JIPMAT 2025 Question Paper

According to question,

In 4 years, an amount of ₹ 6,000 becomes ₹ 8,000 at a certain rate of simple interest.

So, simple interest = ₹ (8,000-6,000) = ₹ 2,000

Now, S.I. = P*r*t/100

so, $$2000=\frac{6000\times\ r\times\ 4}{100}$$

or, r= $$\frac{50}{6}$$%

Now, in second case, simple interest = $$700-525=175$$

So, $$175=525\times\ \frac{50}{6}\times\ \frac{t}{100}$$

or, $$t=175\times\ \frac{12}{525}=4$$

So, time = 4 years

After adding 5 to numerator and denominator of all the fractions,

The fractions will be:

A=$$\dfrac{2+5}{3+5}=\dfrac{7}{8}$$

B=$$\dfrac{3+5}{4+5}=\dfrac{8}{9}$$

C= $$\dfrac{4+5}{5+5}=\dfrac{9}{10}$$

D= $$\dfrac{5+5}{6+5}=\dfrac{10}{11}$$

So, change in value of A = 7/8 - 2/3 = 5/24

change in value of B = 8/9 - 3/4 = 5/36

change in value of C = 9/10 - 4/5 = 5/50

change in value of D = 10/11 - 9/10 =  1/110 = 5/550

Clearly 550>50>36>24

So, 5/550<5/50<5/36<5/24

So, change in value of D<C<B<A

So, option C is the correct answer.

S.P. of glass is ₹ 1,965

Now, loss percent is 25%

Now, S.P. = C.P. - loss

or, $$1965=C.P.\left(1-\dfrac{25}{100}\right)$$

or, $$C.P.=\dfrac{1965}{0.75}=2620$$

Now, in the new case, S.P. = ₹ 3,013

So, profit% = $$\dfrac{3013-2620}{2620}\times\ 100=\dfrac{393}{2620}\times\ 100=15\ \%\ $$

Let's analyse the statements.

Statement A:

Curved surface area of cylinder is given by the formula $$2\pi\ rh$$, where r is the radius of the base of cylinder and h is height

So, we can see in this area is directly proportional to height

So, if height is doubled, area will be doubled

So, statement A is correct.

Statement B:

Total surface area of hemisphere is given by the formula $$3\pi\ r^2$$

So, if radius is doubled, total surface area will become 4 times

So, statement B is also correct.

Statement C:

Let say the radius be r and height be h for both hemisphere and cone

So, curved surface area of cone = $$\pi\ rl$$ = $$\pi\ r\sqrt{\ r^2+h^2}$$

curved surface area of hemisphere = $$2\pi\ r^2$$

So, required ratio is = $$\dfrac{2\pi\ r^2}{\pi\ r\sqrt{\ r^2+h^2}}=\dfrac{2r}{\sqrt{\ r^2+h^2}}$$

From this we can't get a definite numerical value of ratio

So, statement C is wrong.

So, correct answer is option C.

Let, a=3.07-2.15

b=2.15-4.53

c=4.53-3.07

So, a+b+c=3.07-2.15+2.15-4.53+4.53-3.07=0

So, $$\dfrac{(4.53-3.07)^{2}}{(3.07-2.15)(2.15-4.53)}+\dfrac{(3.07-2.15)^{2}}{(2.15-4.53)(4.53-3.07)}+\dfrac{(2.15-4.53)^{2}}{(4.53-3.07)(3.07-2.15)}$$ can be written as

=$$\dfrac{c^2}{ab}+\dfrac{a^2}{bc}+\dfrac{b^2}{ac}$$

=$$\dfrac{a^3+b^3+c^3}{abc}$$

Now, since, $$a+b+c=0$$

So, $$a^3+b^3+c^3=3abc$$ (By property)

So, the value of expression

=$$\dfrac{a^3+b^3+c^3}{abc}$$

=$$\dfrac{3abc}{abc}=3$$

Let's try to calculate each of the values:

List-I:

A. 60% of 264 = 158.4

B. 60% of 28% of 240 = 40.32

C. 85% of 485.5 = 412.675

D. 550% of 250 = 1375

List-II:

I. 20% of 6875 = 1375

II. 50% of 825.35 = 412.675

III. 15% of 1056 = 158.4

IV. 8.4% of 480 = 40.32

So, now we can match the columns,

A-III, B-IV, C-II, D-I

So, option D is the correct answer.

