RRB JE ME (CBT I) Mock Test- 10 - Mechanical Engineering MCQ

= 780

Average of 827, 157 and 529 = (827 + 157 + 528)/3 = 504

HCF of 780 and 504 = 12

a + b + c = 54

b = a + 10

c = a - 10

a + a - 10 + a + 10 = 54

a = 18

second number = b = a + 10 = 18 + 10 = 28

b = √1764 + 2√1089 = 108

c = √(14 x 59 - 67 x 3) = 25

d = √(24 x 22 + 16 x 31) = 32

305 seconds in defective watch = 300 seconds of actual time.

⇒ 305*(54900/305) seconds in defective watch = 300*(54900/305) seconds of actual time

⇒ 54900 seconds in defective watch = 54000 seconds of actual time

= 15 hours of actual time

Actual time is 8 a.m. + 15 hours = 11:00 p.m.

Population in 2018 = 2000(1 + 20/100)² = 2880

Population gained during 2018-2019 = 3456 - 2880 = 576

a = 5b

a: b = 5: 1

(A) Total amount after 4 years when invested at 15% SI = 16000 + [(16000 * 15 * 4)/100] = Rs.25600

(D) Total amount after 2 years when invested at 20% CI = 25600 * (1.2)² = Rs.36864

(B) Cost price of article = Rs.36864 and selling price = Rs.40000

(C) Profit amount earned = 40000 - 36864 = Rs.3136

s₁−s₂=d/t

s₁−6=(450/1000)/(4.5/60)

s₁=6+6=12 km/hr

(6*2) : (4*1) : (1*4) = 12:4:4 = 3:1:1

Let them be 3x km, x km and x km respectively.

Given, 3x + x + x = 60

⇒ 5x = 6

⇒ x = 12

difference between the distance travelled by the slowest Biker and the fastest Biker = 3x - x = 2x = 24 km

Five children in ascending order of their present age be a, (a+3) , (a+6) , (a+9) and (a+12)

Now,

2*(a+12+7) = m - 3

m = 2a + 41 ....(i)

15 years hence,

(a+a+3+a+6+15*3)/3 + 23 = 2a + 41 - 3

a + 18 + 23 = 2a + 38

a = 3

So,

Riya = m = 2a + 41 = 47 years

Youngest = a = 3 years

Here,

→ P gets more votes than U, who gets less votes than R.

So,

R > P > U

Or

P > R > U

Now,

→ T gets more votes than S, who gets less votes than P.

So,

T > P > S

Or

P > T > S

Since,

→ P didn't get more votes than both T and R.

So, we get,

R > P > U and T > P > S

So, combining both we get,

R, T > P > U, S

Since,

→ U didn't get the lowest number of votes and R didn't gets the highest number of votes.

Final arrangement is,

T > R > P > U > S

Hence,

Two candidates P and U get more votes than S, but less votes than R

Given, P(20/100)2 = 800

P = 20,000

Difference between CI and SI in 3 years

= P(R/100)2((300 + R)/100) [symbols have their usual meanings]

= 20,000(10/100)2((300 + 10)/100)

= Rs 620

Then,

2x + 3y = 1/10 and 3x + 2y = 1/8

Solving, we get:

X = 7/200 and y = 1/100

Therefore, (2 men and 1 woman)'s 1 day's work = (2 * (7/200) + 1 * (1/100)) = 16/200 = 2/25

Therefore, (2 men and 1 woman) do the work in 25/2 days.

Hence, option c is correct.

Let the number of balls in a bag be P’ In a bag 40% of balls are red. 20% balls are blue, so green balls will be.

B_Green = 100 - (40 + 20) = 100 - 60 = 40%

According to the given conditions.

Number of red balls in the bag is.

Therefore, the number of red balls in the bag is 30.

Arrangement of price in ascending order:

5, 8, 10, 12, 15, 18, 20, 22, 25, 30

Total number of observations = 10 (even number)

Median = ((n/2)^th term + ((n/2) + 1)^th term)/2

= ((5)^th term + ((6)^th term)/2

= (15 + 18)/2 = Rs.16.5

Mean = (5 + 8 + 10 + 12 + 15 + 18 + 20 + 22 + 25 + 30)/10 = Rs.16.5

Difference = 16.5 - 16.5 = Rs.0

16m² = 9n²

⇒ m²/n² = 9/16

⇒ m/n = √(9/16)

⇒ m/n = 3/4

Now, we want, m³% of n³ = (m³/100)*n³

we cannot calculate the value of (mn)³.

