RRB JE IT (CBT I) Mock Test- 1 - Computer Science Engineering (CSE) MCQ

Let m as maximum marks

Now, according to question

⇒ 0.3 m + 50=320 + 30

⇒0.3 m = 300

m = 1000 marks

Since, the roots of equation 6x² - 25x + p = 0 are real and different. So, its discriminant should be positive.

b² - 4ac > 0

b² > 4ac

(-25)² > 4 * 6 * p

625 > 24p

p < 625/24

p < 26.04

Since the value of p is any integer greater than 18 and less than 26.

So, the possible values of p = 19, 20, 21, 22, 23, 24, 25 and 26.

The total selling price of two bullocks = 9100*2 = Rs 18200

Cost price of 1st bullock = 9100*(100/130) = 7000

According to the question, there is no profit or loss.

Therefore cost price of second bullock = 18200 - 7000 = Rs 11200

Selling price of second bullock = Rs 9100

Therefore loss = 11200 - 9100 = Rs 2100

Hence % loss on second bullock = (2100/11200)*100 = 18.75%

Required Difference = Volume of cylinder - volume of cuboid

= πr²h - lbh

= 22/7 x 6 x 6 x 35 - 35 x 10 x 10

= 3960 - 3500 = 460 cm³

3x6 men = 5x18 women

⇒ 1 man = 5 women

4 men + 10 women = 6 men

M₁D₁ = M₂D₂

⇒ 3x6 = 6D₂

⇒ D₂ = 3 days

Equation of line-

y = mx + c

Where, m = slope and c = y intercept

Required equation of line-

y = (0.8)x + (0.5)

10y = 8x + 5

8x - 10y + 5 = 0

The following series is in the arithmetic progression, as the difference between every two consecutive terms is same. So,

According to the question:

a = 293

d = 305 - 293 = 12

n = 15

an = a + (n - 1)d

= 293 + 14 x 12

= 461

Duration from 11:24 p.m. to 01:36 a.m. = 2 hours 12 minutes = 132 minutes.

22 minutes in faulty watch = 25 minutes of actual time

⇒ 22*(132/22) minutes in faulty watch = 25*(132/22) minutes of actual time

⇒ 132 minutes in faulty watch = 150 minutes of actual time or 2 hours 30 minutes of actual time.

Actual time is 11:24 p.m. + 2 hours 30 minutes = 01:54 a.m.

The Left side of the equilateral triangle is 'a' cm.

Height of equilateral triangle = √[a2 - (a/2)2] = (a√3)/2 = 2√3

a = 4 cm

Area of triangle = (1/2) * side * Height = (1/2) * 4 * 2√3 = 4√3 cm2

Length of circumradius of triangle = abc/4△

= (4 * 4 * 4)/(4 * 4√3)

= (4/√3) cm

Let the no. be N.

Then 87.5% of N - 32.5% of N = 220

⇒ 55% of N = 220

⇒ N = 400

28.5% of 400 = 114

Present age: father =4x son =x

Now, (4x+4)/(x+4) = 3/1

⇒ X= 8.

4 years ago the ages were: father= 32-4=28 son=8-4=4

The required ratio = 28:4 = 7:1

The given geometric progression: 2/3, 2/9, 2/27, 2/81......

a = 2/3

Common ratio, r = (2/9)/(2/3) = 1/3 < 1

Therefore, the sum of infinity is given by:

Sum = a/(1 - r) = (2/3)/(1 - 1/3) = 1

= (12³ - 5⁴ + 2²) ÷ (11² + 2)

= (1728 - 625 + 4) ÷ (123)

= (1107) * (1/123)

= 9

= 3²

= (34 x 289 ÷ 17 + 23 x 22) ÷ (18 x 88 - 79 x 20)

= (578 + 506) ÷ (1584 - 1580)

= 1084 ÷ 4

= 271

Let us assume the fraction as N/D.

So, 7D = 9N + 5 .... (1)

And, 10N = 8D - 8 .... (2)

Solving these two equations we get,

D = 11 and N = 8.

Since we need to find the reciprocal, so the answer is 11/8.

Let X = 6.121212.. (I)

⇒ 100X = 612.121212.. (II)

(II) - (I) gives,

99X = 606

⇒ X = 202/33

The amount after 2(1/4) years

= 25000 * (1 + 20/100)2(1+(20/4)/100)

= Rs 37800

So compound interest= Rs (37800 - 25000) = Rs 12800

Simple interest= (25000 * 20 * (9/4))/100= Rs 11250

Sum of compound and simple interest= Rs (12800 + 11250) = Rs 24050

7 a.m. is directly opposite of 1 p.m.

