Arithmetic(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Question 91.

Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race? (2001)

36 meters
48 meters
60 meters
72 meters

Discuss Question

Answer: Option B. -> 48 meters
:
B

Approach 1: Conventional Approach:
(b) Let the distance of race be x metres which is covered by A in t seconds. Then, in the same time B covers (x-12)metres and C covers (x-18)metres.
$∴ S p e e d o f A = \frac{x}{t} m / s e c,$
$S p e e d o f B = \frac{x - 12}{t} m / s e c,$
$S p e e d o f C = \frac{x - 18}{t} m / s e c .$
Time taken by B to finish the race $= \frac{x}{\frac{x - 12}{t}} = \frac{x t}{x - 12}$ sec.
Now distance travelled by C in this time,
$= x \times t \times \frac{x - 18}{(x - 12) t} = x - 8$
$\to \frac{x (x - 18)}{x - 12} = x - 8 \to x = 48 m .$
Approach 2: Reverse Gear Approach

Going from answer options:

Note the ratio of distance covered should be the same:

Assume option (c) to be correct: then we get the distances travelled to be in the ratio of 60:48:40 and for B and C we get the ratio to be 60:52, here we observe for B and C that the ratio is not the same.

For option B we get A:B:C = 48:36:30 or 8:6:5 for B and C we get 48:40 or 6:5 we see that for both the options we have the same ratio of distance travelled hence this the correct answer option!

Question 92.

A train X departs from station A at 11.00 a. m. for station B, which is 180 km away. Another train Y departs from station B at 11 : 00 a. m. for station A. Train X travels at an average speed of 70 km/hr and not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 km/hr, but has to stop for 15 minutes at station C, which is 60 kms away from station B enroute to station A. Ignoring the lengths of the trains, what is the distance, to the nearest km, from station A to the point where the train crosses each other ? (2001)

112
118
120
None of these

Discuss Question

Answer: Option A. -> 112
:
A
Approach 1:
Option (a) Time taken by B to cover 60 km

= \frac{60}{50} = \frac{6}{5}

hours.
Time taken by Y at station C

= \frac{15}{60} = \frac{1}{4}

hours.
Now distance travelled bt train X in

(\frac{6}{5} + \frac{1}{4}) = \frac{29}{20}

hours

= 70 \times \frac{29}{20} = 101.5

km.
Distance between X and Y when Y starts from station C =180 -(101.5 +60)=18.5 km.
Relative speed =(70+50)=120 km/hr.
Hence, time taken by them in crossing one another

= \frac{18.5}{120} = 0.15

hours. Now, distance travelled by X in 0.15 hour

= 70 \times 0.15 = 10.5

km.
Therefore, distance of X from station A, when they meet =(101.5+10.5)=112 km.
Approach 2:
By using a table to check the distance travelled by each of X and Y:

\begin{matrix} T e x t & X d i s t a n c e & Y d i s t a n c e & T o t a l 12 : 00 & 70 & 50 & 120 12 : 12 & 14 & 10 & 144 12 : 27 & 17.5 & 0 & 161.5 T a l l y & 101.5 \end{matrix}

The last 18.5 kms will have to be divided in the ratio of 70:50. Hence we see that the distance travelled by X is around 112 kms.

Question 93.

A man can walk up a moving 'up' escalator in 30 s. The same man can walk down this moving 'up' escalator in 90 s. Assume that his walking speed is same upwards and downwards. How much time did he take to walk up the escalator, when it is not moving? (1995)

30s
45s
60s
90s

Discuss Question

Answer: Option B. -> 45s
:
B
Let the speed of escalator be y ft per second speed of man's be x ft per second .Let us assume that length of the escalator be 90 ft.
Then,

x + y = \frac{90}{30} \to x + y = 3 . . . . . . . . . . . . (i)

And

x - y = \frac{90}{90} \to x - y = 1 . . . . . . . . . . . . . . . . . . (i i)

\to

x=25 ft/sec.And time taken by the man to walk up bt the escalator when it is not moving

= \frac{90}{2} = 45

second.
Also note : Options a and d can directly be eliminated.

Question 94.

A train approaches a tunnels AB. Inside the tunnel a cat located at a point that is $\frac{3}{8}$ of the distance 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. The speed of the train is greater than the speed of the cat by what order?

