JIPMAT 2023 Question Paper

Statement I

Let the efficiency of the boy be B and the girl be G.
4 boys and 8 girls completed one-third of the work in 5 days. That is, they complete the entire work in 15 days. 
Thus, Total Work = $$15\times(4B+8G)=60B+120G$$
After that, 3 boys and 3 girls increased, and they completed another one-third of the work in 3 days. This means - 
$$3\left(7B+11G\right)=\dfrac{1}{3}\left(60B+120G\right)$$
$$21B+33G=20B+40G$$
$$B=7G$$
Thus, total work = $$60(7G)+120G=540G$$
Now, two-thirds of the work is completed, and the remaining one-third, or 180G, should be finished in 2 days. For this, let's say, we need x more girls. Thus, we have 7 boys, and (11+x) girls, and B = 7G
$$2\left(7B+\left(11+x\right)G\right)=180$$
$$7\left(7G\right)+\left(11+x\right)G=90$$
$$(11+x)G=41G$$
$$x=30$$
Thus, we need 30 more girls to complete the required work in the required number of days.
Statement I is FALSE. 

Statement II

The ratio of time taken by A and C to do the work is 1:2, respectively. Thus, A takes half the time as C to do the same work. Thus, A is twice as efficient as C.
B is 166 (2/3)% more efficient than C. Thus, $$B=\left(1+\dfrac{500}{3\times100}\right)C=\dfrac{8C}{3}$$, and $$A=2C$$
The time taken by A to complete 6% of the work is 6 days. Thus, he will complete 100% work in 100 days. 
Total work = $$A\times100=2C\times100=200C$$
Let the time taken by B and C together to complete the whole work be D days. Then -
$$\left(B+C\right)\times D=200C$$
$$\left(\dfrac{8C}{3}+C\right)\times D=200C$$
$$D=\dfrac{600}{11}=54\dfrac{6}{11}$$days
Statement II is also TRUE. 

Lets assume $$\left(y+\dfrac{\ 1}{y}\right)\ =\ t$$.

So, $$\left(y^{2}+\dfrac{1}{y^{2}}\right)$$ = $$t^2-2$$

So, the given equation can be written as 7(t) - 2($$t^2-2$$) = 9
    7(t) - 2$$t^2$$ = 5 
    Hence, 't' values could be {1} or {5/2}.

Now, $$\left(y+\dfrac{\ 1}{y}\right)\ =\ 1$$ , has no integral solution .
But for $$\left(y+\dfrac{\ 1}{y}\right)\ =\ 5/2$$, we can get y = 2 and y = 1/2.

Therefore, the only integral values of Y possible is {2} . Hence no of integral solutions will be one.

Statement I

The difference, the sum, and the product of two numbers are in the ratio 1:5:12. Let's assume the difference is k, the sum is 5k, and the product is 12.
The product of the two numbers is 18.
$$12k=18$$. Thus, $$k=1.5$$
This means that the difference will be 1.5, and the sum will be 7.5.
But, this is not possible, because the sum or the difference of the two numbers cannot be a decimal value. 
Statement I is FALSE.

Statement II

Unit digit of $$2^{4k+1}=2$$
Unit digit of $$2^{4k+2}=4$$
Unit digit of $$2^{4k+3}=8$$
Unit digit of $$2^{4k}=6$$
Now, Unit digit of $$2^{91}=2^{4\times22+3}=2^{4k+3}$$, where $$k=22$$ will be 8. 
Statement II is FALSE.

As there are total 300 players in the tournament and the goal is to eliminate 299 players to announce the final winner . 
It is given that every single match will eliminate one player as there are no draws . So, in order to eliminate 299 players in the tournament, we need 299 matches to be conducted .

Alternate Solution : 
Round 1 : We can conduct 150 matches among 300 players each playing exactly once , so that we can eliminate all the 150 players who lost their respective matches .
                                       Remaining players in tournament = 150

Round 2 : Now we can conduct 75 matches among 150 players each playing exactly once, so that we can eliminate 75 players who lost their matches.
                            Remaining players = 75

Round 3 : Now we can conduct 37 matches for 74 people and make 1 player sit idle without any match. In this round we can eliminate 37 players.
                            Remaining players = 37+1 = 38.