Angle travelled by hour hand = 180°

Alternate method:

number of hours between 7a.m. and 1p.m. = 6

degrees travelled by hour hand in 1 hour = 30^o

required answer = 6 X 30 = 180^o

Here k = (9 - 4) = (11 - 6) = (15 - 10) = 5

The least number which when divided by 9, 11 and 15 leaves remainders 4, 6 and 10 respectively = (LCM of 9, 11 and 15) - k

LCM of 9, 11 and 15 = 495

Required number = 495 - 5 = 490

Difference between the average marks obtained by Sita and Gita

= (70 + 65 + 90 + 85 + 80)/5 - (60 + 75 + 85 + 90 + 60)/5 = 4

The digit in the unit's place of 2⁵¹ is equal to the remainder when 2⁵¹ is divided by 10.

2⁵ =32 leaves the remainder 2 when divided by 10.

Then 2⁵⁰ = (2⁵)¹⁰ leaves the remainder 2¹⁰ =(2⁵)² which in turn leaves the remainder 2²= 4

Then 2⁵¹= 2⁵⁰x 2, when divided by 10, leaves the remainder 4 x 2=8

F(x) = x⁴ + 6x³ + 7x² - 6x - 8

By hit and trial method, we get -

x = 1, gives F (1) = 0, so, (x - 1) is a factor of F (x)

F (1) = 1 + 6 + 7 - 6 - 8 = 0

Similarly, F(-1)= 0 = 1 + 6(-1) + 7 - 6(-1) - 8 = 1 - 6 + 7 + 6 - 8 = 0

Hence,

(x - 1)(x + 1) i.e. (x² - 1) is a factor of F (x)

On dividing F(x) by (x² - 1), we get,

Hence, (x² + 6x + 8) is other factor = (x+2)(x+4)

Thus, option (c) is correct.

First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Hence, average of all these numbers = (2+3+5+7+11+13+17+19+23+29)/10 = 129/10 = 12.9

Place value of a digit in any number = (face value of the digit) *(value of the place at which digit occurs)

Place value of 7 in '16274529' = 7*10000= 70000

Place value of 3 in '672345' = 300

70000/300 = 700/3

Remainder = 1

The correct order is:

(a) Ratio of milk to water when 30 litres of milk is mixed with 15 litres of water = 30: 15 = 2: 1

(d) Ratio of milk to water when 30 litres of the mixture is mixed with 10 litres of water = [30 * (2/3)]: [30 * (1/3) + 10] = 20: 20 = 1: 1

(c) Ratio of milk to water when 16 litres of the mixture is mixed with 10 litres of pure milk and 4 litres of water = [16 * (1/2) + 10]: [16 * (1/2) + 4] = 18: 12 = 3: 2

(E) Amount of milk and water in the final mixture is 18 litres and 12 litres, respectively.

(b) Percent profit = (12/18) * 100 = 66(2/3)%

Mean = (5 + 8 + 9 + 12 + 12 + 10 + 16 + 8)/8 = 10

Variance = ((5 - 10)² + (8 - 10)² + (9 - 10)² + (12 - 10)² + (12 - 10)² + (10 - 10)² + (16 - 10)² + (8 - 10)²)/8

= 39/4

Standard deviation = √variance = √(39/4) = √39/2

Let the LCM and HCF are p and q, respectively.

So, p + q = 624, p - q = 546

Therefore, p = (624+546)/2 = 585 and q = (624-546)/2 = 39

That means LCM = 585 and HCF = 39

Now, let the numbers be 39a and 39b, where a and b are co-primes.

Therefore 39a + 39b = 312 => a + b = 312/39 = 8

So possible pairs of co-primes, whose sum is 8, are (3,5) and (1,7).

Therefore possible pairs of numbers are

(39*3, 39*5) = (117,195) and

(39*1, 39*7) = (39,273)

Now, LCM*HCF = 585*39 = 22815

Also, 117*195 = 22815 and 39*273 = 10647

Hence, the required numbers are 117 and 195 because the product of two numbers is equal to the product of their LCM and HCF.

After replacing 10 and 12, we get: -

6 + 5 x 4 - 12 x 2 = 9 x 3 - 7 x 5 + 10

6 + 20 - 24 = 27 - 35 + 10

2 = 2.

So, option d is the correct answer.

Number of days from 1^st July to 1^st November = 31+ 31 + 30 + 31= 123

Remainder days = 123 - 119= 4 (119 is divisible by 7)

So,

November 1^st 2012= Sunday + 4= Thursday

Thus, 15^th November = 1+14= Thursday

18^th November = Thursday + 3 = Sunday

Speed ratio = 60 : 90 = 2 : 3

So, time ratio = 3 : 2

So we can assume the time as 3x and 2x.

Hence 3x - 2x = 10 + 5

Since here he is 5 minutes late and 10 minutes early, so the difference is the summation.

So 2x = 30 minutes.

Hence with a speed of 90 km/h, it takes 30 minutes, so the distance is 45 km.