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1. A is 2 13 times as fast as B. If A gives B a start of 80 m, how long should the race course be so that both of them reach at the same time?

A. 170 metre

B. 150 metre

C. 140 metre

D. 160 metre

Here is the answer and explanation

Answer : Option C

Explanation :

---------------------------------------------------------------------
Solution 1
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Speed of A : Speed of B = 73 : 1 = 7 : 3

It means, in a race of 7 m, A gains (7-3)=4 metre

$MF#%\text{If A needs to gain 80 metre, race should be of }\dfrac{7}{4} \times 80\text{ = 140 metre}$MF#%

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Solution 2
---------------------------------------------------------------------

If A is n times as fast as B and A gives B a start of x meters, then the length of the race course , so that A and B reaches the winning post at the same time = $MF#%x\left(\dfrac{n}{n-1}\right)$MF#% metres

[Read more ...]


$MF#%\text{Here, }n = 2\dfrac{1}{3} = \dfrac{7}{3}\text{ , }x = 80$MF#%

$MF#%\text{Length of the race course = }x\left(\dfrac{n}{n-1}\right) \\\\= 80\left(\dfrac{\dfrac{7}{3}}{\dfrac{7}{3}-1}\right) = 80\left(\dfrac{7}{7-3}\right) = 80 \times \left(\dfrac{7}{4}\right) = 20 \times 7 = 140\text{ metre}$MF#%



2. A can run 224 metre in 28 seconds and B in 32 seconds. By what distance A beat B?

A. 36 metre

B. 24 metre

C. 32 metre

D. 28 metre

Here is the answer and explanation

Answer : Option D

Explanation :

---------------------------------------------------------------------------------
Solution 1
---------------------------------------------------------------------------------
Clearly, A beats B by 4 seconds

Now find out how much B will run in these 4 seconds

$MF#%\text{Speed of B = }\dfrac{\text{Distance}}{\text{Time taken by B}} = \dfrac{224}{32} = \dfrac{28}{4} = 7\text{ m/s}\\\\ \text{Distance covered by B in 4 seconds = Speed × time } = 7 \times 4 = 28\text{ metre}$MF#%

i.e., A beat B by 28 metre

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Solution 2
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If A can run $MF#%x$MF#% metre race in t1 seconds and B in t2 seconds, where t1 < t2, then A beats B by a distance $MF#%\dfrac{x}{t_2}(t_2-t_1)$MF#% metres

[Read more ...]

x = 224 metre

t1 = 28 seconds

t2 = 32 seconds

$MF#%\text{Required distance = }\dfrac{224}{32}(32-28) = \dfrac{224}{32} \times 4 = \dfrac{224}{8}\text{= 28 metre}$MF#%



3. In a 100 m race, A can give B 10 m and C 28 m. In the same race B can give C :

A. 20 m

B. 15 m

C. 10 m

D. 25 m

Here is the answer and explanation

Answer : Option A

Explanation :

------------------------------------------------------------------
Solution 1
------------------------------------------------------------------
"A can give B 10 m".
=> In a 100 m race, while A starts from the starting point, B starts 10 meters ahead from the same starting point at the same time.

"A can give C 28 m"
=> In a 100 m race, while A starts from the starting point, C starts 28 meters ahead from the same starting point at the same time.

i.e., in a 100-10=90 m race, B can give C 28-10=18 m

$MF#%\text{=> In a 100 m race, B can give }C \dfrac{18}{90} \times 100\text{ = 20 m}$MF#%

------------------------------------------------------------------
Solution 2
------------------------------------------------------------------
"A can give B 10 m".
=> In a 100 m race, A will have to cover 100 m while B will have to cover only 100-10=90 m.

Similarly, "A can give C 28 m"
=> In a 100 m race, A will have to cover 100 m while C will have to cover only 100-28 = 72 m.

=> When B covers 90 metre, C covers 72 m

$MF#%\text{=> when B covers 100 metre, C covers }\dfrac{72 \times 100}{90}\text{= 80 metre}$MF#%

i.e.,in a 100 m race, B can give C 100-80=20 metre



4. At a game of billiards, A can give B 15 points in 60 and A can give C to 20 points in 60. How many points can B give C in a game of 90?

