RRB JE ME (CBT I) Mock Test- 7 - Mechanical Engineering MCQ

x = 4 must satisfy the given equation

4⁴ - 3(4)³ + 5(4)² - 6(4) + 2k = 0

256 - 192 + 80 - 24 + 2k = 0

⇒ 120 + 2k = 0 ⇒ 2k = -120

⇒ k = -60

5, 7, 7, 8, 12, 14, 16, 18, 21

Total terms = 9(Odd)

Median = [(n + 1)/2]th term

⇒ [(9 + 1)/2]th term = 5th term = 12

Speed of the boatman downstream = 2/18 x 60 = 20/3 km/hr

Rate of current = 1/2 (DownstreamSpeed−UpstreamSpeed)

= (1/2) (20/3 - 6) = 1/3 km/hr

(x+y) = 60/4 = 15

x = 2y

y = 5Km/h

x = 10Km/h

Distance Travelled in 2 hrs going Upstream

= 2 × 5 = 10km

AM = 5√3 cm

Let AB, BC and AC are 5x, 8x and 7x respectively.

s = (a + b + c)/2 = (5x + 8x + 7x)/2 = 10x

Area of △ABC = √[s(s - a)(s - b)(s - c)] = (1/2) * BC * AM

= √[10x(10x - 5x)(10x - 8x)(10x - 3x)] = (1/2) * 8x * (5√3)

= 10x² * √3 = 4x * 5√3

= x = 2

Hence, AB, BC and AC are 10 cm, 16 cm and 14 cm respectively.

Area of △ABC = (1/2) * 16 * 5√3

= 40√3 cm²

Circum-radius = R = (abc)/4△

= (10 * 16 * 14)/(4 * 40√3)

= (14/√3) cm

Compound interest = 1000(1+0.08)3 - 1000 = 1259.712 - 1000 ≈ Rs. 260

(x² + 1) = 4x

(x² - 4x) = -1

(x² - 4x + 6)

-1 + 6 = 5

= 5

New fraction = (4a+8)/(5a-6) = 4/3

⇒ 12a + 24 = 20a - 24

⇒ 8a = 48

⇒ a = 6

Hence the fraction is 24/30

Difference between numerator & denominator = 30 - 24 = 6

= (1/4) of (2/5) of [{(4 − 3) X 2} - 2 + √(21.6 ÷ 3 X 5)]

= (1/4) of (2/5) of [2 - 2 + √(7.2 X 5)]

= (1/4) of (2/5) of [√36]

= (1/4) X (2/5) X 6

= 3/5

(A + B) = 94 ........ (1)

Sum of first four numbers = (167 + 98 + 23 + A)/4 = 84

288 + A = 336

A = 48

From equation (1)-

48 + B = 94

B = 46

Required difference = 48 - 46 = 2

= 9 hours 15 min.= 555 min.

Now, 37/12 min. of this watch = 3 min. of the correct watch

555 min. of this watch

of the correct watch.

Correct time is 9 hours after 7 a.m. i.e . 4 p m.

Let the four parts are 'a', 'a + d', 'a + 2d' and 'a + 3d' respectively.

According to question-

a + (a + 3d) = 12

(2a + 3d) = 12 ...... (1)

a(a + 3d) = 27 .... (2)

a + (27/a) = 12

a² - 12a + 27 = 0

(a - 9)(a - 3) = 0

a = 3 and 9

From equation (1) when a = 3

6 + 3d = 12

d = 2

From equation (2) when a = 9

18 + 3d = 12

d = -2 [Not possible as numbers are in increasing AP]

Required number = a * (a + d) * (a + 2d) * (a + 3d) = 3 * 5 * 7 * 9 = 945

Portion of the cistern filled by pipe of diameter 1cm = 1/61 X (1/2)2 = 1/61 X 1/4

Portion of the cistern filled by pipe of diameter 1(1/3cm) = 1/61 X 1/4 X (4/3)2=1/61 X 4/9

When all the three pipes are open, the portion of the cistern filled is = 1/61+1/61 X 1/4+ 1/61 X 4/9

1/61+1/61 X 1/4+ 1/61 X 4/9 = 1/36

Therefore the time taken is 36 minutes.

