RRB JE CE (CBT I) Mock Test- 2 - Civil Engineering (CE) MCQ

10 x s x s/100 = 1000

s2 = 10,000 s = 100

Each student got 10 x 100/100

=> 10 chocolates

=> CP/SP - 5 / 4…(1)

From equation (1) we can see that CP >SP, therefore there will be loss. Let the CP = 5 and SP = 4 Then Loss = 1

Loss% = loss/CP x 100

= 1/5 x 100

= 20% loss

Overall percentage of marks obtained by Ankit in all the five subjects

= (120 + 100 + 30 + 70 + 100)/700 x 100

= 420/700 x 100

= 60%

Overall percentage of marks obtained by Sujal in all the five subjects

= (180 + 80 + 40 + 90 + 150)/700 x 100

= 540/700 x 100

= 540/7%

Required ratio = 60 : 540/7

= 420 : 540

= 7 : 9

2 x Area of rhombus = (First diagonal x Second diagonal)

Let its second diagonal = (a + b√5) cm

(14 + 13√5) = (3 + 2√5) x (a + b√5)

(14 + 13√5) = (3a + 3b√5 + 2a√5 + 10b)

After comparing-

(3a + 10b) = 14 ...... (1)

(2a + 3b) = 13 ....... (2)

From (1) and (2)-

a = 8 and b = -1

Length of second diagonal = (a + b√5) = (8 - √5) cm

= (35 ÷ 7) + (225 - 210) - 25

= 5 + 15 - 25

= -5

Hence, 7 pipes of first set can fill 1/10th of the vessel in 1 min.

5 pipes of second set can fill 2/5th of the vessel in 5 minutes

Hence, 5 pipes of second set can fill 2/25th of the vessel in 1 min.

10 pipes of third set can empty 1/3rd of the vessel in 10 minutes.

Hence, 10 pipes of third set can empty 1/30th of the vessel in 1 min

So, if all the pipes are open in one minute 1/10 + 2/25 - 1/30 = 11/75th of the vessel is filled up.

Hence, to fill up the whole vessel, time required= 75/11 minutes.

Mass numbers of some common elements-

Neon- 20

Sodium- 23

Silver- 108

20% of x = 1080

x = 5400

SP = 125% of 5400 = 6750

√[(5² + 11) x 10^-6] = 6 x 10^-3 = 0.006

(0.06 ÷ 10²) = 0.0006

√0.000036 = 0.006

According to the 1^st condition; m/n + 11/30= n/m ...(1)

According to the 2nd condition; (m - 1)/n = ⅔

=> 3m - 3 = 2n

=> 3m - 2n = 3 ...(2)

By solving equation 1,

=> 30m² + 11mn - 30n² = 0

=> (6m - 5n)(5m + 6n) = 0

we get the value of m/n = 5/6, -6/5

applying condition(2), m/n = (5 - 1)/6 = 4/6 = 2/3

Hence the required fraction = m/n = 5/6

Hence amount of interest Suresh will get in first year = 500/2 = Rs. 250

If the same amount is invested and interest rate is same then in first year the interest obtained is same whether it is simple interest or compound interest.

Hence compound interest obtained in 1^st year = Rs. 250

Therefore amount of interest obtained in the 2^nd year = Rs. (525 - 250) = Rs. 275

Hence rate of interest = 100 x (275 - 250)/250 = 10%

Average = 150/5 = 30

Average number of presence of employees in company Q on all the five working days = 24 + 16 + 50 + 28 + 62 = 180

Average = 180/5 = 36

Required sum = 30 + 36 = 66

[(0.09)² x (0.3)⁸ x (0.027) x 10¹²] = [3¹² x (0.3)^x]

[(0.3)⁴ x (0.3)⁸ x (0.3)³] = [(0.3)¹² x (0.3)^x]

(0.3)¹⁵ / (0.3)¹² = (0.3)^x

(0.3)³ = (0.3)^x

x = 3

S.P. of pen for Sita = Rs 150 - 90 + 120 = 180

Profit % = [(180 - 100)/100] x 100 = 80%

Hence; 1200/N = A .. (1)

According to the 2nd condition; 1200/(N - 100) = A + 2/5 .. (2)

Equation (2) - Equation (1); 1200/(N - 100) - 1200/N = 2/5

If we will try to solve this equation then it will become too lengthy. Hence such questions we should solve by using the options.

