Instructions

For the following questions answer them individually

Question 51

Five students appeared for an examination. The average mark obtained by these five students is 40. The maximum mark of the examination is 100, and each of the five students scored more than 10 marks. However, none of them scored exactly 40 marks.
Based on the information given, which of the following MUST BE true?

Video Solution

Solution

Option A) False, as it is mentioned no-one scored exactly 40 marks.

Option B) False, as only 1 student or only 2 students can score above 40 and group average can be 40.

Case i:- 44, 39, 39, 39 & 39.

Case ii:- 42, 41, 39, 39 & 39.

Option C) False, student's scores could be 44, 39, 39, 39 & 39.

Option D) False, student's scores could be 44, 43, 42, 32 & 39.

Option E) True, if scores of all students are more than 40. The average will be more than 40.

Question 52

A painter draws 64 equal squares of 1 square inch on a square canvas measuring 64 square inches. She chooses two squares (1 square inch each) randomly and then paints them. What is the probability that two painted squares have a common side?

Video Solution

Solution

From the given information, we get that it is $$8\ inch\times\ 8\ inch$$ square grid.

Total ways of selecting 2 squares out of 64 in $$^{64}C_2$$.

Two squares with a common side can be selected in the following ways.

(i) Horizontal Pairs.

In the first row, R1, we can select 7 pairs of squares with a common side.

They are (R1C1,R1C2), (R1C2,R1C3),.....(R1C7,R1C8).

It applies to other rows as well.

Hence the total number of squares = $$7\times\ 8=56$$

(ii) Vertical Pairs.

In the first column, C1, we can select 7 pairs of squares with a common side.

They are (R1C1,R2C1), (R2C1,R3C1),.....(R7C1,R8C1).

It applies to other columns as well.

Hence the total number of squares = $$7\times\ 8=56$$

The probability of two painted squares having a common side = $$\frac{56+56}{^{64}C_2}$$ = $$\frac{112}{2016}$$.

Option (A) is correct.

Question 53

The addition of 7 distinct positive integers is 1740. What is the largest possible “greatest common divisor” of these 7 distinct positive integers?

Video Solution

Solution

Let 'm' be the Greatest common divisor and A, B, C, D, E, F & G be the 7 unique numbers.

When 'm' divides A, B, ......& G, we get seven unique quotients. Let the quotients be a, b, c, d, e, f, and g.

It is given,

A + B + C + D + E + F + G = 1740

m(a + b + c + d + e + f + g) = 1740

$$\ \ 1740=2^2\times3\times5\times29$$

m(a + b + c + d + e + f + g) = $$\ \ 1740=2^2\times3\times5\times29$$

29 is a prime number and cannot be factorised further.

Therefore, largest possible value m can take is 60.

Least possible sum of 7 unique numbers is 1 + 2 + 3 +...+ 7 = 28

Replacing 7 with 8, we will get 29.

This implies 60(1 + 2 + 3 .....+ 6 + 8) = 1740

Largest possible G.C.D is 60.

The answer is option B.

Question 54

The problem below consists of a question and two statements numbered
1 & 2.
You have to decide whether the data provided in the statements are sufficient to answer the question.

In a cricket match, three slip fielders are positioned on a straight line. The distance between 1st slip and 2nd slip is the same as the distance between 2nd slip and the 3rd slip. The player X, who is not on the same line of slip fielders, throws a ball to the 3rd slip and the ball takes 5 seconds to reach the player at the 3rd slip. If he had thrown the ball at the same speed to the 1st slip or to the 2nd slip, it would have taken 3 seconds or 4 seconds, respectively. What is the distance between the 2nd
slip and the player X?

1. The ball travels at a speed of 3.6 km/hour.
2. The distance between the 1st slip and the 3rd slip is 2 meters.

Video Solution

Solution

Let players at slips 1, 2 and 3 be a, b, and c, respectively.

Let the speed of the ball be m.

The trajectory of the ball is not specified in the question. But let's take it as straight line.

The distance between the 1st and 2nd slips is the same as the distance between the 2nd and 3rd slips.

The length of 'ab' = The length of 'bc'. It implies Xb is the median of triangle Xac.

It takes 3 sec, 4 sec and 5 sec for the ball to reach a, b and c, respectively.

Hence, Xa = 3m, Xb = 4m and Xc = 5m.

Using Apollonius's theorem,

$$2\times\ \left(\left(4m\right)^2+\left(ab\right)^2\right)=\left(3m\right)^2+\left(5m\right)^2$$

$$16m^2+\left(ab\right)^2=17m^2$$

$$ab=m$$

Hence, $$ac=2m$$

Using the properties of triangle, the sum of two sides should be greater than the third side,

But, Xa + ac = Xc

3m + 2m = 5m.

