SSC CGL (Tier II) Practice Test - 3 - SSC CGL MCQ

Putting the values in the given equation,

= (tan 1° tan 89°) (tan 2° tan 88°) × (tan 3° tan 87°) ... (tan 45°)

= (tan 1° cot 1°) (tan 2° cot 2°) ... (tan 45°) = 1.1.1 ... 1 = 1

We have, D = 100 (given)

Old: M x 100 = 100M

New: (M – 10) × (100 + 10) = 100M

Thus, 110(M – 10) = 100M

M = 110

Therefore, Reena's share = (2/5) × 31, 570 = Rs. 12,628

Let its breadth be x.

2(x + x + 4) = 28

4x + 8 = 28

x = 5 cm

So, length of rectangle = 5 + 4 = 9 cm

So, (6/7 - 2/5) of the tank = 160 gallons

16/35th of the tank = 160 gallons

Total volume =

So, 140 gallons means 2/5th of the total volume.

Answer = 2/5 × 4 = 1.6 metres

Total fresher students that appeared for MBA entrance exam in 2008 = 0.32 × 0.61 × 8,00,000 = 1,56,160

Percent = 1,56,160 × = 25.37

Then, W = 12a --- (i) (Total = Average × Number of people)

Let weight of new person be x.

Then, after person leaves and new one joins,

W - 90 + x = 12a + 0.75 × 12 --- (ii)

Putting value of 12a from (i), we get:

W - 90 + x = W + 0.75 × 12

-90 + x = 0.75 × 12

x = 90 + 9

x = 99 kg 108

4/5 × 100 = 80%

Speed = y kmph

Time taken when moving at normal speed - time taken when moving 3 kmph faster = 20 minutes

Now,

Time taken when moving 1 kmph slower - time taken when moving at normal speed = 10 minutes

(ii) ÷ (i) gives:

9y - 9 = 6y + 18

3y = 27

y = 9 km/hr

Let radius of field be r metres.

πr² = 32378.5

(22/7) × r² = 32378.5

r² = 32378.5 x 7/22 = 10302.25

r = 101.5 m

Circumference = 2 × (22/7 )× 101.5 = 638 m

Cost of fencing 1 m = Rs. 154

Cost of fencing 638 m = Rs. 154 638 = Rs. 98,252

So, x = 1, y = 5 and z = 3

Thus, 9x + 10y - 18z = 9 x 1 + 10 x 5 - 18 x 3 = 9 + 50 - 54 = 5

Hence, answer option (d) is correct.

Option (d) is correct.

Let AD = x cm and AC = y cm

∠BCD = ∠BAC

In ΔABC and ΔBDC,

∠ABC = ∠DBC

∠BAC = ∠BCD

⇒ ∠BDC = ∠BCA

ΔABC ~ ΔCBD

x = 7

Also, BC/BD = AC/CD or 12/9 = y/6

y = 8

Perimeter of ΔADC = AC + AD + DC = (8 + 7 + 6) cm = 21 cm

Perimeter of ΔBDC = BD + BC + DC = (9 + 12 + 6) cm = 27 cm

Ratio of perimeter of ΔADC to perimeter of ΔBDC = 7/9

The remainder obtained by diving the given number by second divisor = remainder obtained by dividing the first remainder by the second divisor

Here, 296 ÷ 37 = 8

∴ Required remainder = remainder obtained by dividing 75 by 37 = 1

Principle = Rs. 30,00,000, rate = 12%, time = 1.5 years

Formula used:

CI = principle[(1 + rate / 100)^time - 1]

Calculations:

When compounded half yearly, rate becomes half, time becomes double.

Rate is 12%, when compounded half yearly rate becomes (12 ÷ 2)% = 6% and time is 1.5 years, when compound did half yearly time is (1.5 × 2)years = 3 years

= 24 × 23877

= Rs. 573048

∴ The answer is Rs. 573048

Circumference of the base (C) = 176 cm

Height (h) = 12 cm

Formula used:

Circumference of the base = 2 × π × r

Total surface area (TSA) of cylinder = 2 × π × r × (r + h)

Calculation:

According to the question,

Circumference of the base = 176

176 = 2 × π × r

⇒ 176 = 2 × 22 / 7 × r

⇒ 176 × 7 / 2 × 22 = r

⇒ 8 × 7 / 2 = r

⇒ r = 56 / 2

⇒ r = 28

Now T.S.A.:

2 × π × 28 × (28 + 12)

⇒ 2 × π × 28 × 40

⇒ 2 × π × 1120

⇒ 2240 × π = 7040 cm².

