**CAT Questions on Heights and Distance**

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**Question 1: **Two boats, traveling at 5 and 10 kms per hour, head directly towards each other. They begin at a distance of 20 kms from each other. How far apart are they (in kms) one minute before they collide.

a) 1/12

b) 1/6

c) 1/4

d) 1/3

**Question 2: **At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hr less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 miles round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour?

a) $\frac{7}{3}$

b) $\frac{4}{3}$

c) $\frac{5}{3}$

d) $\frac{8}{3}$

**Question 3: **A truck travelling at 70 kilometres per hour uses 30% more diesel to travel a certain distance than it does when it travels at the speed of 50 kilometres per hour. If the truck can travel 19.5 kilometres on a litre of diesel at 50 kilometres per hour, how far can the truck travel on 10 litres of diesel at a speed of 70 kilometres per hour?

[CAT 2000]

a) 130

b) 140

c) 150

d) 175

**Question 4: **On a 20 km tunnel, connecting two cities A and B, there are three gutters (1, 2 and 3). The distance between gutters 1 and 2 is half the distance between gutters 2 and 3. The distance from city A to its nearest gutter, gutter 1, is equal to the distance of city B from gutter 3. On a particular day, the hospital in city A receives information that an accident has happened at gutter 3. The victim can be saved only if an operation is started within 40 min. An ambulance started from city A at 30 km/hr and crossed gutter 1 after 5 min. If the driver had doubled the speed after that, what is the maximum amount of time would the doctor get to attend the patient at the hospital. Assume 1 min is elapsed for taking the patient into and out of the ambulance?

a) 4 min

b) 2.5 min

c) 1.5 min

d) The patient died before reaching the hospital

**Question 5: **A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is 3/8 of the distance AB measured from the entrance A. When the train whistles the cat runs. If the cat moves to the entrance of the tunnel A, the train catches the cat exactly at the entrance. If the cat moves to the exit B, the train catches the cat at exactly the exit. What is the ratio of speed of train and cat ?

a) 3 : 1

b) 4 :1

c) 5 : 1

d) None of these

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**Question 6: **A man travels three-fifths of a distance AB at a speed 3a, and the remaining at a speed 2b.If he goes from B to A and return at a speed 5c in the same time, then

a) $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$

b) $a + b = c$

c) $\frac{1}{a} + \frac{1}{b} = \frac{2}{c}$

d) None of these

**Instructions**Boston is 4 hr ahead of Frankfurt and 2 hr behind India. X leaves Frankfurt at 6 p.m. on Friday and reaches Boston the next day. After waiting there for 2 hr, he leaves exactly at noon and reaches India at 1 a.m. On his return journey, he takes the same route as before, but halts at Boston for 1hr less than his previous halt there. He then proceeds to Frankfurt.

**Question 7: **If his journey, including stoppage, is covered at an average speed of 180 mph, what is the distance between Frankfurt and India?

a) 3,600 miles

b) 4,500 miles

c) 5,580 miles

d) Data insufficient

**Question 8: **The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is

a) 30

b) 28

c) 32

d) 26

**Question 9: **A boat, stationed at the North of a lighthouse, is making an angle of 30° with the top of the lighthouse. Simultaneously, another boat, stationed at the East of the same lighthouse, is making an angle of 45° with the top of the lighthouse. What will be the shortest distance between these two boats? The height of the lighthouse is 300 feet. Assume both the boats are of negligible dimensions.

a) $300$ feet

b) $\frac{600}{\sqrt{3}}$ feet

c) $\frac{300}{\sqrt{3}}$ feet

d) $600$ feet

e) None of the above

**Question 10: **Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, A and B, are connected to the top of the cone and can fill the empty in 8 hours and 12 hours, respectively. The outlet pipe C, connected to the bottom, can empty a filled cone in 4 hours. When the cone is completely filled with water, all three pipes are opened. Two of the three pipes remain open for 20 hours continuously and the third pipe remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?

a) Pipe A was open for 19 hours.

b) Pipe A was open for 19 hours 30 minutes.

c) Pipe B was open for 19 hours 30 minutes.

d) Pipe C was open for 19 hours 50 minutes.

e) The situation is not possible.

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**Answers & Solutions:**

**1) Answer (C)**

The relative speed is 15 km/hr = 15 km/60 min = 0.25 km/min = 250 m/min.

Therefore, one minute before they collide, they are at a distance of 250m.

**2) Answer (D)**

$12/(R – S) = T$

$12/(R + S) = T – 6$

$12/(2R – S) = t$

$12/(2R + S) = t – 1$

=> $12/(R – S) – 12/(R + S) = 6$ and $12/(2R – S) – 12/(2R + S) = 1$

=> $12R + 12S – 12R + 12S = 6R^2 – 6S^2$ and $24R + 12S – 24R + 12S = 4R^2 – S^2$

=> $24S = 6R^2 – 6S^2 and 24S = 4R^2 – S^2$

=> $6R^2 – 6S^2 = 4R^2 – S^2$

=> $2R^2 = 5S^2$

=> $24S = 10S^2 – S^2 = 9S^2$

=> $S = 24/9 = 8/3$

**3) Answer (C)**

If the truck is being driven at 70 kmph, it takes 1.3 liters of diesel to travel 19.5 km.

