RRB JE IT (CBT I) Mock Test- 2 - Computer Science Engineering (CSE) MCQ

(By mid-point theorem)

DE ∥ BC

DE = 1/2 BC

⇒ ar(∆BDE) = 1/4 × ar(∆ABC)

And ∆BDE = ∆PED

[ ∵ both triangles lie on the same base DE and between two parallel lines DE and BP.]

∴ ar(∆PED) = 1/4 × ar(∆ABC)

Let ∠B be right angle in △ABC

and BD be perpendicular on hypotenuse AC then

1/p² = 1/6² + 1/6²

P² = 6²/2

⇒ p = 3√2

Speed of both trains = 3 : 4

x 3 ↓ : ↓ x 3 → time

Length = 9m : 12m

Time = D/S = = =

= 3 sec.

Daughter’s one-day work = 1/15

Mother’s one-day work = 1/5

Both can complete their work = 1/5+1/15 = (3+1)/15 = 4/15

Hence, required days= 15/4 days .

Total no. of soldiers joining BSF in 2004, 2005, 2006

= 13400

Total no. of soldiers joining in Navy = 10900

Required Percentage= (13400/10900)×100 = 123%

tan (5x – 10°) = cot (5y + 20°)

tan (5x – 10°) = tan (90° – {5y + 20°})

5x – 10° = 90° – (5y + 20°)

5x + 5y = 90° + 10° – 20°

5x + 5y = 80°

x + y = 16°

Speed = 60 km/hr

S = 60 x 5/18 m/sec = 50/3 m/sec.

T = 30 sec.

Length = S x T = (50/3 x 30)m

L = 500m

Let the measure of the angle be x°

Then, the measure of its supplementary angle is (180 - x^o).

It is given that,

180^o - x = 1/4 X x

⇒ (180^o - x) = x

⇒ 720 - 4x = x

⇒ 5x = 720

⇒ x = 144

Thus, the measure of the angle is 144^o and the measure of its supplement

= 180^o - 144^o = 36^o

Total age of 5 members, three years ago

= 17 × 5 = 85 years

Three years hence,

Total age of 5 members = 85+3x5 = 85+15 = 100 years

Sum of present ages of 6 members = 17 × 6 = 102 years

Present age of baby= 102 – 100 = 2 years

Let the length of the ladder be x meter. We have

8² + y² = x² and (y + 2) = x

Hence, 64 + (x - 2)² = x²

⇒ 64 + x² - 4x + 4 = x²

⇒ 68 = 4x ⇒ x = 17 meter

33.33 = 1/3 → 1/2

=

The interests accrued each year on compound interest form a geometric progression with I, as the first-year interest and common ratio of (1 + r/100).

Therefore, interests for 1^st, 2^nd, 3^rd, 4^th and 5^th years will be

respectively

Given, (1 + r/100)³ = 13310 ______________(1)

and (1 + r/100)⁴ = 14641 ______________(2)

(2) divided by (1), we get

1 + r/100) = 14641/13310 = 1.1

Therefore, Sum of the first three years interests

= 13310/(1.1)³ + 13310/(1.1)^2.+ 13310/(1.1)¹

= 13310/1.331 + 13310/1.21 + 13310/1.1

= 10000 + 11000 + 12100

= Rs. 33100

P =

P = Rs. 20,000

cot ? = tan (90 - ?) and cos ? = Sin (90 - ?)

Hence

1 - 1/2 = 1/2

Condition 1- A is twice as fast as

B means their ratio is 2: 1.

Condition 2- B is one third as fast

as C, means their ratio is 1: 3. Then we can say, the ratio of A, B and C is as follows

According to condition 1,

A : B = 2 : 1

According to condition 2;

B : C = 1 : 3

Thus, the ratio of A: B: C = 2: 1 : 3

This ratio denotes the working efficiency of A, B and C which means;

A, B and C can do = (2 + 1 + 3 = 6) work/day

Now we can calculate total work i.e.

