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.
A shopkeeper always weighs 20% less than the correct weight. One day he weighed 20% more than the quantity usually weighed by him. If the profit on the correct weight is 20% what is the effective percentage of profit in the transaction?
Topic : Profit & Loss
Approach :Let the correct weight be 100 kg and its C.P be Rs 100
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
Grab the Unbelievable Offer:
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%
Answer : Option 2
Challenge :Application of percentage to profit & loss transactions
For what minimum value of x will the remainder of (357141)x and (357142)x be the same when divided by 7, where x is a natural number?
Topic : Numbers
Approach : This is just based on understanding the concept that in calculating the remainder obtained when ab is divided by c, simply replace a with the remainder obtained when a is divided by c.
So, the remainders obtained on dividing (357141)x by 7 correspond to the remainders obtained by dividing 1x by 7 because 357141 gives a remainder of 1 on division by 7. Similarly 357142 can be simply replaced by 2.
Now 11 by 7 gives a remainder of 1 and 21 by 7 gives a remainder of 2; 12 by 7 gives a remainder of 1 and 22 by 7 gives a remainder of 4; 13 by 7 gives 1 and 23 by 7 also gives 1.
So x=3
Answer : Option 2
Challenge :Application of divisibility concepts
Given n2 = 123456787654321, find n.
Topic :Numbers
Approach :This question is based on understanding symmetry in numbers.
112 = 121
1112 = 12321
11112 = 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 n2 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
Answer : Option 3
Challenge :Understanding the concept of symmetry in numbers; squares of binary numbers
Let p and q be the roots of the quadratic equation x2-(k-2)x-k-1=0. What is the minimum possible value of p2+q2?
Topic : Quadratic Equations
Approach :Sum of roots = p+q = k-2 (- co-efficient of x/co-efficient of x2 )
Product of roots = pq = -k-1 (constant term/co-efficient of x2)
Now p2 + q2 = (p+q)2 – 2pq
= (k-2)2 – 2(-k-1)
= (k-2)2 + 2(k+1)
= k2 – 2k + 1 + 5
= (k-1)2 + 5
So minimum value is 5
Answer : Option 3
Challenge : Understanding quadratic expressions/equations
CAT Quantitative Aptitude: Preparation Articles
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