Round 4 : Now we can conduct 19 matches for these 38 players , and eliminate 19 players who loose their matches.
                            Remaining players = 19.

Round 5 : Now we can conduct 9 matches for 18 players and make 1 player sit idle in this round. So we can eliminate 9 players here.
                            Remaining players = 9+1 = 10.

Round 6 : Now we can conduct 5 matches for 10 players and eliminate 5 players .
                            Remaining players = 5 .

Round 7 : we can conduct 2 matches and make 1 player sit idle in this round. So we can eliminate 2 players .
                            Remaining players = 2 + 1 = 3.

Round 8 : We can conduct 1 match to eliminate one player .
                            Remaining players = 2.

Round 9 : We can conduct last 1 match to eliminate one among 2.

Therefore, total matches = 150 + 75 + 37 + 19 + 9 + 5 + 2 + 1 + 1 = 299

P weeks = 7(P) days = 7(P) x 24 x 60 x 60 seconds .
P days = (P) x 24 x 60 x 60 seconds.
P hours = (P) x 60 x 60 seconds
P minutes = (P) x 60 seconds.
P seconds = (P) seconds.

Therefore , total no of seconds = P{1 + 60 + 3600 + 86400 + 604800} = 694861 (P)

Sumit has Rs. 90000 with him. He purchases three items, A, B, and C, for Rs. 15000, Rs. 13000, and Rs. 35000, respectively and puts the remaining money in a bank deposit that pays compound interest at 15% per annum.

Thus, he invests Rs $$90000-15000-13000-35000=27000$$ in a compound interest scheme. 

After 2 years, he sells off the three items at 80% of their original price. Thus, the price of items after 2 years is Rs $$0.8\times63000=50400$$

After 2 years, the value compounded will be $$=27000\times\left(1+\frac{15}{100}\right)^2=35707.5$$

Total value after 2 years = Rs 86107.5

%age decrease $$=\dfrac{86107.5-90000}{90000}\times100=-4.325\%$$

7200 has to be prime factorized : 
      7200 = 6 x 1200 
              = 6 x 6 x 200 
              = 2 x 3 x 2 x 3 x 2 x 2 x 5 x 2 x 5
              = $$2^5\times3^2\times5^2$$

Total No of factors = (5+1)(2+1)(2+1) = 6 x 3 x 3 = 54 factors .

So, No of proper divisors are (54 - 2) = 52 . { Out of 54 factors, we need to subtract 1 & 7200 }

Circumference of circle = $$2\pi\ \left(r\right)$$ = $$\dfrac{\ 44}{7}\left(r\right)$$ 

So, the difference between Circumference and radius = $$\left(\frac{\ 44}{7}\left(r\right)-r\right)$$ = $$\dfrac{\ 37}{7}\left(r\right)$$.
It is given this difference as 37 cm in the question : 
                                                    $$\dfrac{\ 37}{7}\left(r\right)$$ = 37
                                                     so, r = 7 cms.

Therefore , circumference = $$\dfrac{\ 44}{7}\left(r\right)$$ = $$\dfrac{\ 44}{7}\left(7\right)$$ = 44 cm. 

P(AUB) = P(A) + P(B) - P(A∩B)
           = 0.25 + 0.50 - 0.14 
           = 0.61
P(A U B) : Represents A or B .
So, the required Neither A nor B would be complementary to P(A U B). 
Hence, we need to calculate = {1 - P(A U B)} 
                                            =  1 - 0.61
                                            =  0.39

Tangent to a circle is defined as a straight line that intersects the circle at exactly one point on the circle . 
Statement A is incorrect as it says two points .
Statement C is correct as it says "No tangent to a circle passes through inside circle" .

Statement B : A circle can have two parallel tangents at the most .
Explanation : For any point on the circle we can draw a tangent and for the point which is diametrically opposite to the initial point, we                           can draw a tangent which will be parallel to the first tangent .
                    Reason : Line joining the centre of circle and tangent should be perpendicular to tangent at that particular point.