RRB JE ECE (CBT I) Mock Test- 3 Free Online Test 2026

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 the right angle in ∆ABC

And BD be perpendicular on hypotenuse AC then

Speed of both trains = 3 : 4

x 3 ↓ : ↓ x 3 → time

Length = 9m : 12m.

= 3 sec.

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 .

= 13400

Total no. of soldiers joining in Navy = 10900

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

Required ratio = 6.5/1.3 = 5/1 = 5:1

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

5x – 10° = 90° – (5y + 20°)

5x + 5y = 90° + 10° – 20°

5x + 5y = 80°

x + y = 16°

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

T = 30 sec.

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

/ = 500m.

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

It is given that,

180° - x = 1/4 * x

⟹ (180° - x) = x

⟹ 720 - 4x = x

⟹ 5x = 720

⟹ x = 144

Thus, the measure of the angle is 144° and the measure of its supplement

= 180° - 144° = 36°.

= 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

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 1st, 2nd, 3rd, 4th and 5th year will be:

I respectively.

(2) divided by (1). we get

∴ Sum of the first three years interests

= 10000 - 11000 + 12100

= Rs. 33100

P = Rs. 20,000

Hence

1 - 1/2 = 1/2

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.

Then a + b = 5 + 5 (∴ ab = 25 and a = b)

= 10

Let observe the value of a, b, c

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

a + b + c = 1

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

3/3 = 1

1 = 1

put the value of a, b, c

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

(1 + 1/3 ) (1 + 1/3 ) (1 + 1/3 )

4/3 x 4/3 x 4/3

64/27 = 2.37

1st discount

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

= 260 - 39 = 221

2nd 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.

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°

S.P. at 10% loss = x × 90/100 = 9x/10

at % gain

S.P. =

According to the question,

⇒ 180x + 1800 = 225x

⇒ 225x – 180x = 1800

⇒ 45x = 1800

⇒ x = $ 40

We can find required answer with the help of given formula :

Here, a = 10%, R = 9, b = 12.5%

C.P. =R × 100/a + b

C.P. =12 × 100/12.5 + 10

C.P. =900/22.5 = $ 40

Volume = 102 cu. cm

1 cu. m = 100 cu. cm

102 cu. m = 102 x 100

= 10200 cu. cm

1000 cu. cm = 1 litre

1 cu. 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 ratio = a²/2a² = 1 : 2

∠ACD = ∠B + ∠A

= 90° + 30°

= 120° (exterior angle)

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

sin² 25° + sin² 65° = sin² (90° – 65°) + sin² 65°

= cos² 65° + sin² 65° = 1.

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.