Here’s and example of a **SMART MATH** problem for **TIME SPEED DISTANCE.**

**Problem**

**Problem**

In a 200 meters race, A beats B by 20 meters, while in a 100 meters race B beats C by 5 meters. By how many meters will A beat C in a kilometer race assuming that speeds of A, B and C do not change in the races?

- 150
- 145
- 130
- 125
- 110

**The Usual Method**

[contentblock id=google-adsense-post]

If in a 200 meters race A beats B by 20 meters, than in a kilometer race (1000 meters), A will beat B by 100 meters.

Similarly if in a 100 meters race B beats C by 5 meters, in a kilometer race, B will beat C by 50 meters.

Hence,

A B C

1000 900

aaaaaaaaaa 1000 950

i.e. A B C

aaaaaa 1000 x 1000 900 x 1000 900 x 950

i.e 1000 900 855

Hence, in a kilometer race if A runs 1000 meters, B runs 900 and C runs 855 meters. Hence, A beats C by 1000 – 855 = 145 meters.

**(Ans: 2)**

*Estimated Time to arrive at the answer = 60 seconds.*

**Using Technique**

[contentblock id=google-adsense-post]

Note that in a kilometer race A beats B by 100 meters and B beats C by 50 meters. Since, B runs 100 meters less than 1000 meters of A (1000 – 100 = 900 meters), C will lag little less than 50 meters behind B in this race where B runs 900 meters. Hence the gap between A and C will be sum of gap between A and B (100 meters) and gap between B and C (little less than 50 meters). Hence, gap between A and C is little less than 150 meters. The only option close to 150 and less than 150 is option ‘2’ i.e 145 meters.

**(Ans: 2)**

*Estimated Time to arrive at the answer = 10 seconds.*

[starrater tpl=10]

[contentblock id=smartmath-blockquote]