A. 22 points

B. 20 points

C. 12 points

D. 10 points

Here is the answer and explanation

Answer : Option D

Explanation :

------------------------------------------------------
Solution 1
-------------------------------------------------------
"A can give 15 points to B in 60".
=> In a game of 60, A starts with 0 points where B starts with 15 points

"A can give 20 points to C in 60".
=> In a game of 60, A starts with 0 points where C starts with 20 points

ie, in a game of 60-15-45, B can give C 20-15=5 points

=> ie, in a game of 90, B can give C 5×2=10 points

------------------------------------------------------
Solution 2
-------------------------------------------------------
"A can give 15 points to B in 60".
=> In a game of 60, A will have to get 60 points while B will have to get only 60-15=45 points

"A can give 20 points to C in 60".
=> In a game of 60, A will have to get 60 points while C will have to get only 60-20=40 points

i.e., when B scores 45 points, C scores 40 points..

$MF#%\text{when B scores 90 points, C scores}\dfrac{40 \times 90}{45}\text{= 80 points}$MF#%

i.e.,in a game of 90, B can give C 90-80=10 points



5. In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:

A. 60 m

B. 20 m

C. 40 m

D. 10 m

Here is the answer and explanation

Answer : Option B

Explanation :

To reach the winning point , A will have to cover a distance of 500 - 140= 360 metre

ratio of the speeds of two contestants A and B is 3 : 4

i.e., when A covers 3 metre, B covers 4 metre.

$MF#%\text{=> When A covers 360 metre, B covers }\dfrac{4}{3} \times 360\text{ = 480 metre}.$MF#%

Remaining distance B have to cover = 500-480= 20 metre

=> A wins by 20 metre



6. In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C will beat B by:

A. 20.25 m

B. 21.5 m

C. 22.75 m

D. 25 m

Here is the answer and explanation

Answer : Option D

Explanation :

It means, when A convers 200 m, B covers only (200-31)=169 m

and C covers only (200-18)=182 m

=> When C covers 182 m, B covers 169 m

$MF#%\text{=> When C covers 350 m, B covers }\dfrac{169}{182}\times 350\text{ = 325 m}$MF#%

Hence, C beats B by 350-325 = 25 metre



7. A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by:

A. 102 m

B. 100 m

C. 75 m

D. 112.5 m

Here is the answer and explanation

Answer : Option B

Explanation :

when B runs 25 m, A runs 22.5 m

$MF#%\text{=> When B runs 1000 metre, A runs }\dfrac{22.5}{25}\times 1000\text{ = 900 m}$MF#%

=> When B runs 1 kilometre, A runs 900 m

Hence, in a kilometre race, B beats A by

(1 kilometre - 900 metre) = (1000 metre - 900 metre) = 100 metre



8. In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by:

A. 21 m

B. 28 m

C. 26 m

D. 29 m

Here is the answer and explanation

Answer : Option B

Explanation :

When A covers 100 m, B covers (100-25)=75 m

When B covers 100 m, C covers (100-4)=96 m

$MF#%\text{=> When B covers 75 metre, C covers }\dfrac{96}{100}\times 75 = 72\text{ m}$MF#%

i.e., When A covers 100 m, B covers 75 m and C covers 72 m

=> In a 100 m race, A beat C by (100-72)=28 m



9. In a 300 m race A beats B by 22.5 m or 6 seconds. B's time over the course is:

A. None of these

B. 80 sec

C. 76 sec

D. 86 sec

Here is the answer and explanation

Answer : Option B

Explanation :

B runs 22.5 m in 6 seconds

$MF#%\text{=> B runs 300 m in }\dfrac{6}{22.5} \times 300\text{ = 80 seconds}$MF#%

i.e., B's time over the course = 80 seconds



10. A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:

A. 4.4 kmph

B. 4.25 kmph

C. 4.14 kmph

D. 5.15 kmph

Here is the answer and explanation

Answer : Option C

Explanation :

$MF#%\text{Speed of A = 5 kmph = }5\times \dfrac{5}{18} = \dfrac{25}{18} \text{m/s}\\\\ \text{Time taken by A to cover 100 m} = \dfrac{\text{distance}}{\text{speed}} = \dfrac{100}{\left(\dfrac{25}{18}\right)}\text{ = 72 seconds}$MF#%

It is given that, A gives B a start of 8 m and still beats him by 8 seconds.