So, answer cannot be determined.

⇒ CP of 36 metres + profit of 36 metres = SP of 36 metres

⇒ But profit of 36 metres= SP of 12 metres

CP of 36 metres + SP of 12 metres= SP of 36 metres

CP of 36 metres = SP of 24 metres

⇒ Profit = (36-24)/24*100

⇒ Profit = 12*100/24

⇒ Profit = 50%

Number when divided by 3, 5 and 7 leaves remainder 2 in each case.

So, the number = 105 + 2 = 107

Remainder when '107' is divided by '9' = 107/9 = (99 + 8)/9 = 8

Number of girl pupils initially = 1342 x = 732

Required ratio = (610 -120): (732 + 80) = 490: 812 = 35:58

Let the number of Females be 'x'

Number of Males = 10800 - x

Number of Female literate people = (48/100)*x = 12x/25

Number of Male literate people = 1620 + 12x/25

Given, 1620 + 12x/25 + 12x/25 = 5940

⇒ 24x/25 = 4320

⇒ x = 4500

Number of Males = 10800 - 4500 = 6300

Gender Ratio = 6300:4500 = 7:5

LCM(3k, 4k, 6k) = 36

⇒ k*LCM(3, 4, 6) = 36

⇒ 12k = 36

⇒ k = 3

Difference between the largest and the smallest number = 6k - 3k = 3k = 9

Mean = 954/70 = 13.63

Variance = 24696/70 - 13.632 = 167.02

= (1/9) of [13 X 3 + {36 + (0.5 X 12)}] X (1/8) of [40 + {20 + (0.25 X 16)}]

= (1/9) of [39 + {36 + 6}] X (1/8) of [40 + {20 + 4}]

= (1/9) of [39 + 42] X (1/8) of [40 + 24]

= (1/9) of 81 X (1/8) of 64

= 9 X 8

= 72

(c₁, c₂) = (a₁ + b₁)/2, (a₂ + b₂)/2

(x, y) = (4 - 0)/2, (3 + 7)/2

(x, y) = (2, 5)

Line of equation = y = mx + c [where m = 3 and line passes through (2, 5)]

5 = 3 * 2 + c

c = -1

Required equation of line = y = 3x - 1

3x - y = 1

x² + 4x - 21 = 0

x² +7x - 3x - 21 = 0

x = -7, 3

a = -7, b = 3

Now, (2a + 3b) = 2 x (-7) + 3 x 3 = -5

(2b - 3a) = 2 x 3 - 3 x (-7) = 27

New quadratic equation will be:

x² - (-5 + 27)x + (-5) x 27 = 0

x² - 22x - 135 = 0

Speed of Metro

Let the length of the bridge be x m.

Therefore

9(120 + x) = 200

9x = 2000 - 1080

9x = 920

x = 102.22

1/a of [{6.5 X (√100 − √4)} ÷ 13 X {2.5 X (√625 + √9)} ÷ 7] + 1/3 of √(15 ÷ 3 X 8 ÷ 4 - 1) = 6

⇒ 1/a of [{6.5 X (10 − 2)} ÷ 13 X {2.5 X (25 + 3)} ÷ 7] + 1/3 of √(5 X 2 - 1) = 6

⇒ 1/a of [{6.5 X 8} ÷ 13 X {2.5 X 28} ÷ 7] + 1/3 of √9 = 6

⇒ 1/a of [52 ÷ 13 X 70 ÷ 7] + 1/3 of 3 = 6

⇒ 1/a of [4 X 10] + 1 = 6

⇒ 1/a of 40 = 5

⇒ a = 40/5 = 8

As per given, l² + b² = 89 cm² and lb = 40 cm²

(l +b)² = l² + b² +2lb = 89 + 80 = 169

l +b =13 cm

l = 13-b

As lb = 40

(13-b) x b =40

13b -b² =40

b² - 13b +40 = 0

b = 8 or 5

If b = 8, then l = 5 and if b = 5, then b = 8

as length is greater than breadth. So, b = 5.

Area of square having side equal to breadth of rectangle= 25 cm².

⇒ a - 5 = − 1/2a²

∴ (a - 5)⁴⁹ + 1/2⁴⁹a⁹⁸

= −1/2⁴⁹a⁹⁸ + 1/2⁴⁹a⁹⁸ = 0

Sum of first n odd numbers = n²

Required sum = 80 x 80 = 6400.