Therefore, with 10 liters of diesel, the truck can travel 10/1.3 * 19.5 km = 150 km.

**4) Answer (C)**

Let the distance between gutter 1 and A be x and between gutter 1 and 2 be y.

Hence, x + y + 2y + x = 20 => 2x+3y=20

Also x = 30kmph * 5/60 = 2.5km

Hence, y = 5km

After the ambulance doubles its speed it goes at 60kmph i.e. 1km per min. Hence, time taken for the rest of the journey = 15*2 + 2.5 = 32.5

It takes 1 min to load and unload the patient.

Hence, total time = 5 + 32.5 + 1 = 38.5 mins

So, the doctor would get 1.5 min to attend to the patient.

**5) Answer (B)**

Let the length of the tunnel be x and distance of the train to entrance A be y. Let the speeds of train and cat be t and c respectively.

Hence, when the cat runs 3x/8, the train covers y.

=> (3x/8)/c = y/t — (1)

When the cat runs 5x/8 to the other end, the train covers x+y

=>(5x/8)/c = (x+y)/t —(2)

Taking ratio of (1) to (2)

3/5 = y/(x+y) => 3x = 2y —(3)

Substituting (3) in (1)

(2y/8)/c = y/t

=> t = 4c

Hence the ratio t:c is 4:1.

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**6) Answer (C)**

$\frac{\frac{3x}{5}}{3a} + \frac{\frac{2x}{5}}{2b} = \frac{2x}{5c}$

(Where x is distance between A and B; $\frac{\frac{3x}{5}}{3a}$ = time taken to cover the distance with speed 3a ; $\frac{\frac{2x}{5}}{2b}$ = time taken to cover the distance with speed 2b; $\frac{2x}{5c}$ = time taken to cover the distance x from B to A then return.)

$\frac{\frac{3x}{5}}{3a} + \frac{\frac{2x}{5}}{2b} = \frac{2x}{5c}$

Or $\frac{1}{a} + \frac{1}{b} = \frac{2}{c}$

**7) Answer (B)**

X leaves Frankfurt at 6 PM and reaches Boston at 10 AM, which is 6 AM in Frankfurt => 12-hour journey

Leaves Boston at 12 PM ad reaches India at 1 AM, which is 11 PM in Boston => 11-hour journey

=> Total time = 12 + 2 + 11 = 25 hours

Average speed = 180 mph

=> Distance = 25 * 180 = 4500 miles

**8) Answer (C)**

Let the average height of 22 toddlers be 3x.

Sum of the height of 22 toddlers = 66x

Hence average height of the two toddlers who left the group = x

Sum of the height of the remaining 20 toddlers = 66x – 2x = 64x

Average height of the remaining 20 toddlers = 64x/20 = 3.2x

Difference = 0.2x = 2 inches => x = 10 inches

Hence average height of the remaining 20 toddlers = 3.2x = 32 inches

**9) Answer (D)**

One boat is stationed to the North of the light house and the other boat is stationed to the East of the light house.

The boat stationed to the East subtends an angle of 45 degrees and the boat stationed to the North subtends an angle of 30 degrees.

Now, distance between the boat stationed to the East and the light house,d1 = tan 45

$300/d1 = 1$

=> $d1 = 300$ feet

Distance between the boat stationed to the North and the light house,d2 = tan 30

$300/d2 = 1/\sqrt{3}$

=> $d2=300\sqrt{3}$

Shortest distance between the 2 boats = $\sqrt{300^2+(300\sqrt{3})^2}$

= $300*\sqrt{4}$

= $600$ feet.

Therefore, option C is the right answer.

**10) Answer (C)**

Height of cone comes down to 50%, => it becomes $\frac{1}{2}$

=> Volume would become $\frac{1}{8}$ as radius will also become half by similar triangles.

Let the capacity of cone = 24 litres

Volume of water run-off = $24 – \frac{1}{8} \times 24 = 21$ litres

Volume of water left in the cone = $ \frac{1}{8} \times 24 = 3$ litres

Pipe A’s efficiency = $\frac{24}{8} = 3$ litres/hr

Pipe B’s efficiency = $\frac{24}{12} = 2$ litres/hr

Pipe C’s efficiency = $\frac{24}{-4} = -6$ litres/hr

All will run 19 hours simultaneously (going by the options)

=> Net effect = $(3 + 2 – 6) \times 19 = -19$ litres

This means that after 19 hours, 19 litres of water has been removed, we need to remove 2 more litres as per the requirement. Thus, C will definitely run for another hour.

If we run A and C together for the 20th hour, net effect = $(3 – 6) \times 1 = -3$ litres

Run B for 30 minutes => $2 \times \frac{1}{2} = 1$ litres

$\therefore$ Volume of water removed = $-19 -3 + 1 = -21$ litres

Thus, Pipe B was open for 19 hours 30 minutes.

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