The good thing about CAT is, that it is not a mathematics test. So, it helps that you know the process and the steps but you should be smart and try to minimize the work. This is why there are options to help you. A smart use of options will save a lot of time and help you increase your score.
Some examples
Two different two-digit natural numbers are written beside each other
such that the larger number is written on the left. When the absolute
difference of the two numbers is subtracted from the four-digit number
so formed, the number obtained is 5481. What is the sum of the two
two-digit numbers?
(a) 70 (b) 71 (c) 72 (d) 73
Now it is pretty much obvious the number is of the form 55xy
To find xy we can just subtract 55 from the options which gives xy= 15, 16, 17, 18
If you are lucky you will start with 73 and you will know it is right, else even if you start with 15, you will soon reach 18 and get your answer.
The direct method will be a bit long.
Let n be the total number of different 5-digit numbers with all distinct digits, formed using 2, 3, 4, 5 and 6 and divisible by 4. What is the value of n?
1] 44 2] 32 3] 36 4] 38 5] 40
Permutations & Combinations is probably the most ‘hated’ topic. However, if you understand the basics and use logic, then it is the most fun-filled topic of all. Let us get to this question. As mentioned, we need to find out the 5 digit numbers divisible by 4 formed by the digits given in the question. To be divisible by 4, the last 2 digits should be divisible by 4. So to arrive at the answer, the first step is to find out combinations of the 2 digits from 2,3,4,5 & 6 that are divisible by 4 – eg: 24. Then, for each such combination, the last 2 digits are fixed. The remaining 3 digits can be arranged in 3! Ways = 6 ways. So the answer would be - 6 multiplied by the total no. of combinations of 2 digits divisible by 4. The answer necessarily should be a multiple of 6 and therefore the answer is 36 – option [3]. We just got lucky here, by the way, since there is just one option that is divisible by 6!
![\begin{align*}\gcd (m^3, n^2) & = 2^2 \cdot 3^2,\\ \text{LCM} [m^2, n^3] & = 2^4 \cdot 3^4 \cdot 5^6,\end{align*}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uqXZSj3qfON8WQzY_4kLQgQ-92Gs-WKwvZD-54RNY_pJArv7UFfknUeocAWyVUwevp7JAbxqwb26ZaftJ_2dj3J98Yz502VujH62nHfz8vw8rYMhQvosjNXwMzvXM_J2TzQG4UlWwD59Rsna4CrjCQf6Q7xWvd4caAH2KvKSrBom6ExEb5EA=s0-d "\begin{align*}\gcd (m^3, n^2) & = 2^2 \cdot 3^2,\\ \text{LCM} [m^2, n^3] & = 2^4 \cdot 3^4 \cdot 5^6,\end{align*}")
![\text{LCM}[a, b]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vsfyeKgCx9mJFzVOYN0dX3xxm8fNAIsCXC7eftnDW6EOGUeB36wl2A-KsXZc68jmPsEW0PcObpkc-M3bwiCGumYuaOtSvsvqUWkKW--9sWBU5JBwOW2y_shff_J4FUrwQIdn7rTiXvgHo6LQEJZKAFMn26oVUun9BW_YrDgXuDvv21I2ca0A=s0-d "\text{LCM}[a, b]")
