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.

The three possibilities on the basis of the second condition are: 1^{23}, 2^{46}, 3^{69}

However, 2^{46}does not end in 1, hence ruled out.

Out of 1^{23}and 3^{69} , 1^{23} definitely ends in 1.

We just need to check for 3^{69}.

Now, an even digit (2,4,6,8) raised to the power of 2 or more will always be a multiple of 4. So, 3^{69}=34k

Apply cyclicity concept. Cycle for 3 is 3,9,7,1

Thus 34k ends in 1. Thus 3^{69} also ends in 1.

Grab the Unbelievable Offer:

Now 44^{2}= 1936, 43^{2}= 1849

If he was 43 years in the year 1849, then in 1957 he would be more than 100 years, hence it is ruled out. Thus he was 44 years in the year 1936. Thus his year of birth is 1936-44=1892

Now 44+m=d^{2}, where m is the serial number of the birth month and d is the date.

Now m has to be less than equal to 12. Thus m=5, d=7

The birth date is 7th May, 1892

x/2 - 2250 +720

And A’s income will be:

x/2 + 2250 -720

Thus x/2 - 2250 +720 = ½ (x/2 + 2250 -720)

Solve for x.

X = 9180

Then y= (75n-(56+52) +x)/(n-2+1) = (75n-108+x)/(n-1)

Rearranging the terms, the equation can be written as:

y= (75 (n-1) + (x-33) )/(n-1) = 75 + (x-33)/(n-1)

Now check for options

If x=81, then y = 75 + 48(n-1). For y to be prime, it has to be a natural number. Thus n-1 has to be a factor of 48 and can take values 2,3,4,6,8,12,16,24,48. When n-1 is 6, y is 83, which is a prime number. Hence x=81 is a possible value.

Note that other options will not satisfy this condition. For example,

If x=87, then y=75+ 54/(n-1). For n-I to be a factor of 54 it can take values 2,3,6,9,18,27,54. For none of these, y is a prime number.
__Answer : __Option 1
__Challenge : __Application of averages to questions with higher levels of difficulty, more in the context of multiple possibilities and elimination of ineligible options
**The speed of a railway engine is 42 km/hr when no compartment is attached, and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 km/hr when 9 compartments are attached, what is the maximum number of compartments that can be carried by the engine?**
**1. 49**
**2. 48**
**2. 0**
**4. None of these**
__Topic : __Variation & Proportion
__Approach: __The reduction in speed r is directly proportional to the square root of the number of compartments attached. Thus r = k n^{1/2}
__Answer:__ Option 2
__Challenge:__ Ability to apply the concept of variation and understanding the fact that n_{max} cannot be 49, because at 49, the engine ceases to move. Here it is more about managing the emotion of instant gratification as reaching the value of 49 is not much of a difficulty, but marking 48 as the right answer is the deciding point.

Free Master Classes:

Now 42- kn^{1/2}is the net speed when n compartments are attached

Also, 42- k9^{1/2}= 24

Therefor k = 6.

So net speed is 42 – 6n^{1/2}

Now 42 -6 n_{max}^{1/2}is greater than zero

Thus n_{max} is less than 49

Or n_{max} = 48