=> B takes (72+8)=80 seconds to cover (100-8)=92 metre

$MF#%\text{Speed of B =} \dfrac{\text{distance}}{\text{time}} = \dfrac{92}{80} \text{m/s} = \dfrac{92}{80} \times \dfrac{18}{5}\text{ kmph}\text{ = 4.14 kmph}$MF#%



11. A runs 1 23 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

A. 200 m

B. 270 m

C. 300 m

D. 160 m

Here is the answer and explanation

Answer : Option A

Explanation :

Speed of A : Speed of B = 53 : 1 = 5 : 3

i.e., In a race of 5 m, A runs 5 m and B runs 3 m

i.e., A gains 2 m over B in a race of 5 m

$MF#%\text{=> A gains 80 m over B in a race of }\dfrac{5}{2}\times 80\text{ = 200 m}$MF#%

Hence, winning point should be 200 m away from the starting point.



12. In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C:

A. 40 points

B. 10 points

C. 14 points

D. 8 points

Here is the answer and explanation

Answer : Option B

Explanation :

In a game of 100 points, A scores 100 points, B scores (100-20)=80 points and

C scores (100-28)=72 points

i.e., when B scores 80 points, C scores 72 points

$MF#%\text{=> When B scores 100 points, C scores }\dfrac{72}{80} \times 100\text{ = 90 points}$MF#%

i.e., In a game of 100 points, B can give C (100-90)=10 points



13. In 100 m race, A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by:

A. 9 m

B. 25 m

C. 22.5 m

D. 20 m

Here is the answer and explanation

Answer : Option D

Explanation :

In 100 m race, A covers the distance in 36 seconds and B in 45 seconds.

Clearly, A beats B by (45-36)=9 seconds

$MF#%\text{Speed of B = }\dfrac{\text{Distance}}{\text{Time}} = \dfrac{100}{45}\text{ m/s}\\\\ \text{Distance Covered by B in 9 seconds = Speed × Time = }\dfrac{100}{45} \times 9\text{ = 20 metre}$MF#%

i.e., A beats B by 20 metre



14. In a 100 m race, A beats B by 10 m and C by 13 m. In a race of 180 m, B will beat C by:

A. 5.4 m

B. 5 m

C. 4.5 m

D. 6 m

Here is the answer and explanation

Answer : Option D

Explanation :

While A runs 100 m, B runs (100-10)=90 m and C runs (100-13)=87 m

i.e., when B runs 90 m, C runs 87 m

$MF#%\text{=> when B runs 180 m, C runs }\dfrac{87}{90} \times 180 \text{= 174 m}$MF#%

Hence, in a 180 m race, B will beat C by (180-174)=6 m



15. In a 200 metres race A beats B by 35 m or 7 seconds. A's time over the course is:

A. 33 sec

B. 40 sec

C. 47 sec

D. None of these

Here is the answer and explanation

Answer : Option A

Explanation :

B runs 35 m in 7 sec.

$MF#%\text{=> B runs 200 m in }\dfrac{7}{35}\times 200\text{ = 40 seconds}$MF#%

Since A beats B by 7 seconds, A runs 200 m in (40-7) = 33 seconds

Hence, A's time over the course = 33 seconds



16. A, B and C are the three contestants in one km race. If A can give B a start of 40 metres and A can give C a start of 64 metres. How many metres start can B give C?

A. None of these

B. 20 m

C. 25 m

D. 35 m

Here is the answer and explanation

Answer : Option C

Explanation :

While A covers 1000 m, B covers (1000-40)=960 m and C covers (1000-64)=936 m

i.e., when B covers 960 m, C covers 936 m

$MF#%\text{When B covers 1000 m, C covers }\dfrac{936}{960} \times 1000\text{ = 975 m}$MF#%

i.e., B can give C a start of (1000-975) = 25 m



17. In a game of 90 points A can give B 15 points and C 30 points. How many points can B give C in a game of 100 points?

A. 140

B. 20

C. 300

D. 50

Here is the answer and explanation

Answer : Option B

Explanation :

While A scores 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points

i.e., when B scores 75 points, C scores 60 points

$MF#%\text{=> When B scores 100 points, C scores }\dfrac{60}{75}\times 100\text{ = 80 points}$MF#%

i.e., in a game of 100 points, B can give C (100-80)=20 points



18. In a 100 metres race. A runs at a speed of 2 metres per seconds. If A gives B a start of 4 metres and still beats him by 10 seconds, find the speed of B.

