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

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.

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

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

Profit = Rs. 250

SP = Rs. 350

New CP = Rs. 125

New SP = Rs. 350

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

Required percentage

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

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]

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

⇒ 25x + 70 = 24x + 84

⇒ x = 84 - 70 = 14

∴ B's present age

= 5x + 7 = 5 × 14 + 7 = 77 years

i.e. b = 2

∠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)

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

= 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.

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

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

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.

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.

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

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.

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.

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.

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

Rate of interest applied = 10% p.a.

Formula used:

(1.) C.I_n+1 = (1 + R%) × C.I_n

Where,

C.I_n = Compound interest earned for nth year

C.I_n+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.

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.

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.

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.

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.

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.

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

(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

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