Out of the given options, option (c) satisfies above equation.

Hence; N = 600 km/hr

=> (12x² - 1)/3x = 9

=> 12x² - 1 = 27x ------- (i)

Now,

(144x⁴ - 1)/(27x² + 2x)

= [(12x² - 1)(12x² + 1)]/[x(27x + 2)] ------- [(a² - b² = (a - b)(a + b)]

= [27x(27x -1 + 1 +1)]/[x(27x + 2)] ---- {Putting the value of equation (i)]

= [27x(27x + 2)]/[x(27x + 2)]

= 27

= (cos³ x/sin³ x) sin² x + cos² x cot³ x

= cot³ x sin² x + cos² x cot³ x

= cot³ x (sin² x + cos² x)

= cot³ x

=> 5A + 5B = 6A - 6B => A = 11B

=> Let p% of B is equal to A

=> p/100 x B = A

=> But B = 11A

[Sum of opposite angle of a cyclic quadrilateral is 180⁰.]

∠ROQ = 180^∘ − ∠POR = 180⁰ − 115^∘ = 65^∘

In △QOR, ∠OQR = ∠ORQ [Angles opposite or same side] ∠ROQ + ∠OQR + ∠ORQ = 180^∘

65⁰ + ∠ORQ + ∠ORQ = 180^∘

∠ORQ = (180⁰ − 65^∘)/2 = 57.5^∘

According to questions:

a(a + d) = 2160 ...... (1)

(a - d) + a + (a + d) = 135

a = 45

From equation (1)

45(45 + d) = 2160

(45 + d) = 48

d = 3

Required ratio = (a - d) : (a + d) = 42 : 48 = 7 : 8

= 3[{(√9 + √8) - (√9 - √8)}/(√9 - √8)(√9 + √8)] + (2 - 3√2)²

= 3[2√8] + (2 - 3√2)²

= 6√8 + (4 + 18 - 12√2)

= 12√2 + (22 - 12√2)

= 22

294 + A = 343

A = 49 kg

Arrange the weights in ascending order = 42, 45, 48, 49, 50, 54, 55

Median of their weights = 4^th term in series = 49 kg

22x + 22y = 682

x + y = 31 → (1)

22x x 22y = 2376 x 22

xy = 108 → (2)

Solving (1) and (2), we get

x = 27 and y = 4

Difference = 22x - 22y

= 22(27 - 4)

= 22 x 23

= 506

(A) Amount of milk and water in first mixture is 60 x (7/12) = 35 litres and 60 x (5/12) = 25 liters respectively.

(D) Amount of milk and water in second mixture is 40 x (3/8) = 15 liters and 40 x (5/8) = 25 liters respectively.

(B) Total amount of milk and water is final mixture is (35 + 15) = 50 litres and (25 + 25) = 50 liters respectively.

(E) Cost price of final mixture = 50 x 1 = 50 and selling price of final mixture = 100 x 1 = 100

(C) Per cent profit earned = [(100 - 50)/50] x 100 = 100%

⇒ cosec θ - cot θ = 1/k

On adding, we get

2 cosec θ = k + 1/k = 2(x + 1/4x) = k + 1/k

⇒ x + 1/2x = k + 1/k ⇒ k = 2x

⇒ present age of K is 63 - x

7 years later

R's age = x + 7

K's age = 63 - x + 7= 70 - x

⇒ (x + 7) / (70 - x) = 7 / 4

⇒ 4x + 28= 490 - 7x

⇒ 11x = 462

⇒ x = 42 years

therefore present age of R = 42 years

Days taken by 1 girl to embroider a dress = 9 x 5 days

Days taken by 3 girls to embroider a dress = 9 × 5/3 days = 15 days