# Height and Distance Question & Answers

## Heights and Distances

Quantitative aptitude questions are asked in many competitive exams and placement exam.'Height and Distance' is a category in Quantitative Aptitude. Quantitative aptitude questions given here are extremely useful for all kind of competitive exams like Common Aptitude Test (CAT), MAT, GMAT, IBPS Exam, CSAT, CLAT, Bank Competitive Exams, ICET, UPSC Competitive Exams, CLAT, SSC Competitive Exams, SNAP Test, KPSC, XAT, GRE, Defense Competitive Exams, L.I.C/ G. I.C Competitive Exams, Railway Competitive Exam, TNPSC, University Grants Commission (UGC), Career Aptitude Test (IT Companies) and etc., Government Exams etc.

We have a large database of problems on "Height and Distance "answered with explanation. These will help students who are preparing for all types of competitive examinations.

If an object travels at five feet per second, how many feet does it travel in one hour?

 A) 30 B) 3000 C) 18 D) 1800

Explanation:

If an object travels at 5 feet per second it covers 5x60 feet in one minute, and 5x60x60 feet in one hour.

$\inline \fn_cm \therefore$Answer = 1800

Subject: Height and Distance - Quantitative Aptitude - Arithmetic Ability
Exam Prep: GRE

6

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?

 A) Data inadequate B) 8 units C) 12 units D) None of these

Explanation:

One of AB, AD and CD must have given.

Subject: Height and Distance - Quantitative Aptitude - Arithmetic Ability

5

An observer 1.6 m tall is $\inline \fn_cm 20\sqrt{3}$ away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:

 A) 21.6 m B) 23.2 m C) 24.72 m D) None of these

Explanation:

Draw BE ${\color{Black}&space;\bot}$ CD

Then, CE = AB = 1.6 m,

BE = AC = ${\color{Black}&space;20\sqrt{3}m}$

${\color{Black}\frac{DE}{BE}=tan30^{0}=\frac{1}{\sqrt{3}}}$

${\color{Black}\Rightarrow&space;DE=\frac{20\sqrt{3}}{\sqrt{3}}m=20m}$

${\color{Black}\therefore&space;}$ CD = CE + DE = (1.6 + 20) m = 21.6 m.

Subject: Height and Distance - Quantitative Aptitude - Arithmetic Ability

6

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:

 A) 173 m B) 200 m C) 273 m D) 300 m

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

${\color{Black}&space;Then,AB=100m,&space;\angle&space;ACB=30^{\circ}&space;and&space;\angle&space;ADB=45^{\circ}}$

${\color{Black}&space;\frac{AB}{AC}=\tan&space;30^{\circ}=\frac{1}{\sqrt{3}}\Rightarrow&space;AC=AB\times&space;\sqrt{3}=100\sqrt{3}m}$

${\color{Black}&space;\frac{AB}{AD}=\tan&space;45^{\circ}=1\Rightarrow&space;AD=AB=100&space;m}$

${\color{Black}&space;\therefore&space;CD=(AC+AD)=(100\sqrt{3}+100)m=100(\sqrt{3}+1)=100\times&space;2.73=273&space;m}$

Subject: Height and Distance - Quantitative Aptitude - Arithmetic Ability

3

Jack takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?

 A) 6 B) 8 C) 9 D) 10

Explanation:

Average speed = total distance / total time

Total distance covered = 6 miles; total time = 45 minutes = 0.75 hours

Average speed = 6/ 0.75 = 8 miles/hour