3 : 1
4 : 1
5 : 1
None of these

Discuss Question

Answer: Option B. -> 4 : 1
:
B

Approach 1: Conventional Approach :
Let the speed of the train be T, and let it be at a distance of X from the tunnel AB.
Let the speed of the cat be C, and let the distance of the tunnel be Y, hence the cat is at a distance of $\frac{3}{8} Y$ from A.
Now 2 conditions are given:
When the train meets the cat at B it would have travelled a distance of X+Y while the cat will travel $\frac{5}{8} Y$ at its respective speeds,
$\frac{X + Y}{T} = (\frac{5}{8}) \frac{Y}{C} . . . . . . . (1)$
When the train meets the cat at A it would have travelled a distance of X while the cat will travel $\frac{3}{8} Y$ at its respective speeds.
Also, $\frac{X}{Y} = \frac{(\frac{5}{8}) Y}{C} . . . . . (2)$
From (1)
$\frac{X}{T} + \frac{Y}{T} = \frac{5 Y}{5 C}$
Sub form (2)
$\frac{(\frac{5}{8}) Y}{C} + \frac{Y}{T} = \frac{5 Y}{8 C}$
$= \frac{Y}{T} = \frac{2 Y}{8 C}$
or $T : C = 4 : 1$

Approach 2: Shortcut- Assumption

Assume the lengths of the tunnel to be 8 km and the speed of the cat be 8 km/hr

Time taken for the cat to reach A is 3 hr.

Time taken for the cat to reach B is 5 hr.

The difference in time of A and B gives the time taken by the train, which is 2 hours.

Hence the cat takes 8 hours while the train takes 2 hours,

Hence T:C = 4:1

Question 95.

A man travels A to B at a speed x km/hr. He then rests at B for x hours. He then travels from B to C at a speed 2x km/hr and rests for 2x hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hr. If distances A-B, B-C and C-D are all equal to 12 km, the time for which he rested at B could be:

3 hr
6hr
2hr
4hr

Discuss Question

Answer: Option A. -> 3 hr
:
A

Using answer options :
Option (a) total time taken from A to D.
$\frac{12}{x} + x + \frac{12}{2 x} + 2 x + \frac{12}{4 x} = 16$
$\to \frac{21}{x} + 3 x = 16$
$\to 3 x^{2} - 16 x + 21 = 0$
$\to x = 3, \frac{3}{7}$
Alternatively:
Going by the answer options:
Let us assume the answer to be 2 hrs.

$\begin{matrix} FROM-TO & Speed & Time taken to cover 12 kms & Rest time & Cumulative time A - B & 2 & 6 & 2 & 8 B - C & 4 & 3 & 4 & 7 C - D & 8 & 1.5 & 0 & 1.5 T o t a l t i m e t a k e n & 16.5 \end{matrix}$
Let us assume the answer to be 3 hrs.
$\begin{matrix} FROM-TO & Speed & Time taken to cover 12 kms & Rest time & Cumulative time A - B & 3 & 4 & 3 & 7 B - C & 6 & 2 & 6 & 8 C - D & 12 & 1 & 0 & 1 T o t a l t i m e t a k e n & 16 \end{matrix}$

Question 96.

If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?

11 km/hr
12 km/hr
13 km/hr
14 km/hr

Discuss Question

Answer: Option B. -> 12 km/hr
:
B

Using $\frac{1}{x} & \frac{1}{x + 1}$
We know that the speed is increasing by $\frac{1}{2}$ hence the time has to decrease by $\frac{1}{3}$ (using the $\frac{1}{x} & \frac{1}{x + 1})$
shortcut (refer demo tutorial) but the time is actually decreasing by 2 hrs.
Hence the actual time of travel is 6 hrs hence he leaves at 7:00 AM.
The total distance travelled is $10 \times 6 = 60 k m s$
To reach at noon he will have to cover 60 km in 5 hours. i.e., he has to travel at 12 km/hr.
Approach 2: Conventional Approach:
Let the person take t hrs to cover the distance at 10 km/hr.
If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m, which means he takes (t-2)hr.
Let the distance covered be D.
So we have $15 \times (t - 2) = 10 \times (t)$
T=6 , or he started 6 hours before 1, i.e., at 7.
Hence the total distance now is $10 \times 6 = 60$ km.
To reach at noon he will have to cover 60 km in 5 hours. He was to travel at 12 km/hr.

