount returned = Rs. 75000

Formula used:

SI = (P × R × T)/100

Calculation:

According to question,

Loan taken of Rs.500000

SI on loan for x years = (500000 × 5 × x)/100

⇒ 25000x

Total amount to be paid = 500000 + 25000x

After investing Rs.500000 amount gained,

⇒ 500000 + 20% of 500000

⇒ Rs. 600000

Difference between total amount after gain and amount returned = Rs. 75000

⇒ 600000 - (500000 + 25000x) = 75000

⇒ 100000 - 25000x = 75000

⇒ 25000x = 25000

⇒ x = 1 year

∴ The correct answer is 1 year.

SSC CGL (Tier II) Practice Test - 5 - Question 17

In the given figure find the ratio of sum of inradius and circumradius for smaller triangle and bigger triangle, if the side of smaller equilateral triangle is 25% of the side of bigger equilateral triangle ?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 17
Given

Side of smaller equilateral triangle = 25% side of bigger equilateral triangle

Formula used:

1) Circumradius = abc/4Δ

2) Inradius = Δ/s

Where, s = (a + b + c)/2

And Δ = Area of triangle

3) Area of equilateral triangle = √3/4 × a2

Calculation:

According to question,

Side of smaller equilateral triangle = 25% side of bigger equilateral triangle

Side of smaller equilateral triangle/Side of bigger equilateral triangle = ¼

Sum of circumradius and inradius for smaller triangle,

⇒ 1 × 1 × 1/4 × √3/4 × (1)2 + √3/4 × (1)2/(1 + 1 + 1)/2

⇒ 1/√3 + 1/2√3

⇒ 1/√3 × (1 + 1/2)

⇒ 3/2√3

Similarly,

Sum of circumradius and inradius for bigger triangle,

⇒ 4 × 4 × 4/ 4 × √3/4 × (4)2 + √3/4 × (4)2/ (4 + 4 + 4)/2

⇒ 4/√3 + 2√3/3

⇒ 4/√3 + 2/√3

⇒ 6/√3

Ratio = 3/2√3 : 6√3

⇒ 3/2 : 6

⇒ 3 : 12

⇒ 1 : 4

∴ The correct answer is 1 : 4.

SSC CGL (Tier II) Practice Test - 5 - Question 18

Raman and Tapan starts a business and invested Rs. 10,000 and Rs. 50,000 respectively. After six months Raman added Rs. 10,000 more and Tapan withdrew Rs. 30,000. At the end of the year if business profited Rs. 10,000 then find the profit earned by Tapan.

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 18
Given

Raman and Tapan starts a business and invested Rs. 10,000 and Rs. 50,000 respectively.

After six months Raman added Rs. 10,000 more and Tapan withdrew Rs. 30,000.

At the end of the year if the business profited Rs. 10,000.

Concept used:P1 : P2 = (I × T)1 : (I × T)2

Where,

P1 and P2 are the profit earned by the first person and second person respectively.

(I × T)1 is the capital invested or product of the amount of investment done and the time of investment for the first person.

(I × T)2 is the capital invested or product of the amount of investment done and the time of investment for the second person.

Calculation:

According to the question,

For Raman:

The amount invested for the first six months = Rs. 10,000

The amount invested for the last six months = Rs. 10,000 + Rs. 10,000 = Rs. 20,000

For Tapan:

The amount invested for the first six months = Rs. 50,000

The amount invested for the last six months = Rs. 50,000 - Rs. 30,000 = Rs. 20,000

Now,

Capital invested by Raman = (Rs. 10000 × 6) + (Rs. 20000 × 6)

Capital invested by Tapan = (Rs. 50000 × 6) + (Rs. 20000 × 6)

Now,

The required ratio for the profit = (Rs. 10000 × 6) + (Rs. 20000 × 6) : (Rs. 50000 × 6) + (Rs. 20000 × 6)

The required ratio for the profit = 3 : 7

Again according to the question,

The profit earned by Tapan

SSC CGL (Tier II) Practice Test - 5 - Question 19

A temple of a conical roof of radius 2 m and height is twice that of the radius. If the cost of painting per sq. m is Rs. 512, then find the total approximate cost of painting the external surface of the conical roof. (in Rs.)

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 19
Given:

A temple of a conical roof of radius 2 m and height is twice that of the radius.

Cost of painting per sq. m = Rs. 512

Formula used:

(1.) Lateral surface area of cone = πrl

Where,

r = radius of cone

l = slant height

(2.) Slant height = √(r2 + h2)

Where,

h = height of the cone

Calculation:

According to the question,

The radius of the cone = 2 m

The height of the cone = 4 m

Let l be the slant height of the cone.