Hence, this arrangement is not possible.

As this question has ambiguity, XAT officials awarded full marks to all candidates for this question.

Question 55

Find the value of
$$\frac{\sin^{6}15^{\circ} + \sin^{6}75^{\circ} + 6\sin^{2}15^{\circ}\sin^{2}75^{\circ}}{\sin^{4}15^{\circ} + \sin^{4}75^{\circ} + 5\sin^{2}15^{\circ}\sin^{2}75^{\circ}}$$

Video Solution

Solution

$$Let\ \sin^215^{\circ}\ =a\ and\ \sin^275^{\circ}\ =\ b\ $$

Then the given equation becomes,

$$\ \frac{\ a^3+b^3+6ab}{a^2+b^2+5ab}$$

=$$\ \frac{\ \left(a+b\right)^3-3ab\left(a+b\right)+6ab}{\left(a+b\right)^2-2ab+5ab}$$ $$\ \therefore\ \ a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)$$ , $$\ \therefore\ \ a^2+b^2=\left(a+b\right)^2-2ab$$

Using $$\sin\left(\theta\right)\ =\cos\left(90-\theta\ \ \right)$$

we get, $$\sin^275^{\circ\ }=\cos^215^{\circ\ }$$

now, a+b = $$\sin^215^{\circ\ }+\sin^275^{\circ\ }$$ = $$\sin^215^{\circ\ }+\cos^215^{\circ\ }$$ = 1.

The equation becomes, $$=\ \frac{\ 1-3ab+6ab}{1-2ab+5ab}\ =\ \frac{\ 1+3ab}{1+3ab}\ =1\ $$

Question 56

The problem below consists of a question and two statements numbered
1 & 2.
You have to decide whether the data provided in the statements are sufficient to answer the question.

Rahim is riding upstream on a boat, from point A to B, at a constant speed. The distance from A to B is 30 km. One minute after Rahim leaves from point A, a speedboat starts from point A to go to point B. It crosses Rahim’s boat after 4 minutes. If the speed of the speedboat is constant from A to B, what is Rahim’s speed in still water?
1. The speed of the speedboat in still water is 30 km/hour.
2. Rahim takes three hours to reach point B from point A.

Video Solution

Solution

Let 'a' and 'b' be the speeds of Rahim and the speed boat in still water, respectively

Let 'c' be the speed of the stream.

Speed boat starts after 1 minute of Rahim's departure and crosses him after 4 minutes.

It implies that the speed boat covered the same distance in 4 minutes that Rahim took 5 minutes.

Hence, the ratio of upstream speeds of Rahim and Speed boat = $$\frac{a-c}{b-c}=\frac{4}{5}$$

Using Statement 1, we get $$b=30\ kmph$$. But, it's not sufficient to get the value of 'a'.

Using Statement 2, we get $$a-c=\frac{30}{3}=10\ kmph$$. But it's also not sufficient to get the value of 'a'.

using both statements, we get

$$\ \frac{\ a-c\ }{b-c}=\frac{4}{5}$$

$$\ \frac{\ 10\ }{30-c}=\frac{4}{5}$$

$$\ \ 50\ =120-4c$$

$$\ \ c\ =17.5$$

$$\ \ a=17.5+10=27.5\ kmph$$

Question 57

The Guava club has won 40% of their football matches in the Apple Cup that they have played so far. If they play another n matches and win all of them, their winning percentage will improve to 50. Further, if they play 15 more matches and win all of them, their winning percentage will improve from 50 to 60. How many matches has the Guava club played in the Apple Cup so far? In the Apple Cup matches, there are only two possible outcomes, win or loss; draw is not possible.

Video Solution

Solution

Let 'm' be the number of matches Guava played till now. They won '0.4m' matches.

After playing another 'n' matches and wining all of them, their winning percentage will improve to 50.

i.e,

$$\ \frac{\ 0.4m+n}{m+n}=0.5$$

$$m=5n$$

Playing 15 more matches and wining all of them, their winning percentage will improve from 50 to 60.

i.e $$\ \frac{\ 0.4m+n+15}{m+n+15}=0.6$$

Solving, we get $$6+0.4n\ =\ 0.2m$$

Substituting $$m=5n$$,

we get $$6+0.4n\ =\ n$$

$$n=10\ \&\ m\ =50$$

Hence, Guava club played 50 matches so far.

The answer is option A.