(a+b)/100 x (a-b) = 2/100 x (a+b)/2........(i)

Or

a-b = 1 .............(ii)

also

a² = 1.44 b² ..........(iii)

So,

a = 1.2 b ............(iv)

Using (ii) and (iv)

0.2b = 1

b = 5

a = 6

So, a+b = 11

Required percentage = (11 - 1)/1 x 100 = 1000%

B:C = 1/5:1/3 = 3:5

C:D = 1/6:1/7 = 7:6

A:B:C = 8:3:5

C:D = 7:6

A:B:C:D = (8*7):(3*7):(5*7):(6*5) = 56:21:35:30

Therefore (A+B+D):(B+C+D) = (56+21+30):(21+35+30) = 107:86

Junior Year students in College A = (5/9)*1800 = 1000

Total number of students in College B = 20% of 12000 = 2400

Senior Year students in College B = (2/5)*2400 = 960

Required sum = 1000 + 960 = 1960

So total population = 5 X 4p = 20p

And number of adults = 3/5 of 20p = 12p

Total number of females = 7/12 of 12p = 7p

So the required ratio = 4p : 7p = 4 : 7

x = 6

The sum of their squares = 6² + 18² + 30² = 1260

Half the sum of their square = 630

Required answer = 630

Then, 27a + 27b = 216

⇒ a + b = 8

Now, co-primes with sum 8 are (1, 7) and (3, 5).

∴ Required number are (27 X 1, 27 X 7) and (27 X 3, 27 X 5) i.e.

(27, 189) and (81, 135);

Out of these, the one given in the answer choices is the pair (27, 189).

b² = (-21)² = 441

4ac = 4 * 4 * 35 = 560

Since, b² < 4ac. So, roots of this equation are imaginary.

Similarly, roots of the equation: Ax² + 2x + C = 0 will also be imaginary.

b² < 4ac

(2)² < 4 * A * C

AC > 1

Hence total runs scored from Rohit by running between wickets = (132 - 66) runs = 66 runs.

Therefore required % = 66/132 X 100 = 50%

⇒ 3cos A + 2(1 - cos²A) = 0

⇒ 2cos²A - 3cos A - 2 = 0

⇒ 2cos²A + cos A - 4cos A - 2 = 0

⇒ cos A(2cos A + 1) - 2(2cos A + 1) = 0

⇒ (2cos A + 1)(cos A - 2) = 0

∴ cos A = −1/2 or 2

cos A ≠ 2 [−1≤cos A≤1]

∴ cos³A = (−1/2)³ = −1/8

The correct order is:

(C) Amount of discount given on A = 15% of 400 = Rs.60

(A) Amount of discount given on B = 2 * 60 = Rs.120

(D) Total cost price = 300 + 300 = Rs.600 and Total selling price = (400 - 60) + (500 - 120) = Rs.720

(B) Profit per cent = [(720 - 600)/600] * 100 = 20%

= [5 X {25 ÷ 5}] ÷ 2.5 X 2 - 4 + 8

= [5 X 5] ÷ 2.5 X 2 - 4 + 8

= 25 ÷ 2.5 X 2 - 4 + 8

= 10 X 2 - 4 + 8

= 20 - 4 + 8

= 24

Together six men and four boys can complete a piece of work in 8 days.

Thus the total work = 8(6m + 4b)

Also the same piece of work can be completed by two men in 'd' days only and by eight boys in (d + 5) days only.

Thus 8(6m + 4b)/8b - 8(6m + 4b)/4m = (d + 5) - d

or (6m/b + 4) - (12 + 8b/m) = 5

or 6(m/b) - 8(b/m) = 13

On solving, we get the value of m/b = 8/3

Thus the efficiency of a boy is 3/8 times that of a man, and hence a boy's efficiency is 62.5% less than that of a man.

∛(5² * 135) = ∛3375 = 15/1 = (Rational)

(√43.2)/5 = (√43.2/25) = √1.728 = √1.2³ = 1.2√1.2 = (Irrational)

√(8² + 3² + 2³) = √(64 + 9 + 8) = √81 = 9/1 = (Rational)

Let maximum marks be 100

Hence his average = 60

Let his marks be 10x, 9x, 8x, 7x, 6x respectively in 5 subjects

i.e. 10x + 9x + 8x + 7x + 6x=60% of 5 x 100

= 60/100 x 500 = 60 x 5 ......(i)

Solving we get his marks were 75, 67.5, 60, 52.5, 45 Since passing marks are 50

He passed in 4 subjects

Ages of A and B after 2 years will be in the ratio 2: 3 respectively, i.e.

(a + 2): (b + 2) = 2: 3

a = (2b - 2)/3

Ages of B and C before two years were in the ratio 7: 8 respectively, i.e.

(b - 2): (c - 2) = 7: 8

c = (8b - 2)/7

Present ages of A and C are in the ratio 5: 9 respectively, i.e.

a: c = 5: 9

9 x (2b - 2)/3 = 5 x (8b - 2)/7

b = 16 years