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

They are 9 in total.

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

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.

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

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.

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

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.

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

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

[∵ (a – b)(a² + ab + b²) = a³ – b³]

1/0.25 = 100/25

= 4

So, square root of 4 is 2.

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

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

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.

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

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

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

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

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

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

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)

∴ The answer is 67(2/7).

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 1^st and 2^nd 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

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%

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 1^st 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.

Given:

Outer radius = b

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

And internal radius of cylinder = Outer radius - thickness

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

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%

∴ The answer is 1%.

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.

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.

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.

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%.

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 .

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