∴ The Total Surface Area of the cylinder will be 7040 cm².

Given:

Compound Interest (CI) - Simple Interest (SI) = Rs. 97.2

Principle (P) =?

Rate of interest (R) = 9% per annum

Time period (T) = 2 years

Concept Used:

(CI - SI) = P(R)T/(100)T.

Calculations:

According to the question,

⇒ (97.2) = P(9)²/(100)².

⇒ 81P = 972000

⇒ P = 972000/81 = Rs. 12000

∴ The sum invested is Rs. 12000.

Tyres produced by company ‘D’ in 2017 = 15% of 1,80,000 = 1,80,000 × (15/100) = 27,000

Rakesh got 92 % and

Kiran got 96 % out of 375 marks

Concept :

X % = X/100

Calculations :

Marks obtained by Rakesh

⇒ 92% of 375

⇒ (92/100)× 375

⇒ 345

Marks obtained by Kiran

⇒ 96 % of 375

⇒ (96/100) × 375

⇒ 360

Now, the sum of marks obtained by Rakesh and Kiran

⇒ (345 + 360)

⇒ 705

∴ The required sum is 705.

Given:

The ratio of the length and breadth of the rectangle is 24 : 7.

The diagonal length of the rectangle is 50 cm

Formula used:

(1) The diagonal length = d

d² = l² + b²

(2) The area of the rectangle = l × b

Where,

l = length

b = breadth

Calculation:

The length of the rectangle = 24k

The breadth of the rectangle = 7k

The diagonal length of the rectangle = 50 cm

Therefore,

⇒ (24k)² + (7k)² = 502

⇒ 576k² + 49k² = 2500

⇒ 625k² = 2500

⇒ k² = 4

⇒ k = ± 2

Neglect the negative value of k.

Now,

The length of the rectangle = 24 × 2 = 48 cm

The breadth of the rectangle = 7 × 2 = 14 cm

Therefore,

The area of the recatangle = 48 × 14 = 672 cm²

∴ The required answer is 672 cm².

Earning on Sunday = 100

Earning on Monday = 500

Earning on Tuesday = 300

Earning on Wednesday = 600

Earning on Thursday = 650

Earning on Friday = 300

Earning on Saturday = 350

∴ The difference in earnings was highest between Sunday and Monday.

Given:

The total number of white balls and pink balls is five times the number of blue balls.

The total number of pink balls and blue balls is twice the number of white balls.

There are 33 pink balls.

Calculation:

Let W, P and B are the numbers of white balls, pink balls and blue balls.

According to the question,

The number of pink balls, P = 33

⇒ W + P = 5B

⇒ W + 33 = 5B ----(1)

⇒ P + B = 2W

⇒ 33 + B = 2W

⇒ B = 2W - 33 ----(2)

Now, substitute the equation (2) in equation (1).

⇒ W + 33 = 5(2W - 33)

⇒ W + 33 = 10W - 165

⇒ 9W = 198

⇒ W = 22

Substitute the above value in equation (1).

⇒ 22 + 33 = 5B

⇒ 5B = 55

⇒ B = 11

Therefore,

The number of blue balls = 11

The total number of balls in the bag = 33 + 22 + 11 = 66

∴ The required answer is 66.

Given:

The batsman scored 84 runs in his 12^th innings, thereby improving his average score per innings by 4 runs.

Concept used:

Total = Average × Number of entities

Calculation:

Let the average of the batman in the first 11 innings be N.

