RRB JE EEE (CBT I) Mock Test- 10 - Electrical Engineering (EE) MCQ

Minimum required straight lines are - AB, AC, BC, AH, BD, CD, EF, EG, FH, GH, FG, FM, MG, IK, KL and LJ.

Total = 16

1.96 of 1.2 of a = 3.43 of 1.6 of b

1.96 x 1.2 x a = 3.43 x 1.6 x b

3a = 7b

b: a = 3: 7

= 1500 + 1000 + 1000 + 1800 + 2000 + 1500 = 8800

number of Lee Cooper clothes sold

1200 + 1500 + 1800 = 4500

Difference = 8800 - 4500 = 4300

x ÷ [(5/12) of (12/25) of 125 ÷ {√(21 + √16)/2}] = x + 3

⇒ x ÷ [(5/12) of (12/25) of 125 ÷ {√(21 + 4)/2}] = x + 3

⇒ x ÷ [(5/12) of (12/25) of 125 ÷ {√(25)/2}] = x + 3

⇒ x ÷ [(5/12) of (12/25) of 125 ÷ {5/2}] = x + 3

⇒ x ÷ [25 ÷ 2.5] = x + 3

⇒ x/10 = x + 3

⇒ x = 10x + 30

⇒ x = −10/3

There are 9 triangles in the given figure.

Triangles are - ABE, ADE, BLM, BCM, DEJ, DJI, FHM, FGM, HLM.

Mean = (3 + 6 + 8 + 1 + 4 + 6 + 4 + 2 + 2 + 4)/10 = 4

Now, distance of each numbers from their mean are:

4 - 3 = 1

6 - 4 = 2

8 - 4 = 4

4 - 1 = 3

4 - 4 = 0

6 - 4 = 2

4 - 4 = 0

4 - 2 = 2

4 - 2 = 2

4 - 4 = 0

Mean deviation = (1 + 2 + 4 + 3 + 0 + 2 + 0 + 2 + 2 + 0)/10 = 1.6

Variance = (12 + 2² + 4² + 3² + 0² + 2² + 0² + 2² + 22 + 0²)/10 = 4.2

Difference = 4.2 - 1.6 = 2.6

Cyclicity chart:

(Note: when remainder is 0, power = cyclicity)

= (254+145)⁴²

= 399⁴²

Unit digit of 399 is 9.

From the cyclicity chart, the cyclicity of 9 is 2.

So, we have to divide the power of 399 i.e.42, by 2.

After dividing 42 by 2, remainder = 0. So, power = cyclicity = 2

Unit digit of 9²=1

Hence, the unit of (254+145)⁴² is 1.

(11/5) + (9/4) + (7/2)

= (44 + 45 + 70)/20

= 159/20

(132 - 10)

= 169 - 10

= 159

According to the question-

(159/20) ÷ 159

= (159/20) * (1/159)

= 1/20

= 0.05

It will show the correct time again when it gains 24 hours.

It gains 1 hour in 6 hours

It gains 24 hours in 6*24 hours or 6 days.

(6*5) + (4*2) = 38

Similarly,

(7*6) + (1*8) = 50

D = 0 = (-4√3)² - 4 x 3 x m

m = 4

Therefore,

= 3m² - 5m + 16

= 3 x 4 x 4 - 5 x 4 + 16

= 44

1. 6th root of 2187 = 3 x 6th root of 3 (irrational)

2. 5th root of 4096 = 4 x 5th root of 4 (irrational)

3. √(28 x 23 - 283) = 19

4. √(34 x 35 - 23 x 22) = 6√19 (irrational)

5. √4802 = 49√2 (irrational)

= Product of (0.003 x 25 + 4.65) and (0.05 x 24 - 0.2)

= 4.725 x 1

= 4.725

Option (2):

= Sum of (23 x 0.005 + 4 x 0.003) and 3.598

= 0.127 + 3.598

= 3.725

Option (4):

= 2 x 3 x 5.6 x 0.005 + 45.57 x 0.1

= 4.725

Hence, option (2) is odd among all four options.

This is half the work. It means the remaining half is 1260 men weeks.

Work done by 140 men in 5 weeks = 700 men weeks.

So remaining work = 1260 - 700 = 560 men weeks which has to be completed over 7 weeks.

Required number of men = 560/7 = 80

Quantity of Papaya bought is less than that of Oranges as well as Mango.

We get,

Case I:

Papaya < Oranges < Mango

Case II:

Papaya < Mango < Oranges

Now,

Quantity of Apples bought is less than both Kiwi and Banana.

Apple's quantity is more than that of oranges.

We get,

Oranges < Apples < Kiwi/banana < Kiwi/Banana

Since,

Quantity of Mango is more than only one fruits.

