XAT 2022 Question Paper

The pace of Jim's watch is slower by 3 minutes in comparison with Kim's watch for every one hour. The difference increases as the hours pass by.

The time and "am" and "pm" of the watch coincide when the difference between the two clocks reduces to 24 hours. The two clocks display the same time including am and pm.

For a difference of 24 hours, the clock needs to lag by 1440 minutes.

For every one hour, the clock lags by 3 minutes. Hence in order to have a difference of 1440 minutes, it takes $$\frac{1440}{3}=\ 480\ hours$$. This is equivalent to 20 days of time.

Hence for every twenty days, they display the same time.

In the period of Jan 1 and June 1, there are 150 days which includes 7, 20-day intervals. Along with the 7 times once on Jan 1st and once on June 1st, they display the same time.

A total of 9 times.

From the given conditions :

Considering the three-digit number to be a b c.

With the given conditions :

a = 2b, c = 3b.

Hence the number is of the form : 2b b 3b.

Since all three of the values must be less than 10 and non-negative:

This takes values : b = 1, b = 2, b = 3.

Hence the possible numbers are : (213, 426, 639) :

The interchanged number must be greater than the original by 198.

Hence the different rearrangements for the three numbers are :

213 : (312, 321, 132, 123, 231).

426 : (462, 624, 642, 246, 264)

639 : ( 693, 963, 936, 396, 369)

The only possible value which is higher than the original by 198 is :

(426, 624).

The middle digit is 2.

The total number of ways of selecting 3 notes from the :

five 10-rupee notes, three 20-rupee notes, and two 50-rupee notes = 10 notes in total.

$$10_{C_3}$$

= 120

The possibilities for the value of the three notes combined is at least  90 :

Rs 50 + Rs 20 + Rs 20 : The possibilities for this selection is  :

$$2_{C_1}\cdot3C_2$$. Selection of one Rs 50 note from the two and selection of 2 Rs 20 notes from the three.

Rs 50 + Rs 50 + Rs 10 :

$$\left(2_{C_2}\right)\cdot\left(5_{C_1}\right)$$ : Selection of two Rs 50 notes from the two and selection of 1 Rs 10 notes from the five.

Rs 50 + Rs 50 + Rs 20 :

$$\left(2_{C_2}\right)\cdot\left(3C_1\right)$$ : Selection of two Rs 50 notes from the two and selection of 1 Rs 20 notes from the three.

A total of 6+5+3 = 14 possibilties

The probability is $$\frac{14}{120}$$ = 7/60.

The number of oranges, apples, guavas, and pears = 126, 162, 198, and 306.

Each box must contain an equal number of fruits with only one type of fruit. The additional condition provided is that there should be a minimum number of boxes in total.

The distribution is possible in multiple ways in such a way that distribution in each box is placed in such that each box contains a certain number of fruits n which is a factor for all the four given number of fruits :

Arrangement of 1 fruit of one kind in a basket.

2 is a factor of 126, 162, 198, and 306. So we can place 2 fruits of a particular kind in a basket.

Since we were asked for the minimum number of boxes this is possible when a maximum number of fruits of a kind are placed in a box.

Hence each box must contain the Highest common factor for the four numbers :

The prime factorization for the four numbers :

126 : $$2\cdot7\cdot9$$, 162 : $$2\cdot9\cdot9$$, 198 : $$2\cdot9\cdot11$$, $$306\ =\ 2\cdot9\cdot17$$

The HCF is 18.

The number of boxes required for each :

$$\frac{126}{18},\ \frac{162}{18},\ \frac{198}{18},\ \frac{306}{18}$$

7+9+11+17 = 44.

The function $$f(x)=\frac{\log{(3x-7)}}{\sqrt{2x^{2}-7x+6}}$$ is only defined when both the numerator and the denominator of the function are defined are the denominator is not equal to zero.

The logarithm of the function is only defined for positive values :

Hence 3x-7 is greater than zero. Hence 

x > 7/3.

The value inside square root are defined for positive values. The value of the quadratic equation in the square root must be positive.

Hence $$2x^2-7x+6\ =\ 0$$ has the roots :

$$\frac{\left(7+\sqrt{\ 49-48}\right)}{4},\ \frac{\left(7-\sqrt{\ 49-48}\right)}{4}$$    : 2, 3/2

The quadratic equation is positive for :

$$\left(-\infty\ \frac{3}{2}\right)\ U\ \left(2,\ \infty\ \right)$$

Since in order to be a part of the domain the values of x must be greater than 7/3  and 7/3 is greater than 2 all values of x which are greater than 7/3 must be a part of the domain for x.