⇒ l2 = r2 + h2

⇒ l2 = 22 + 42

⇒ l2 = 4 + 16

⇒ l = 2√5 m

Let A be the lateral surface area of the cone.

⇒ A = πrl

⇒ A = 3.14 × 2 × 2√5

⇒ A ≈ 28 m2

Since the cost of painting per sq. m = Rs. 512

Therefore, the total cost of painting a conical roof = Rs. 512 × 28 = Rs. 14336

Therefore, 'Rs. 14336' is the required answer.

SSC CGL (Tier II) Practice Test - 5 - Question 20

If Vikash walks at the speed of 10 kmph then he misses the bus by 5 min and when he increases his speed to 20 kmph then he reaches the bus station 10 min earlier. Find the distance from Vikash's destination to the bus stop. (in km)

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 20
If Vikash walks at the speed of u kmph then he misses the bus by t1 min and when he increases his speed to b kmph then he reaches the bus station t2 min earlier.

The distance from Vikash's destination to the bus stop

Where,

u and v is in kmph

t1 and t, is in min

According to the question,

The distance from Vikash's destination to the bus stop

Therefore, '5 km' is the required answer.

SSC CGL (Tier II) Practice Test - 5 - Question 21

A man sells an article at 10% above its cost price. If he had bought it at 15% less than what he paid for it and sold it for Rs. 33 less, he would have gained 10%. Find the cost price of the article.

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 21
Given

Markup percentage = 10%

Profit percentage = 10%

Concept used:

Selling price = Cost price × (1 + profit %)

Calculation:

Let the cost price be 100x

Selling price = 100x × 110/100 = 110x

Reduced Cost price = 100x × 85/100 = 85x

Reduced selling price = 85x × 110/100 = 187x/2

According to the question, 110x - 187x/2 = Rs. 33

⇒ 33x/2 = Rs.33 ⇒ x = Rs.2

∴ The cost price of the article = 100 × 2 = Rs.200.

SSC CGL (Tier II) Practice Test - 5 - Question 22

Find the compound interest earned in the 9th year, if the compound interest earned on the principal amount Rs. 10000 at the 8th year is Rs. 1948.72 and the rate of interest applied is 10% p.a.

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 22
Given:

Compound interest earned on the 8th year = Rs. 1948.72

Rate of interest applied = 10% 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

Calculation:

According to the question,

⇒ C.I9 = (1 + 10%) × C.I8

⇒ C.I9 = (1 + 0.1) × 1948.72

⇒ C.I9 = 1.1 × 1948.72

⇒ C.I9 = Rs. 2143.59

Therefore, 'Rs. 2143.59' is the required answer.

SSC CGL (Tier II) Practice Test - 5 - Question 23

Study the following graph and answer the question given below.

The below Histogram shows the data of the annual rainfall (in cm).

Find the difference in the number of times the rainfall above 130 cm and the number of times the annual rainfall below 130 cm.

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 23
Given

The number of time the annual rainfall 100-110 cm = 5

The number of time the annual rainfall 110-120 cm = 6

The number of time the annual rainfall 120-130 cm = 7

The number of time the annual rainfall 130-140 cm = 2

The number of time the annual rainfall 140-150 cm = 9

Calculation:

The number of times the annual rainfall above 130 cm = 2 + 9 = 11

The number of times the annual rainfall below 130 cm = 5 + 6 + 7 = 18

The difference in the number of times the rainfall above 130 cm and the number of times the annual rainfall below 130 cm = 18 - 11 = 7

Therefore, '7' is the required answer.

SSC CGL (Tier II) Practice Test - 5 - Question 24

Study the below graph and solve the following question.

The below chart shows the percentage wise sale of number of cars on different days.

The total number of cars sold = 700

If the total number of cars sold on Saturday is 50% of the total sale of the number of cars on Monday, then find the total number of cars sold on Thursday and Saturday.

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 24
Given

The total number of cars sold = 700

Calculation:

The total number of cars sold on Monday = 20% of 700 = 140

The total number of cars sold on Thursday = 10% of 700 = 70

The total number of cars sold on Saturday = 50% of 140 = 70

The total number of cars sold on Thursday and Saturday = 70 + 70 = 140

Therefore, '140' is the required answer.

SSC CGL (Tier II) Practice Test - 5 - Question 25

Which of the following statement is correct regarding the number 434434?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 25
Given

The given number = 434434

Concept used:

(1.) abcabc is divisible by 1001.

(2.) If x is divisible by y, then x is also divisible by the factors of y.

