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.

Profit on the correct weight is 20% , therefore S.P = 120

If he weighs 20% less than the given, it is 100*0.8= 80 kg

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Now 20% more than 80 = 80*1.2= 96 kg

So 96 kg is sold for Rs 120

But C.P of 96 kg is Rs 96

Therefore, profit % = (120-96)/96 * 100 = 25%

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So, the remainders obtained on dividing (357141)^{x} by 7 correspond to the remainders obtained by dividing 1^{x} by 7 because 357141 gives a remainder of 1 on division by 7. Similarly 357142 can be simply replaced by 2.

Now 1^{1} by 7 gives a remainder of 1 and 2^{1} by 7 gives a remainder of 2; 1^{2} by 7 gives a remainder of 1 and 2^{2} by 7 gives a remainder of 4; 1^{3} by 7 gives 1 and 23 by 7 also gives 1.

So x=3

11^{2} = 121

111^{2} = 12321

1111^{2} = 1234321

Applying trend analysis, we observe that the value of the square asymptotes till a number which is the number of 1s in the number being squared.

Since n^{2} has a value which increases symmetrically till 8 and then drops in the same fashion till 1, the number of 1s in n should be 8

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Product of roots = pq = -k-1 (constant term/co-efficient of x^{2})

Now p^{2} + q^{2} = (p+q)^{2} – 2pq

= (k-2)^{2} – 2(-k-1)

= (k-2)^{2} + 2(k+1)

= k^{2} – 2k + 1 + 5

= (k-1)^{2} + 5

So minimum value is 5