Question 58

ABC is a triangle with BC=5. D is the foot of the perpendicular from A on BC. E is a point on CD such that BE=3. The value of $$AB^2 - AE^2 + 6CD$$ is:

Video Solution

Solution

Given that, $$AD\bot\ BC$$

$$BE=3$$

$$BC=5$$

Using Pythagoras' theorem,

$$AD^2+BD^2=AB^2$$ .....(1)

$$AD^2+DE^2=AE^2$$......(2)

$$AD^2+DC^2=AC^2$$......(3)

(1) - (2) gives

$$BD^2+DE^2=AB^2-AE^2$$

$$x^2-\left(3-x\right)^2=AB^2-AE^2$$

$$AB^2-AE^2=6x-9$$

$$AB^2-AE^2+6CD=6x-9+6\left(5-x\right)$$ ($$CD=\left(5-x\right)$$)

$$AB^2-AE^2+6CD=6x-9+30-6x$$

$$AB^2-AE^2+6CD=21$$

Question 59

Jose borrowed some money from his friend at simple interest rate of 10% and invested the entire amount in stocks. At the end of the first year, he repaid 1/5th of the principal amount. At the end of the second year, he repaid half of the remaining principal amount. At the end of third year, he repaid the entire remaining principal amount. At the end of the fourth year, he paid the last three years’ interest amount. As there was no principal amount left, his friend did not charge any interest in the fourth year. At the end of fourth year, he sold out all his stocks. Later, he calculated that he gained Rs. 97500 after paying principal and interest amounts to his friend. If his invested amount in the stocks became double at the end of the fourth year, how much money did he borrow from his friend?

Video Solution

Solution

Let the amount Jose borrowed be 'x'. The rate of interest is 10%

Interest occurred on 'x' is = $$\ \frac{\ x\times1\times\ \ 0.1\ }{5}+\ \frac{\ 2x\ \times2\times\ \ 0.1}{5}\ +\ \ \frac{2x\times\ 3\times\ 0.1\ }{5}$$

= $$\ \frac{\ 1.1x\ }{5}$$

The invested amount doubled at the end of the fourth year, i.e. 2x.

Jose's profit after paying principal and interest amounts to his friend at the end of the fourth year is Rs 97500.

i.e., $$\ 2x-x-\frac{\ 1.1x\ }{5}=97500$$

$$\frac{\ 3.9x}{5}=97500$$

$$\ \ \ \ x=125000$$

The amount that Jose borrowed is Rs 1,25,000.

Option (D) is correct.

Question 60

Separately, Jack and Sristi invested the same amount of money in a stock market. Jack’s invested amount kept getting reduced by 50% every month. Sristi’s investment also reduced every month, but in an arithmetic progression with a common difference of Rs. 15000. They both withdrew their respective amounts at the end of the sixth month. They observed that if they had withdrawn their respective amounts at the end of the fourth month, the ratio of their amounts would have been the same as the ratio after the sixth month.
What amount of money was invested by Jack in the stock market?

Video Solution

Solution

Let the amount invested by Jack and Sristi be $$'x'$$.

Jack's amount after four months will become $$\frac{x}{2^4}$$ and after six months it will become $$\frac{x}{2^6}$$.

Sristi's amount after four months will become $$x-\left(4\times\ 15000\right)$$ and after six months it will become $$x-\left(6\times\ 15000\right)$$

From the given information,

$$\ \frac{\ \ \frac{\ x}{2^4}}{x-\left(4\times\ 15000\right)}=\ \frac{\ \ \frac{\ x}{2^6}}{x-\left(6\times\ 15000\right)}$$

$$\ \ \left(\ x-\left(4\times\ 15000\right)\right)=\ 2^2\times\ \left(x-\left(6\times\ 15000\right)\right)$$

$$3x\ =\ 300000$$

$$x\ =\ 100000$$.

The amount invested by Jack and Sristi is 100000.

Option 'A' is correct.

CAT Content

XAT 2023 Question Paper

A painter draws 64 equal squares of 1 square inch on a square canvas measuring 64 square inches. She chooses two squares (1 square inch each) randomly and then paints them. What is the probability that two painted squares have a common side?

The addition of 7 distinct positive integers is 1740. What is the largest possible “greatest common divisor” of these 7 distinct positive integers?

Find the value of
$$\frac{\sin^{6}15^{\circ} + \sin^{6}75^{\circ} + 6\sin^{2}15^{\circ}\sin^{2}75^{\circ}}{\sin^{4}15^{\circ} + \sin^{4}75^{\circ} + 5\sin^{2}15^{\circ}\sin^{2}75^{\circ}}$$

ABC is a triangle with BC=5. D is the foot of the perpendicular from A on BC. E is a point on CD such that BE=3. The value of $$AB^2 - AE^2 + 6CD$$ is:

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