Total work = Total days × Per day work

Total work = 30 × 6 = 180

A can do the same work in;

= (Total work / Efficiency of A) = 180 / 2 = 90 days

B can do the same work in;

= (Total work / Efficiency of B)

180 / 1 = 180 days

C can do the same work in;

= (Total work / Efficiency of C) = 180 / 3 = 60 days

Thus, A, B and C can individually complete the same work in 90, 180 and 60 days, respectively.

Minimum value of (a+ b) can be obtained when a= b

Then a + b = 5 + 5 (∴ ab= 25 and a=b)

=10

Given, a + b + c = 1

Let observe the value of a, b, c

a = 1 / 3, b = 1 / 3, c = 1 / 3

a + b + c

1 / 3 + 1 / 3 + 1 / 3 = 1

3 / 3 = 1

1 = 1

Put the values of a, b, c

(1 + a) (1 + b) (1 + c)

4 / 3 x 4 / 3 x 4 / 3

64 / 27

= 2.37

1^st discount

S.P. = Rs. 260 - 15% Rs.260

= 260 - 39 = 221

2^nd discount

S.P. = Rs.221 - 20% Rs. 221

= 221 - 44.2 = 176.8

Sales tax = 5 / 100 x 1768 / 10

= 8840 / 1000

= Rs. 8.84

Amount paid by buyer

= Rs. 176.8 + Rs. 8.84

= Rs. 185.64

Area of surface to be cemented = 2 x (l+b) x h + (lxb)

i.e, area of four walls + area of the floor

= 2 x (21) x 4 + (106.25)

= 274.25 m²

∴ cost of cementing = 24 x 274.25 = Rs. 6582

Let the goat be tied at P.

The area over which the goat could graze = sum of the areas of the regions A and B = Area of a

The sector of radius 14 m and central angle 270°

=

=

=

= 462 sq.m.

Let the cost price of the article = Rs. x

S.P. at 10% loss = Rs. 0.9x

According to question if it is sold by Rs 9 more there will be gain of 25 / 2 %

0.9x + 9 =

0.9x + 9 =

⇒ 180x + 1800 = 225x

⇒ 225x - 180x = 1800

⇒ 45x = 1800

⇒ x = Rs. 40

Alternate Solution,

Let the CP of the article = 100

x X 22.5 = 9

x = 9 / 22.5

= 18 / 25

Required,

⇒ CP X x

⇒ 100 x 18 / 45

⇒ 40

Volume = 6 x 5 x 3.4

Volume = 102 cu.cm

1 cu.m = 100 cu.cm

102 cu.cm = 102 x 100

= 10200 cu.cm

1000 cu.cm = 1 litre

1cu.cm = 1 / 1000

10200 cu.cm = (1 / 1000) x 10200

10.2 litres

Its area = a²

Area of square on the diagonal = (√2a)² = 2a²

Required ration= a² / 2a² = 1: 2

=

=

=

=

=

=

= sin θ

∠ACD = ∠B + ∠A

= 90° + 30°

= 120° (exterior angle)

C.I. =

⇒ 1261 =

⇒ 1261 =

⇒ 1261 =

= 1261x / 8000

⇒ x =

= Rs. 8000

2^x-1 + 2^x+1 = 320

2^x-1(1 + 2²) = 320

2^x-1 = 320

2^x-1 = 320 / 5 = 64

2^x-1 = 2⁶

x - 1 = 6

x = 7

sin²25^o + sin²62^o = sin²25^o + sin²(90^o - 25^o)

= sin²25^o + cos²25^o = 1

As per the question, A does twice the work as done by B.

So A:B = 2:1

Also (A+B)'s one-day work = 1/14

To get days in which B will finish the work,

let's calculate work done by B in 1 day

=(1/14) x (1/3) = 1/42

So B will finish the work in 42 days and A will finish the work in 21 days.