Question 97.

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

$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$
a + b = c
$\frac{1}{a} + \frac{1}{b} = \frac{2}{c}$
None of these

Discuss Question

Answer: Option C. ->

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

:
C

Approach 1: Conventional Approach :
option (c) time taken to cover AC $= \frac{3 x}{5} \times 3 a = \frac{x}{5 a}$ hr.
Time taken to cover CB $= \frac{x}{5 b}$ hr.
Time taken to cover BA and back AB $= \frac{2 x}{5 c}$
Given $\frac{x}{5 a} + \frac{x}{5 b} = \frac{2 x}{5 c} \to \frac{1}{a} + \frac{1}{b} = \frac{2}{c}$
Approach 2: Shortcut using Assumption !
As the questions in terms of variables we can assume values for all the given variables .
Let the distance AB be 5 kms:
The man travels three-fifth of a distance AB at a speed 3a. If we take a=1 we see that he covers a distance of 3 km. The remaining distance of 2 kms is covered in 2b time we get b=1. Also, he goes from B to A and returns at a speed 5c in the same time, we see that c=0.5. Sub these values into the options we find that only (c) fits in. (all the other options get eliminated)

Question 98.

Directions: Answer the questions based on the following information.
A road network (shown figure)connects cities A, B, C and D all road segments are the straight line. D is the mid-point of the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB, the segment AB is 100km long.

Ms.X and Mr Y leave A at 8.00 a.m .., take different routes to city C and reach at the time. X takes the highway from A to B to C and travels at an average speed of 61.875 km/hr. Y takes the direct route AC and travels at 45 km/hr on segment AD. Y's speed on segment DC is 55 km/hr.
The total distance travelled by Y during the journey is approximately:

105 km
150 km
130 km
Cannot be determined

Discuss Question

Answer: Option A. -> 105 km
:
A
(a) Since, X and Y reach C at the same time ,

\frac{100 + B C}{61.875} = \frac{A C}{49.5}

Also,

B C^{2} = A C^{2} - 100^{2}

Now we can assume from answer options, the value of AC.

(105)^{2} - (100)^{2} = 1025 ∴ B C = 32

\frac{100 + B C}{61.875} = \frac{105}{49.5}

Even in this calculation we see that BC is 32.Hence this is the right option.

Question 99.

Directions: Answer the questions based on the following information.
A road network (shown figure)connects cities A, B, C and D all road segments are the straight line. D is the mid-point of the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB, the segment AB is 100km long.

Ms.X and Mr Y leave A at 8.00 a.m .., take different routes to city C and reach at the time. X takes the highway from A to B to C and travels at an average speed of 61.875 km/hr. Y takes the direct route AC and travels at 45 km/hr on segment AD. Y's speed on segment DC is 55 km/hr.
What is the average speed of Y?

47.5 km/hr
49.5 km/hr
50 km/hr
52 km/hr

Discuss Question

Answer: Option B. -> 49.5 km/hr
:
B
(b) Time taken by Y to cover AC

= \frac{x}{45} + \frac{x}{55} = \frac{x \times 100}{55 \times 45}

Average speed of Y between AC.

= \frac{2 x \times 55 \times 45}{x \times 100} = 49.5

km/hr.
or just take the harmonic mean of 45 and 55 that will give 49.5 which is the answer.

Question 100.

Directions: Answer the questions based on the following information.
A road network (shown figure) connects cities A, B, C and D all road segments are the straight line. D is the mid-point of the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB, the segment AB is 100km long.

Ms.X and Mr Y leave A at 8.00 a.m .., take different routes to city C and reach at the time. X takes the highway from A to B to C and travels at an average speed of 61.875 km/hr. Y takes the direct route AC and travels at 45 km/hr on segment AD. Y's speed on segment DC is 55 km/hr.
What is the length of road segment BD?

50 km
52.5 km
55 km
Cannot be determined

Discuss Question

Answer: Option B. -> 52.5 km
:
B

BD is half of AC. Solve for AC. You will get AC = 105.
Hence BD will be 52.5 or option (b)

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