Calculation:

According to the question,

Let N = 434434

It can be written as:

⇒ 434434 = 434000 + 434

⇒ 434434 = 434 × (1000 + 1)

⇒ 434434 = 434 × (1001)

It can be interpreted as 434434 is divisible by 434 and 1001.

Now,

Factors of 1001 = 7 × 11 × 13

Since 434434 is divisible by 1001,

Therefore, 434434 is also divisible by 7, 11 and 13.

Therefore, 'Option C' is the correct answer.

SSC CGL (Tier II) Practice Test - 5 - Question 26

In a ΔDEF where right-angle at E, If tan D = 15/8 and ED = 16 cm, then find the value of EF. (in cm)

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 26
Given

In a right-angle triangle ΔDEF,

tan D = 15/8 and ED = 16 cm

Calculation:According to the question, the required image is:

Now,

Therefore, '30 cm' is the required answer.

SSC CGL (Tier II) Practice Test - 5 - Question 27

What should come in the place of (?) for the expression given below?

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 27
Given

Formula used:(secx - tanx)(secx + tanx) = sec2x - tan2x = 1

Calculation:According to the question,

⇒ ? = (2secx + tanx - secx)(secx - tanx)

⇒ ? = (secx + tanx)(secx - tanx)

⇒ ? = sec2x - tan2x = 1

Therefore, '1' is the required answer.

SSC CGL (Tier II) Practice Test - 5 - Question 28

What should come in the place of (?) for the question given below?

? = 240% of 400 × 0.6 + (262 - 162) ÷ 2 - 30

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 28
Given? = 240% of 400 × 0.6 + (262 - 162) ÷ 2 - 30

Concept used:

Calculation:

According to the question,

⇒ ? = 240% of 400 × 0.6 + (262 - 162) ÷ 2 - 30

⇒ ? = 240% of 400 × 0.6 + (676 - 256) ÷ 2 - 30

⇒ ? = 240 × 4 × 0.6 + 210 - 30

⇒ ? = 576 + 210 - 30

⇒ ? = 756

SSC CGL (Tier II) Practice Test - 5 - Question 29

If (a + b) : (b + c) : (c + a) = 5 : 7 : 8 , Then find the value of ab : bc : ca

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 29
Given

(a + b) : (b + c) : (c + a) = 5 : 7 : 8

Calculation:

Let, a + b = 5k , so (b + c) and (c + a) is 7k and 8k respectively.

As per the question,

(a + b) + (b + c) + (c + a) = 5k + 7k + 8k

⇒ 2 (a + b + c) = 20

⇒ (a + b + c) = 10k

So,(a + b + c) - (b + c) = 10k - 7k

⇒ a = 3k

(a + b + c) - (c + a) = 10k - 8k

⇒ b = 2k

(a + b + c) - (a + b)) = 10k - 5k

⇒ c = 5k

So, ab : bc : ca = 3k × 2k : 2k × 5k : 5k × 3k

⇒ 6k2 : 10k2 : 15k2

⇒ 6 : 10 : 15

∴ The value is 6:10:15

SSC CGL (Tier II) Practice Test - 5 - Question 30

What will be the cost of painting the curved surface area of a plastic cylindrical toy with a radius of 42 cm and a height of 3.5 cm? If the cost of painting is at the rate of Rs 70 per cm2.

Detailed Solution for SSC CGL (Tier II) Practice Test - 5 - Question 30
Given

The radius of the cylindrical toy = 42 cm.

The height of the cylindrical toy = 3.5 cm

Cost of painting = Rs 70 per cm2.

Formula used:

The curved surface area of the cylindrical toy = 2πrh

( r = radius , h = height)

Calculation:

As per the question,

The curved surface area of the cylindrical toy = 2 × π × 42 × 3.5

⇒ 2 × (22/7) × 42 × 3.5 [as we know π = 22/7]

⇒ 2 × 22 × 42 × 0.5

⇒ 2 × 22 × 21

⇒ 44 × 21

⇒ 924 cm2

So, the total cost = 924 × 70

⇒ Rs 64680

∴ The cost of painting the curved surface area is Rs 64680

## SSC CGL Tier II Mock Test Series 2024

66 docs|110 tests
Information about SSC CGL (Tier II) Practice Test - 5 Page
In this test you can find the Exam questions for SSC CGL (Tier II) Practice Test - 5 solved & explained in the simplest way possible. Besides giving Questions and answers for SSC CGL (Tier II) Practice Test - 5, EduRev gives you an ample number of Online tests for practice

## SSC CGL Tier II Mock Test Series 2024

66 docs|110 tests