A. 1.6 m/sec.

B. 4 m/sec.

C. 2.6 m/sec.

D. 1 m/sec.

Here is the answer and explanation

Answer : Option A

Explanation :

Speed of A = 2 m/s

$MF#%\text{Time taken by A to run 100 m }\dfrac{\text{distance}}{\text{speed}}=\dfrac{100}{2}\text{ = 50 seconds}$MF#%

A gives B a start of 4 metres and still A beats him by 10 seconds

=> B runs (100-4)=96 m in (50+10)=60 seconds

$MF#%\text{Speed of B = }\dfrac{\text{distance}}{\text{time}}=\dfrac{96}{60}\text{ = 1.6 m/s}$MF#%



19. P runs 1 km in 3 minutes and Q in 4 minutes 10 secs. How many metres start can P give Q in 1 kilometre race, so that the race may end in a dead heat?

A. 210 metre

B. 180 metre

C. 220 metre

D. 280 metre

Here is the answer and explanation

Answer : Option D

Explanation :

P run 1 km in 3 minutes

Q run 1 km in 4 minutes 10 secs

$MF#%\text{=> Q runs 1 km in }\dfrac{25}{6}\text{ minutes}\\\\ \text{=> Q runs }\left(1 \times \dfrac{6}{25} \times 3\right)= \dfrac{18}{25} = 0.72\text{ km in 3 minutes}$MF#%

Hence, in a 1 km race, P can give Q (1 - 0.72)=0.28 km = 280 metre



20. A runs 1 38 times As fast as B. If A gives B a start of 90 m and they reach the goal at the same time. The goal is at a distance of

A. 330 m

B. 440 m

C. 120 m

D. 280 m

Here is the answer and explanation

Answer : Option A

Explanation :

Speed of A : Speed of B = 118: 1 = 11 : 8

Let Speed of A = 11k

and Speed of B = 8k

Let the distance to the goal = x

$MF#%\begin{align}&\text{Time taken by A = time taken by B}\\\\&\Rightarrow \dfrac{\text{Distance travelled by A}}{\text{Speed of A}} = \dfrac{\text{Distance travelled by B}}{\text{Speed of B}}\\\\ &\Rightarrow \dfrac{x}{11k} = \dfrac{x-90}{8k}\\\\&\Rightarrow \dfrac{x}{11} = \dfrac{x-90}{8}\\\\&\Rightarrow 8x = 11x - 990\\\\&\Rightarrow 3x = 990\\\\&\Rightarrow x = 330\end{align}$MF#%

Distance to the goal = 330 metre



21. Two boys A and B run at 4 15 and 8 km an hour respectively. A having 150 m start and the course being 1 km, B wins by a distance of

A. 325 m

B. 60 m

C. 120 m

D. 275 m

Here is the answer and explanation

Answer : Option A

Explanation :

A has a start of 150 m. So A has to run 1000-150=850 metre while B has to run 1000 metre.

$MF#%\begin{align}&\text{Speed of A = }4 \dfrac{1}{5}\text{ km/hr }= \dfrac{21}{5}\text{ km/hr }= \dfrac{21}{5} × \dfrac{5}{18} = \dfrac{21}{18} = \dfrac{7}{6}\text{ m/s}\\\\ &\text{Speed of B = 8 km/hour = }8 × \dfrac{5}{18} = \dfrac{20}{9}\text{ m/s}\\\\ &\text{Time taken by B to travel 1000 metre = }\dfrac{\text{distance}}{\text{Speed of B}} = \dfrac{1000}{\left(\dfrac{20}{9}\right)}= \text{ 450 second}\\\\ &\text{Distance travelled by A by this time = time × speed of A = }450 \times \dfrac{7}{6}= 525\text{ metre}\end{align}$MF#%

Hence, A win by 850-525=325 metre



22. In a 100 metres race, A can beat B by 10 metres and B can beat C by 5 metres. In the same race, A can beat C by

A. 14.5 metre

B. 14 metre

C. 15.5 metre

D. 15 metre

Here is the answer and explanation

Answer : Option A

Explanation :

While A runs 100 metre, B runs 100-10=90 metre

While B runs 100 metre, C runs (100-5)=95 metre

$MF#%\text{=> While B runs 90 metre, C run }\dfrac{95}{100} \times 90 = \dfrac{95 \times 9}{10}\text{ = 85.5 metre}$MF#%

ie, when A run 100 metre, B run 90 metre and C run 85.5 metre

Hence, A beat C by (100-85.5)= 14.5 metre



23. X, Y and Z are the three contestants in one km race. If X can give Y a start of 52 metres and X can also give Z a start of 83 metres, how many metres start Y can give Z?