Let, $$\sqrt{\ 2+\sqrt{\ 3}}=x$$

So, the given expression, $$\sqrt{2+\sqrt{3}}\times\sqrt{2+\sqrt{2+\sqrt{3}}}\times\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\times\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$$ can be written as:

=$$x\sqrt{\ 2+x}\sqrt{\ 2+\sqrt{\ 2+x}}\sqrt{\ 2-\sqrt{\ 2+x}}$$

=$$x\sqrt{\ 2+x}\sqrt{\ \left(2\right)^2-\left(\sqrt{\ 2+x}\right)^2}$$

=$$x\sqrt{\ 2+x}\sqrt{\ 4-\left(2+x\right)}$$

=$$x\sqrt{\ 2+x}\sqrt{\ 2-x}$$

=$$x\sqrt{\ 2^2-x^2}$$

=$$x\sqrt{4-x^2}$$

=$$x\sqrt{4-\left(\sqrt{\ 2+\sqrt{\ 3}}\right)^2}$$

=$$x\sqrt{4-\left(\ 2+\sqrt{\ 3}\right)}$$

=$$x\sqrt{2-\sqrt{\ 3}}$$

=$$\sqrt{\ 2+\sqrt{\ 3}}\sqrt{2-\sqrt{\ 3}}$$

=$$\sqrt{\left(2\right)^2-\left(\sqrt{\ 3}\right)^2}$$

=$$\sqrt{\ 4-3}$$

=1

According to question, it is given that,

The traffic lights change after every 48 sec, 72 sec and 108 sec.

So, all three lights will again change simultaneously after a time interval = lcm(48,72,108)

Now, $$48=2^4\times\ 3$$

$$72=2^3\times\ 3^2$$

$$108=2^2\times\ 3^3$$

So, lcm(48,72,108)=$$2^4\times\ 3^3$$ = $$432$$ sec

Also, 432 sec = 7 min and 12 sec

So, they will again change simultaneously after 7 min and 12 sec

Initial time was 08:20:00 hours

So, final time = 08:27:12 hours

Option B is the correct answer

Given, 

$$x^{3}+y^{3}=468$$ and x+y=12

We know,

$$x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)$$

Now, $$12\left(x^2+y^2-xy\right)=468$$

or, $$x^2+y^2-xy=39$$ -->(1)

or, $$\left(x+y\right)^2-2xy-xy=39$$

or, $$12^2-3xy=39$$

or, $$3xy=144-39=105$$

or, $$3xy=105$$

or, $$xy=35$$

Now, from equation (1),

$$x^2+y^2-35=39$$

or, $$x^2+y^2=\ 39+35\ =\ 74$$

So, $$x^4+y^4=\ \left(x^2+y^2\right)^2-2x^2y^2=74^2-2\left(35\right)^2$$

or, $$x^4+y^4=74^2-2\left(35\right)^2$$

or, $$x^4+y^4=5476-2450$$

or, $$x^4+y^4=3026$$

Let A receives ₹a, B receives ₹b, C receives ₹c.

From the question, it can be inferred,

$$\dfrac{a}{b}=\dfrac{4}{5}$$

Also, $$\dfrac{b}{c}=\dfrac{2}{3}$$

We can write $$\dfrac{a}{b}=\dfrac{8}{10},\dfrac{b}{c}=\dfrac{10}{15}$$

So, a:b:c = 8:10:15

Let a= 8x, b=10x, c=15x

Now, a+b+c=11550

So, 8x+10x+15x=11550

or, 33x=11550

or, x=350

Now, c-a = 15x-8x = 7x = 7*350 = ₹2450