RRB JE CE (CBT I) Mock Test- 3 - Civil Engineering (CE) MCQ

DE ∥ BC

DE = 1/2 BC

⇒ (∆BDE) = 1/4 × (∆ABC)

And ∆BDE = ∆PED

[ ∵ both triangles lie on the same base DE and between two parallel lines DE and BP.]

∴ (∆PED) = 1/4 × (∆ABC)

Let ∠B be right angle in △ABC

and BD be perpendicular on hypotenuse AC then

1 / P² = 1 / 6² + 1 / 6²

P² = 6² / 2 => p = 3√2.

×3 ↓ : ↓ ×3 → time

Length = 9m : 12m

Time = D / S = L₁ + L₂ / S₁ + S₂ = 9 + 12 / 4 + 3 = 2 / 14 + 3

= 3 Sec

Mother’s one day work = 1/5

Both can complete their work = 1/5 + 1/15 = (3 + 1)/15 = 4/15

Hence, required days= 15/4 days .

= 13400

Total no. of soldiers joining in Navy = 10900

Required Percentage = (13400/10900) × 100 = 123%

tan (5x – 10°) = tan (90° – {5y + 20°})

5x – 10° = 90° – (5y + 20°)

5x + 5y = 90° + 10° – 20°

5x + 5y = 80°

x + y = 16°

S = 60 × 5 / 18m/sec.= 50 / 3m⋅/sec

T = 30sec

Length = S × T = (50 / 3 × 30)m

I = 500m

Then, the measure of its supplementary angle is (180 − x^∘). It is given that, 180^∘ − x = 1 / 4 × x

⇒ (180^∘ − x) = x

⇒ 720 − 4x = x

⇒ 5x = 720

⇒ x = 144

Thus, the measure of the angle is 144^∘ and the measure of its supplement 180^∘ - 144^∘ = 36^∘

= 17 × 5 = 85 years

Three years hence, Total age of 5 members = 85 + 3 x 5 = 85 + 15 = 100 years

Sum of present ages of 6 members = 17 × 6 = 102 years

Present age of baby = 102 – 100 = 2 years

Let the length of the ladder be x meter. We have

8² + y² = x² and (y + 2) = x

Hence, 64 + (x − 2)² = x²

⇒ 64 + x² − 4x + 4 = x²

⇒ 68 = 4x ⇒ x = 17 meter

33.33 = 1 / 3 → 1 / 2 = 50% [1 / 2 = 1 / 3 - 1]

I, I(1 + r/100)¹, I(1 + r/100)², I(1 + r/100)³, I(1 + F/100)⁴ respectively

Given, (1 + r/100)³ = 13310

and (1 + r/100)⁴ = 14641

2) divided by (1), we get (1 + r/100) = 14641 / 13310 = 1.1

∴ Sum of the first three years interests = 13310 / (1.1)³ + 13310 / (1.1)² +13310 / (1.1)¹

= 13316 / 1.331 + 13316 / 2.21 + 13316 / 2.1

= 10000 + 11000 + 12100

= Rs. 33100

Hence

1 - 1/2 = 1/2

B, means their ratio is 2 : 1.

Condition 2- B is one third as fast

as C, means their ratio is 1 : 3. Then we can say, the ratio of A, B and C is as follows According to condition 1, A : B = 2 : 1

According to condition 2;

B : C = 1 : 3

Thus, Ratio of A : B : C = 2 : 1 : 3

This ratio denotes the working efficiency of A, B and C which means;

A, B and C can do = (2 + 1 + 3 = 6) work/day

Now we can calculate total work i.e.

Total work = Total days × Per day work

Total work = 30 × 6 = 180

A can do the same work in;

= (Total work / Efficiency of A) = 180 / 2 = 90 days

B can do the same work in; = (Total work / Efficiency of B)

180 / 1 = 180 days

C can do the same work in;

= (Total work / Efficiency of C) = 180 / 3 = 60 days

Thus, A, B and C can individually complete the same work in 90, 180 and 60 days, respectively.

Then a + b = 5 + 5 (∴ ab = 25 and a = b)

= 10

Given, a + b + c = 1 Let observe the value of a, b, c

a = 1 / 3, b = 1 / 3, c = 1 / 3

a + b + c = 1

1 / 3 + 1 / 3 + 1 / 3 = 1

3 / 3 = 1

1 = 1

put the value of a, b, c (1 + a)(1 + b)(1 + c)

(1 + 1 / 3)(1 + 1 / 3)(1 + 1 / 3)

4 / 3 × 4 / 3 × 4 / 3

64 / 27 = 2.37

= 260 − 39 = 221

2^nd discount

S.P = Rs.221 − 20%Rs.221

= 221 − 44.2 = 176.8

Sales tax = 5 / 100 × 1768 / 10

= 8840 / 1000

= Rs⁡.8.84

Amount paid by buyer

= Rs.176.8 + Rs.8.84

= Rs. 185.64

i.e, area of four walls + area of the floor

= 2 x (21) x 4 + (106.25)

= 274.25 m²

∴ cost of cementing = 24 x 274.25 = Rs. 6582

Let the goat be tied at P.

The area over which the goat could graze = sum of the areas of the regions A and B = Area of a sector of radius 14 m and central angle 270°

S.P. at 10% loss = Rs0.9x

According to question if it is sold by Rs 9 more there will be gain of 25 / 2%

0.9x + 9 = x ×

0.9x + 9 = 225x / 200

⇒ 180x + 1800 = 225x

⇒ 225x − 180x = 1800

⇒ 45x = 1800

∴ x = Rs.40

Alternate solution, Let the CP of the article =100

x × 22.5 = 9

x = 9 / 22.5

= 18 / 45

Required, ⇒ CP × x

⇒ 100 × 18 / 45

⇒ 40

Volume = 102cu.cm

1cu.m = 100cu.cm

102cu.m = 102 × 100

100cu⋅cm = 1 litre

1cu⋅cm = 1 / 1000

10200cu⋅cm = (1 / 1000) × 10200

= 10.2 litres

Let the side of the square be ‘α’

Its area = a²

Area of square on the diagonal = (√2a)² = 2a²

Required ratio = a² / 2a² = 1 : 2

= 90° + 30°

= 120° (exterior angle)

2^{x − 1}(1 + 2²) = 320

2^{x − 1} = 320

2^{x − 1} = 320 / 5 = 64

2^{x − 1} = 2⁶

x − 1 = 6

X = 7

sin²⁡25^∘ + sin²⁡65 = sin²⁡25^∘ + sin²⁡(90^∘ − 25^∘)

= sin²⁡25^∘ + cos^2⁡25^∘ = 1

So A : B = 2 : 1

Also (A + B)'s one-day work = 1 / 14

To get days in which B will finish the work,

let's calculate work done by B in 1 day

= (1 / 14) x (1 / 3) = 1 / 42

So B will finish the work in 42 days and A will finish the work in 21 days.