A. 33.3 m

B. 33 m

C. 32 m

D. 32.7 m

Here is the answer and explanation

Answer : Option D

Explanation :

While X runs 1000 metre, Y runs (1000-52)=948 metre and Z runs (1000-83)=917 metre

i.e., when Y runs 948 metre, Z runs 917 metre

$MF#%\text{=> When Y runs 1000 metre, Z runs }\dfrac{917}{948} \times 1000\text{ = 967.30 metre}$MF#%

i.e., Y can give Z (1000-967.30) = 32.7 metre



24. In a race of 600 m. A can beat B by 60 m and in a race of 500 m. B can beat C by 50 m. By how many metres will A beat C in a race of 400 m?

A. 76 metre

B. 74 metre

C. 72 metre

D. 78 metre

Here is the answer and explanation

Answer : Option A

Explanation :

While A runs 600 m, B runs (600-60)=540 metre

$MF#%\text{=>While A runs 400 m, B runs }\dfrac{540}{600} \times 400\text{ = 360 metre}$MF#%

While B runs 500 m, C runs (500-50)=450 metre

$MF#%\text{=>While B runs 360 m, C runs }\dfrac{450}{500} \times 360\text{ = 324 metre}$MF#%

i.e., When A runs 400 metre, B runs 360 metre and C runs 324 metre.

Hence, A beats C by (400-324)=76 metre in a race of 400 m



25. A can run 220 metres in 41 seconds and B in 44 seconds. By how many seconds will B win if he has 30 metres start?

A. 8 sec

B. 4 sec

C. 2.5 sc

D. 3 sec

Here is the answer and explanation

Answer : Option D

Explanation :

B has a start of 30 metre

=> A has to run 220 metre and B has to run (220-30)=190 metre

Given that A takes 41 seconds to cover this 220 metre

B takes 44 seconds to cover 220 metre

$MF#%\text{=> B takes }\dfrac{44}{220}\text{ seconds to cover 1 metre}\\\\ \text{=> B takes }\dfrac{44}{220} \times 190 = \dfrac{4}{20} \times 190 \text{ = 38 seconds to cover 190 metre}$MF#%

i.e., A beats B by (41-38)= 3 seconds



26. In one km race A beats B by 4 seconds or 40 metres. How long does B take to run the kilometer?

A. 112 sec

B. 110 sec

C. 101 sec

D. 100 sec

Here is the answer and explanation

Answer : Option D

Explanation :

This means, B takes 4 seconds to run 40 metres

$MF#%\text{=> B takes }\dfrac{4}{40} = \dfrac{1}{10}\text{ seconds to run 1 metre}\\\\ \text{=> B takes }\dfrac{1}{10} \times 1000 = 100 \text{ seconds to run 1000 metre}$MF#%



27. In a game, A can give B 20 points, A can give C 32 points and B can give C 15 points. How many points make the game?

A. 120 points

B. 90 points

C. 80 points

D. 100 points

Here is the answer and explanation

Answer : Option D

Explanation :

Let x points make the game

A can give B 20 points, A can give C 32 points
=> When A scores x points, B scores (x-20) points and C scores (x-32) points

B can give C 15 points
=> When B scores x points, C scores (x-15) points

$MF#%\text{=> When B scores 1 point, C scores }\dfrac{(x-15)}{x}\text{ points}\\\\ \text{=> When B scores (x-20) point, C scores }\dfrac{(x-20)(x-15)}{x}\text{ points}$MF#%

