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

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

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

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

0 = 1(13 + 3P)

3P = -13

P = -13/3

Original ⇒ 24 + 40 = 64 litres

According to the question

24 + ? = 40

? = 40 - 24

? = 16 litres

⇒ x : y = 3 : 2 and y : z = 3 : 2

Thus, x : z = = 9 : 4

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

Probability = 8/20 = 2/5

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

Speed of current = y kmph

From equations (i) and (ii),

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

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

Profit = $1120 - 1100 = $20

Required probability =

The given data can be represented diagrammatically as follows.

Triangles ABC, ADB and ADC are right angled triangles.

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

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.

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

Let, the side of equilateral triangle is = a

Then AB = BC = CA = a

BD = DC = a/2

AB² = AD² + BD²

AC² = AD² + CD²

Add. Eq. (i) & (ii)

AB² + AC² = 2AD² + BD² + CD²

Add BC2 both sides

AB² + AC² + BC²

= 2AD² + BD² + CD² + BC²

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

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

⇒ Average marks = 260/5 = 52

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

The sixth question is the last wrong question

⇒ 2 errors are in the first 5 questions

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

⇒ ⁵C₂ / ¹⁰C₃

∴ 1/12

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

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

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

s² – 33s + 244 = 0

Formula Used:

(x + y)² = x² + 2xy + y²

Calculation:

Suppose,

Taking square both sides, we have –

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.

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.

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

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.

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.