 # What’s the CATch? – Part 1

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. A company offered its 350 employees a bonus of Rs 10 to each male and Rs 8.15 to each female. All the females accepted but a certain percentage of males refused to accept. The total bonus paid was independent of the number of males. What was the total amount paid to the females (in rupees)?
1. 1672.50
2. 1222.50
3. 2852.50
4. None of these
Topic : Equations
Approach :Let ‘m’ be the number of males, and let ‘x’ be the fraction of males refusing the bonus. Then the total bonus paid T= 8.15(350-m)+10(1-x)m
= 2852.5+m(1.85-10x),which will be independent of m only if x=0.185; so T=2852.5
Now, number of males refusing bonus= 0.185m, which has to be a +ve integer
Therefore 185m/1000 is a +ve integer or 185/(1000/m) is a +ve integer. 1000/m =37 or 5. 37 not possible; so 1000/m=5, which gives m=200. Thus females=150. Bonus paid to females= 8.15*150=1222.5
Challenge: Ability to conceptualize the correlation between ‘m’ and ‘x’ and understanding the underlying algebra
2. Ram bought ‘a’ flowers for ‘b’ rupees, where ‘a’ and ‘b’ are integers. If he had bought 10 more flowers he would have got all for Rs. 2 and saved 80 paisa a dozen. What are the respective vales of ‘a’ and ‘b’?
1. 5,1
2. 4,2
3. 3,2
4. 6,1
Approach :a flowers are for b rupees; and a+10 flowers are for 2 rupees
Therefore, b < 2 and since b is an integer, so b=1
(100/a) – (200/a+10)= 80/12
a = 5
Challenge : Ability to arrive at the value of ‘b’ and proceeding further on
3. The integers 49966 and 52231 when divided by a three digit number ‘n’ give the same remainder. What is the value of n?
1. 367
2. 453
3. 462
4. 298
Topic : Numbers (Remainders)
Approach :52231-49966=2265
2265 should be exactly divisible by the three digit number (if x=kn+m, y=ln+m, the remainder being m when x and y are divided by n, then x-y= n(k-l))
Now 2265 = 453*5
Thus n= 453
Challenge : Understanding that the difference of two natural numbers will also be divisible by the same divisor
4. The HCF of two numbers is 21 and their LCM is 4641. If one of the numbers is between 200 and 300, find the other number.
1. 221
2. 273
3. 357
4. 441
Topic : Numbers (H.C.F, L.C.M)
Approach :Let the numbers be x and y
Then 4641*21 = x * y
Numbers can also be expressed as multiples of the HCF. Thus, x= 21a and y = 21b
Now 4641 * 21 = 21a * 21b
Or 13*17*21 *21= 21a *21b
Comparing LHS and RHS, a and b are 13*21 (=273) and 17*21(=357) in any order.
Challenge :Applying the basics of H.C.F, L.C.M and using ab-initio methods to solve questions
5. What is the least number of digits in the number formed from only zeroes and ones such that the number is divisible by 225?
1. 4
2. 9
3. 11
4. 7
Topic : Numbers (Remainders)
Approach :225 = 9 * 25
For the number to be divisible by 225 it has to be divisible by both 9 and 25
For divisibility by 25, the last two digits should be zeroes or they should form a number which is multiple of 25. Hence we need 2 zeroes
For divisibility by 9, the sum of digits should be a multiple of 9. Only ones contribute to sum,and the least multiple of 9 is 9 itself; hence 9 ones
So, the number should have 9 ones and 2 zeroes or 11 digits