$MF#%\begin{align}&\text{i.e., } (x-32) = \dfrac{(x-20)(x-15)}{x}\\\\ &\Rightarrow x(x-32) = (x-20)(x-15)\\\\ &\Rightarrow x^2 - 32x = x^2 - 35x + 300\\\\ &\Rightarrow - 32x = - 35x + 300\\\\ &\Rightarrow 3x = 300\\\\ &\Rightarrow x = 100\end{align}$MF#%

i.e., 100 points make the game



28. In a game A can give B 20 points in 60 and C 18 points in 90. How many points can C give B in a game of 120?

A. 20 points

B. 22 points

C. 18 points

D. 40 points

Here is the answer and explanation

Answer : Option A

Explanation :

A can give C 18 points in 90
=> While A scores 90 points, C scores (90-18)=72 points

$MF#%\text{=> While A scores }\dfrac{90}{72}\text{ points, C scores 1 point.}\\\\ \text{=> While A scores }\dfrac{90}{72} \times 120 =\dfrac{90}{6} \times 10\text{ = 150 points, C scores 120 point.}$MF#%

A can give B 20 points in 60

=> While A scores 60 points, B scores (60-20)=40 points

$MF#%\text{=> While A scores 1 point, B scores }\dfrac{40}{60} = \dfrac{2}{3}\text{ points}\\\\ \text{=> While A scores 150 points, B scores }\dfrac{2}{3} \times 150 = 100\text{ points}$MF#%

i.e., while C scores 120 points, B scores 100 points

Hence, in a 120 race, C can give B (120-100)=20 points



29. In a km race A can beat B by 100 m and B can beat C by 60 m. In the same race A can beat C by

A. 144 m

B. 164 m

C. 144 m

D. 154 m

Here is the answer and explanation

Answer : Option D

Explanation :

In a km race A can beat B by 100 m

=> While A run 1000 m, B run (1000-100)=900 m

In a km race B can beat C by 60 m

=> While B runs 1000 m, C runs (1000-60)=940 m

$MF#%\text{ => While B run 1 m, C run }\dfrac{940}{1000}\text{ m}\\\\ \text{ => While B runs 900 m, C runs }\dfrac{940}{1000} \times 900 = 94 \times 9 = 846\text{ m}$MF#%

i.e., while A run 1000 m, C run 846 m

Hence, A can beat C by (1000-846)= 154 m



30. In a flat race, A beats B by 15 metres and C by 29 metres. When B and C run over the course together, B wins by 15 metres. Find the length of the course

A. 225 m

B. 125 m

C. 220 m

D. 256 m

Here is the answer and explanation

Answer : Option A

Explanation :

Let x be the length of the course

A beats B by 15 metres and C by 29 metres

=> When A runs x metre, B runs (x-15) metre and C runs (x-29) metre

When B and C run over the course together, B wins by 15 metres

=> when B runs x metre, C run (x-15) metre

$MF#%\begin{align}&\text{=> when B runs 1 metre, C run }\dfrac{x-15}{x}\text{ metre}\\\\ &\text{=> when B runs (x-15) metre, C run }\dfrac{x-15}{x} \times (x-15) = \dfrac{(x-15)^2}{x}\text{ metre}\\\\ &\text{i.e., }(x-29) = \dfrac{(x-15)^2}{x}\\\\ &\Rightarrow x^2 - 29x = x^2 - 30x + 225 \\\\ &\Rightarrow - 29x = - 30x + 225 \\\\ &\Rightarrow x = 225\end{align}$MF#%

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Comments(7) Sign in (optional)
showing 1-7 of 7 comments,   sorted newest to the oldest
Teesta
2014-03-13 08:20:48 
south-west
(0) (0) Reply
srk
2013-12-08 08:52:18 
3. south west
(0) (0) Reply
VISHAL
2013-11-22 09:04:37 
I am Standing at the centre of a circular field. I go down south to the edge of the field and then turning left I walk along the boundary of the field equal to three-eights of its length. Then I turn west and go right across to the opposite point to the boundary. In
which direction am I from the starting point?

(1) North-west     (2)North                       
(3) South-west     (4)West
(5) None of these
(0) (0) Reply
saurabh tripathi
2015-01-23 20:07:48 
north west is the correct answer
(0) (0) Reply
p
2014-09-01 17:31:27 
north west

(0) (0) Reply
Sam
2013-11-24 16:41:23 
                      |
                      |
    |                 |
    |                 |
    |                 |
    ___________

Hence, answer is north
(0) (0) Reply
Sam
2014-03-14 14:25:55 
Ok, South West seems to be the answer
(0) (0) Reply
 
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