JIPMAT 2023 Question Paper

For pair of linear equations : $$a_1x+b_1y=c_1\ \&\ a_2x+b_2y=c_2$$ to have an infinite no of solutions , the below given condition has to hold true .
                                             $$\dfrac{a_1\ }{a_2}=\dfrac{\ b_1}{b_2}=\dfrac{\ c_1}{c_2}$$

So, it implies :   $$\dfrac{2 }{2p}=\dfrac{3}{p+q}=\dfrac{7}{28}$$

Therefore, P = 4, Q = 8.

In a right angled triangle ABC with right angle at B : $$AB^2+BC^2=AC^2$$.

given, BC = 5 , (AC-AB) = 1 .
    Hence, we can write :  $$AB^2+5^2=(1+AB)^2$$
                                         24 = 2 (AB)
    so, AB = 12 cm.

AC = 1 + AB = 13 cm. 
BC = $$\sqrt{13^2-12^2}=\ 5$$

Hence, Sin(C) = $$\dfrac{\ AB}{AC}$$ = $$\dfrac{\ 12}{13}$$
           
           Cos(C) = $$\dfrac{\ BC}{AC}$$ = $$\dfrac{\ 5}{13}$$     

Therefore, $$\dfrac{1 + \sin(C)}{1 + \cos(C)}$$ = $$\dfrac{25}{18}$$ .                 

Only English = 18

English but not Hindi = 23. This implies that students studying English and German but not Hindi will be = 23 - 18 = 5. 

English and German = 8. Now, students studying English and German but not Hindi = 5, thus students studying all three will be = 8 - 5 = 3

English = 26, and students studying English and German = 8 and only English = 18; thus students studying English and Hindi but not german = 26 - 18 - 8 = 0

German and Hindi = 8. Students studying all three languages = 3, thus students studying German and Hindi, but not English = 8 - 3 = 5.

German = 48. Thus, students studying only German will be = 48 - 5 - 3 - 5 = 35

No language = 24.

Total number of students = 100

Number of students studying only Hindi = 100 - (18 + 5 + 35 + 3 + 5 + 10 + 24) = 10 

A. Number of students who were studying Hindi = 18. TRUE 
B. Number of students who were studying English and Hindi = 3. TRUE
C. Number of students who were studying English, Hindi and German = 3. TRUE

220 = $$2^2\times5\times11$$
So, the Sum of all divisors  = $$(2^0+2^1+2^2)(5^0+5^1)(11^0+11^1)$$
                                         = (7)(6)(12) 
                                         =  504
Proper divisors are all the divisors except number itself .
So, the sum of all proper divisors = $$504-220$$ = 284.

Now to check if the (220,284) are amicable pair of numbers, we have to verify it with checking for sum of proper divisors for 284 : 
284 = $$2^2\left(71\right)$$
So, the Sum of all divisors = $$(2^0+2^1+2^2)(71^0+71^1)$$
                                        = (7)(72)
                                        = 504

Therefore, no of proper divisors of 284 are = 504-284 = 220.

Hence, (220,284) are amicable pair of numbers.

Two numbers a,b can be either odd or even : 

Statement A : (a+b) 
odd + odd = even
even + even = even
even + odd = odd 
odd + even = odd
                             Hence, the probability for summation being even = $$\dfrac{\ 2}{4}$$ = 0.5

Statement B : (ab)
odd x odd = odd
even x even = even
even x odd = even
odd x even = even
                              Hence, the probability for product being even = $$\dfrac{\ 3}{4}$$ = 0.75

Let f(x) = $$2x^{3}+ax^{2}+bx+3$$ 
It is given that (x-1) is a factor of f(x) , so it means x = 1 is a root of f(x)
    f(1) = a + b + 5 = 0
     this implies , a + b = -5 .... Equation 1

It is said that f(x) leaves remainder of 15 when it is divided by (x+2) , so it implies : 
     f(x) = $$\left[2\left[(x+\alpha\right)\left(x\ +\ \beta\right)(x+2)\right]$$ + 15
     $$f(x)-15$$ = $$\left[2\left[(x+\alpha\right)\left(x\ +\ \beta\right)(x+2)\right]$$ 

So, (x+2) is a factor of "f(x)-15" .
  at x = -2, the value of f(x) - 15 = 0
Hence, 15 = -16 + 4a -2b + 3
this implies,  4a - 2b = 28 .... Equation 2

Therefore, by solving both equations 1 & 2 , we get a = 3 & b = -8

Let the upstream and downstream distances be D. Let the speed of the boat in still water be B, and the speed of the current be R.

Aman takes half the time rowing a certain distance downstream than upstream.

$$\dfrac{D}{B+R}=\dfrac{1}{2}\times\dfrac{D}{B-R}$$

$$2B-2R=B+R$$

$$B=3R$$

Statement I

Ratio of speed of boat in still water to the speed of current = B:R = 3R:R = 3:1

Thus, statement I is FALSE

Statement II

Speed of boat in still water (B) = 6 km/h

Speed of river (R) = B/3 = 6/3 = 2 km/h

Thus, statement II is TRUE

There are three cases for exactly two solving it right : 
Case I : A, B solving correct and D going wrong 
                Probability = (1/3)(1/3)(3/4) = 1/12

Case II : A, D solving correct and B going wrong 
                 Probability = (1/3)(2/3)(1/4) = 1/18

Case III : B,D solving correct and A going wrong
                Probability = (2/3)(1/3)(1/4) = 1/18

Therefore, total probability = 1/12 + 1/18 + 1/18 = 1/12 + 1/9 = 7/36

Statement 1 : $$(x^{2}+3x+1)=(x-2)^{2}$$
                    $$X^2+3X+1=X^2-4X+4$$
                     X = 3/7
Therefore, it is not a quadratic equation.

Statement 2 : $$x^{2}+2x\sqrt{3}+3=0$$
                      Discriminant of quadratic equation = $$b^2-4ac$$
                                                                           = $$\left(2\sqrt{3}\right)^2-12$$
                                                                           = 0
Hence, the quadratic equation has real and equal roots.

When two identical cubes are joined side by side , the breadth of the cuboid becomes 2 times the side length of cube.
            So, l = 5, b = 2(5) , h = 5.
So, the surface area of cuboid = 2(lb + bh +lh)
                                              = 2(50 + 50 + 25)
                                              = 250.