In this article certain questions have been randomly taken from different topics of the quantitative section of CAT. Each question has been selected keeping in mind certain challenges you need to manage as a test taker. These challenges may range from concept basics to application orientation and test taking strategies.

1. 0

2. 62.33

3. 100.33

4. Indeterminate

a_{3} = a_{2}-a_{1}

a_{4} = a_{3}-a_{2} = a_{2}-a_{1}-a_{2} = -a_{1}

a_{5} = a_{4}-a_{3} = -a_{1}-(a_{2}-a_{1})= -a_{2}

a_{6} = a_{5}-a_{4} = -a_{2}-(-a_{1}) = a_{1}-a_{2}

a_{7} = a_{6}-a_{5} = a_{1}-a_{2}+a_{2}= a_{1}

a_{8} = a_{7}-a_{6} = a_{1}-(a_{1}-a_{2}) = a_{2}

After a_{6} there is a repetition in a cyclic way, so we just need to focus on the first 6 terms a_{1} to a_{6}.

Now a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}= a_{1}+a_{2}+(a_{2}-a_{1})-a_{1}-a_{2}+a_{1}-a_{2}= 0

Thus, sum of first 6000 terms will be 0

Sum of 6002 terms = 6001^{th} term+6002^{th} term = a_{1}+a_{2}= 81.33-19=62.33

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1. 0

2. 1

3. -1

4. None of these

From the given information,

11/2 (2a+10d) = 19/2 (2a+18d)

Rearranging, we get 16a + 232d= 0

Or 2a + 29d= 0

Now, sum of first 30 terms = 30/2 (2a+29d)= 15 *0 =0

1. 5 times

2. 10 times

3. Once

4. None of these

Since they meet after 30 seconds, l/(S_{a}-S_{b}) =30

Or S_{a}/l – S_{b}/l = 1/30…………….I

Also, l/S_{b} – l/S_{a} = 5………..II (as A takes 5 seconds less than B to complete l metres)

Let S_{a}/l = x and S_{b}/l = y

Then from I and II, we get x-y=1/30 and 1/y -1/x=5

Or xy = (x-y)/5 = 1/150

Now (x+y)^{2}=(x-y)^{2}+4xy= (1/30)^{2}+4(1/150)= 1/36

Or x+y=1/6

Thus S_{a}/l +S_{b}/l = 1/6, or (S_{a}+S_{b})/l = 6, Thus they will meet every 6 seconds if they run in opposite directions. So they will meet 10 times in a minute.

1. A wins by 100m

2. B wins by 100m

3. A wins by 200m

4. B wins by 200m

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Ratio of Number of steps/second for A and B= 1:2

Thus, ratio of speeds of A and B – (3*1): (2*2) = 3:4

Now B gives A a lead of 400m in a 2000m race, so A has to cover 1600m while B has to cover 2000m

If B wins the race then he will cover 2000m while A in that time will cover 2000*3/4= 1500m

But A already has a lead of 400m, thus A will have covered 1900 m from the start point.

Thus B wins the race by 100m.