Total run scored by the batsman in the first 11 innings = 11 × N = 11N

Total run scored by the batsman in all 12 innings = 11N + 84

New average score after 12^th innings = N + 4

According to the question,

(11N + 84)/12 = (N + 4)

⇒ (11N + 84) = (N + 4) × 12

⇒ 11N + 84 = 12N + 48

⇒ 12N - 11N = 84 - 48

⇒ N = 36

Now, N + 4 = 36 + 4

⇒ 40

∴ 40 is the average score per innings of the batsman after the 12^th innings.

Total production of type E cars over the year = 20 + 42 + 40 + 35 + 43 = 180

Production of type E cars in 2013 = 42 thousands

180 thousands represents = 360°

∴ 42 thousands will represent = [360°/180] × 42 = 84°

Given:

x(x + y + z) = 30 -----(1)

y(x + y + z) = 64 -----(2)

z(x + y + z) = 50 -----(3)

Where x, y, z > 0.

Calculation:

By adding the equations (1), (2), and (3) we get

[x (x + y + z)] + [y (x + y + z)] + [z (x + y + z)] = 144

Taking (x + y + z) common,

⇒ (x + y + z) (x + y + z) = 144

⇒ (x + y + z)² = 144

⇒ (x + y + z) = 12

Multiplying 2 on both sides,

⇒ 2 (x + y + z) = 24

∴ The value of 2(x + y + z) is 24.

Given:

Length of the train = 225 m

Speed of the train = 145 km/h

Speed of the man = 17 km/h

Concept used:

Time = Distance/speed

Relative speed = When two objects move opposite to each other then their relative speed is equal to the sum of their individual speeds.

When a train of x m crosses a man or a pole then it will cross its own length.

km/h × 5/18 = m /sec

Calculation:

Relative speed of the train in km/h = 145 + 17

⇒ 162 km/h

Relative speed of the train in m/sec = 162 × 5/18

⇒ 45 m/sec

According to the concept, the train will cover its own length

So, time = 225/45

⇒ 5 sec

Time taken by the train to cover the man = 5 sec

∴ In 5 sec it will pass the man.

Given:

(x + 1) ∶ (x + 5) ∶∶ (x + 17) ∶ (x + 53)

Concept used:

Mean proportion of two numbers x and y = √(x × y)

Calculation:

Here,

(x + 1)/(x + 5) = (x + 17)/(x + 53)

⇒ x² + 53x + x + 53 = x² + 17x + 5x + 85

⇒ x² + 53x + x - x² - 17x - 5x = 85 - 53

⇒ 32x = 32

⇒ x = 32/32

⇒ x = 1

Now,

(x + 5) and (9x – 1) = (1 + 5), (9 × 1 - 1)

⇒ 6, 8

Now,

mean proportion = √(6 × 8)

⇒ √48

⇒ 4√3

∴ Required mean proportion is 4√3.

Given:

36 ÷ (8 × 3) - [3 ÷ {4 × {3 × 4 ÷ (5 - 9) + 6}}]

Concept used:

Calculation:

36 ÷ (8 × 3) - [3 ÷ {4 × {3 × 4 ÷ (5 - 9) + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × {3 × 4 ÷ (- 4) + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × {3 × (- 4/4) + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × {3 × (- 1) + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × {- 3 + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × 3}]

⇒ 36 ÷ 24 - [3 ÷ 12]

⇒ 36 ÷ 24 - (1/4)

⇒ 3/2 - 1/4

⇒ (6 - 1)/4

⇒ 5/4

⇒ 1.25

So, this is lies in between 1 and 1.3

∴ Required answer is Option B.

Given:

Total quantity of the initial mixture = 90 ml

Sugar percentage = 38%

Sugar percentage of the new mixture = 17.1%

Calculation:

In the mixture of 90 ml amount of sugar = 90 × 38%

⇒ 34.2 ml

Now, % of the sugar in the new mixture becomes 17.1

According to the question,

In the new mixture 34.2 ml of sugar = 17.1% of the mixture [∵ We added water so amount of sugar will unchanged]

So, 17.1% = 34.2

⇒ 1% = 2

⇒ 100% = 200

So, total quantity of new mixture = 200 ml

Water added = 200 - 90

⇒ 110 ml

∴ Required amount of water to be added is 110 ml.