Mettl Time, Speed and Distance, Boats and Streams Quiz 1

Reqd. Distance

=[2 + \frac{2\times 2}{3} + \left ( \frac{2\times 2}{3} \right )^{2}+\left ( \frac{2\times 2}{3} \right )^{3} +  .... \infty ]

If S∞ represents the sum of the ∞ terms of a geometric series

So, S∞ = a + ar + ar2 + ar3 + ar4 +.... + ∞

The sum to infinity is given by

S_{\infty} = \frac{a}{1 - r} only if  –1 < r < 1

Here a = 2 and common ratio r = 2/3. On putting the values, we get

=2\times \frac{1}{1-\frac{2}{3}} = 2\times 3 = 6km

Let the speed of the car from Patna be x and the speed of the car from Gaya be y.

Then,\frac{110}{x + y} = 1

So, x + y = 110....(i)

And,\frac{110}{x - y}= 11

∴ x – y = 10 ....(ii)

From equation (i) and (ii), we get

x = 60 kmph, y = 50 kmph

So, the speed of car from Gaya = 50 kmph

1 Bike + 1 Walk = 5 hours 40 minutes

2 Bike = 5 hours 40 minutes – 3 hours = 2 hours 40 minutes

Hence, 1 way Bike journey takes time = 2 hours 40 minutes ÷ 2 = 1 hour 20 minutes

So, 1 way walk should take (5 hours 40 minutes) – (1 hour 20 minutes) = 4 hours 20 minutes

2 way walk would take time = 4 hours 20 minutes × 2 = 8 hours 40 minutes

During the period 11:00 am to 11:30 am,

Distance covered by Prem = 72\times \frac{1}{2} = 36km

(∵  Speed of Prem is 72 kmph)

Then in 4 hours Shyam requires to increase the speed to catch up Prem by

=\frac{36}{4} = 9 kmph

∴ Shyam's speed = 72 + 9 = 81 kmph.

As the distances are in Arithmetic Progression

∵ Sum of n terms in A.P.=\frac{n}{2}{2a + (n – 1)d}[/altex]<br /> <br /> ∴ Total Distance = \frac{n}{2}{2a + (n – 1)d}

Given, n = 10, a = 56, d = 6

=\frac{10}{2}{2 × 56 + (10 – 1)6}

= 5(112 + 54) = 830 km

Downstream speed = Speed in still water + Speed of stream

7 = Speed in still water + 1

Speed in still water = 6

Upstream speed = Speed in still water – Speed of stream

Upstream speed = 6 – 1 = 5 km/hr

Downstream Speed of boat = 10 + 3 = 13 kmph

Time taken =\frac{\text{ Distance }}{\text{ Speed }}= \frac{52}{13} = 4 hrs.

Given,

Downstream rate per hour= \frac{30}{6}= 5 km/h

Upstream rate per hour = \frac{12}{6}= 2 km/h

By the short trick approach, we get

Speed of the current = \frac{1}{2}

[DOWNSTREAM rate – UPSTREAM rate]

\frac{1}{2}(5 - 2) = \frac{3}{2}

Given,

Downstream time = 3 h 45 mins = \frac{15}{4}

So, Speed of downstream =\frac{22.5}{15/4} = \frac{22.5\times 4}{15}[/latex ]= 6 km/hr<br /> <br /> Upstream time = 2 h 30 mins = \frac{5}{2}h

Now, Speed of upstream =\frac{10}{5/2} = \frac{10\times 2}{5} = 4 km/hr

By the short trick approach, we get

Speed of the current =\frac{1}{2}[DOWNSTREAM rate – UPSTREAM rate]

=\frac{1}{2}\times (6 – 4) = \frac{2}{2}= 1km/hr

Let speed of the stream be x.

Rate downstream = 6 + x

Rate upstream = 6 – x

\frac{36}{6 -x} -

⇒ 6 + x - 6 + x = 36 - x2

⇒ 2x = 36 - x2

⇒ x2 + 2x – 36 = 0

By Sridharacharya's formula, roots of quardatic eq. ax2 + bx + c = 0 will be

\frac{\pm b \pm \sqrt{b^{2} - 4ac}}{2a}

Here, a = 1, b = 2 and c = – 36

Therefore,

x = \frac{-2 \pm \sqrt{(2)^{2} - 4\times 1\times (-36)}}{2\times 1}

=\frac{-2 \pm \sqrt{148}}{2}

=\frac{-2\pm 2\sqrt{37}}{2}

\frac{2(-1 \pm \sqrt{37})}{2} = -1 \pm \sqrt{37} = \sqrt{37} - 1