And, that of oranges is more than at least two fruits.

Thus, case I becomes invalid.

We get,

Papaya < Mango < Oranges < Apples < Kiwi/banana < Kiwi/Banana

So, Orange's quantity is less than that of three fruits.

Since, P and Q divide AB and AC respectively in the same ratio,

PQ is parallel to BC

In ∆APQ and ∆ABC,

∠APQ = ∠ABC and ∠AQP = ∠ACB [Corresponding angles]

So, ∆APQ and ∆ABC are similar,

Corresponding sides are proportional,

BC/PQ = AB/AP

⇒ 18/PQ = (AP + PB)/AP

⇒ 18/PQ = 1 + PB/AP

⇒ 18/PQ = 1 + ½

⇒ PQ = 18*(2/3) = 12 cm

Total employees = 588*(100/14) = 4200

Number of employees who work in Senior Management

= 6% of 4200 = 252

Area of circle = πR2

R² =706.5/3.14 = 225

R = 15 cm.

Let radius is increased by S cm.

Increase in circumference = (2π x (15 +S)) - (2π x15)

= 2πS

2πS = 31.4

S = 5 cm

New area of circle = π x (20)² = 3.14 x 400 = 1256 cm².

In 1 second, the second hand travels 6°

Time duration betweenthe 02:37:42 p.m. and 02:39:54 p.m.

= 2 minutes 12 seconds.

In every minute, the second hand comes back to its original position.

Angle between initial and final position of second hand

= 12*6 = 72°

(C) Initial amount invested in scheme A and B is 60% of 60000 = Rs.36000 and 40% of 60000 = Rs.24000 respectively.

(E) Interest from scheme A and scheme B after 4 years = [(36000 * 15 * 4)/100 = Rs.21600] and [(24000 * 25 * 4)/100 = Rs.24000] respectively.

(A) Amount invested in scheme A and scheme B for the second time is Rs.60000 and [21600 + 24000 = Rs.45600] respectively.

(D) Interest from scheme A and scheme B after 2 years = [(60000 * 15 * 2)/100 = Rs.18000] and [(45600 * 25 * 2)/100 = Rs.22800] respectively.

(B) Required difference = (21600 + 18000) ~ (24000 + 22800) = Rs.7200

Let X = 0.112112112.. (I)

⇒ 1000X = 112.112112112.. (II)

(II) - (I) gives,

999X = 112

⇒ X = 112/999

⇒ 70/100 x (a - b) = 40/100 x (a + b)

⇒ 7a - 7b = 4a + 4b

⇒ 7a - 4a = 4b + 7b

⇒ 3a = 11b

⇒ a/b = 11/3

⇒ a:b = 11:3

Also, B + C = 15 => B + A = 15

So, (B - A) cannot be determined.

(2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √(80 ÷ √(100 ÷ 8 X 2))}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √(80 ÷ √(12.5 X 2))}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √(80 ÷ √(25))}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √(80 ÷ 5)}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √16}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {25 − 4}]

= (2/3) of (6/7) of (14/15) of [63 ÷ 21]

= (2/3) X (6/7) X (14/15) X 3

= 8/5

let a = 7m , b = 7n

According to the question:

10290 = 7m x 7n x 21

10 = m x n

There are two possible values for m and n.

Case 1: when, m = 10, n = 1

a = 7m = 7 x 10 = 70

b = 7n = 7 x 1 = 7 < 10

So, this case is not valid.

Case 2: when, m = 5, n = 2

a = 7m = 7 x 5 = 35

b = 7n = 7 x 2 = 14

LCM of 35 and 14 = 70

= 13 + 27 - 48 + 4

= - 4

SP of 45 sweets = Rs 40

SP of 1 sweets = 40/45

L = 20%

CP = 100/80 * 40/45 = 10/9

SP₂ = (100 + 20) / 100 * CP

SP₂ = 120/100 * 10/9 = 4/3

SP of Rs 4/3 sweets for 1 sweets

SP of Rs 24 for = 24/ (4/3) = (24 * 3) / 4 = 18 sweets

Then, the arrangement of weights in ascending order are:

a, a + 5, a + 10, a + 15, a + 20

Mean = 30

5 x 30 = a + a + 5 + a + 10 + a + 15 + a + 20

a = 20 kg

Weight of heaviest item = 20 + 20 = 40 kg

Sum of weight of lightest and heaviest item = 20 + 40 = 60 kg

Let the amount lent at SI be Rs x.

For 2 years, SI = x*15*2/100 = 30*x/100

Amount lent at CI = 2000 - x

For 2 years, CI = (2000 - x)*(1 + 10/100)2 - (2000 - x) = (2000 - x)(21/100)

30x/100 + (2000-x)*21/100 = 528

After solving we get x = Rs 1200