JIPMAT 2025 Question Paper

Given, A is 40% less than B

So, $$A=\left(1-0.4\right)B=0.6B$$

Also, C is 40% of sum of A and B

So, $$C=0.4\left(A+B\right)$$

or, $$C=0.4\left(0.6B+B\right)$$

or, $$C=0.4\times\ 1.6B$$

or, $$C=0.64\ B$$

Now, $$B-A=B-0.6B=0.4B$$

So, required percentage = $$\dfrac{0.4B}{C}\times\ 100=\dfrac{0.4B}{0.64B}\times\ 100\ \%\ =62.5\ \%\ $$

Let's calculate the perimeters.

Figure A:

Area = 36 sq. cm

So, side of the square = $$\sqrt{\ 36}$$ cm = 6 cm

So, perimeter of square = 4(side) = $$6\times\ 4=24$$ cm

Figure B:

Perimeter of equilateral triangle = 3(side) = $$3\times\ 9=27$$ cm

Figure C:

Length of rectangle = 10 cm

Area of rectangle = 40 sq. cm

So, breadth of rectangle = (Area)/(Length) = 40/10 = 4 cm

So, perimeter of rectangle = 2(length + breadth) = $$2\left(10+4\right)\ =\ 28$$ cm

Figure D:

Radius of circle is 4 cm

So, perimeter of circle = $$2\pi\ r=2\times\ \dfrac{22}{7}\times\ 4=25.14$$ cm

So, clearly we can see: C>B>D>A is the descending order

Since, the solid brass sphere is converted to right circular cylindrical rod, there volumes must be same

Now volume of sphere = $$\dfrac{4}{3}\pi\ r^3$$ = $$\dfrac{4}{3}\pi\ \left(21\right)^3$$

Volume of cylinder = $$\pi\ r^2l$$ = $$\pi\ r^2\left(28\right)$$

So, $$\dfrac{4}{3}\pi\ \left(21\right)^3$$ = $$\pi\ r^2\left(28\right)$$

or, $$r^2=441$$

or, $$r=21$$ cm

So, basically we can see radius of both the cylinder and sphere are same.

So, total surface area of rod = $$2\pi\ rl+2\pi\ r^2$$

Also, total surface area of sphere = $$4\pi\ r^2$$

So, required ratio = $$\dfrac{2\pi rl+2\pi\ r^2}{4\pi\ r^2}=\dfrac{l+r}{2r}=\dfrac{21+28}{2\times\ 21}=\dfrac{49}{42}=\dfrac{7}{6}$$

Given, ratios of copper to zinc in alloys A and B are 3:4 and 5:9,respectively.

So, ratio of copper to the total amount in alloy A = $$\dfrac{3}{3+4}=\dfrac{3}{7}$$

Ratio of copper to the total amount in alloy B = $$\dfrac{5}{5+9}=\dfrac{5}{14}$$

Now, if A and B are taken in ratio 2:3,

Ratio of copper to the total amount in alloy C = $$\dfrac{2\left(\dfrac{3}{7}\right)+3\left(\dfrac{5}{14}\right)}{2+3}=\dfrac{\dfrac{6}{7}+\dfrac{15}{14}}{5}=\dfrac{\dfrac{12}{14}+\dfrac{15}{14}}{5}=\dfrac{\dfrac{27}{14}}{5}=\dfrac{27}{14\times\ 5}=\dfrac{27}{70}$$

So, ratio of copper to zinc in alloy C = $$\dfrac{27}{70-27}=\dfrac{27}{43}$$

Arranging the data set in ascending order, it is: 8,9,13,13,14,14,14,14,15,15,15,16,16,17,17

So, total number of terms = 15

So, total number of terms are odd

So, median = $$\left(\frac{n+1}{2}\right)^{th}$$ term = $$\left(\frac{15+1}{2}\right)^{th} = 8^{th}$$ term = 14

So, x=14

Now, if 8 is replaced by 18, then the terms when arranged in ascending order will be like: 9,13,13,14,14,14,14,15,15,15,16,16,17,17,18

So, median = $$\left(\frac{n+1}{2}\right)^{th}$$ term = $$\left(\frac{15+1}{2}\right)^{th} = 8^{th}$$ term = 15

So, y=15

So, sum of x and y = 14+15 = 29

Let AB be the lighthouse

Also, let C and D be the points where the ships were present such that angle of elevation of the top of the lighthouse are 30° and 45°respectively.

So, $$\angle\ ACD=30^{\circ\ }$$ and $$\angle\ ADC=45^{\circ\ }$$

Now, $$AB=100$$ m (given)

Triangle ABC is a right angled triangle, so $$BC=\dfrac{AB}{\tan30^{\circ\ }}=\dfrac{100}{\dfrac{1}{\sqrt{\ 3}}}=100\sqrt{\ 3}$$ m

Triangle ABD is a right angled triangle, so $$BD=\dfrac{AB}{\tan45^{\circ\ }}=\dfrac{100}{1}=100$$ m

So, distance between two ships = CD = $$100\sqrt{\ 3}+100$$ m = $$100\left(\sqrt{\ 3}+1\right)=100\left(1.732+1\right)=100\left(2.732\right)=273.2$$ m

So, option C is the correct approximate answer.

Let, cost price of article B be ₹ x

So, cost price of article A = ₹ (x+200)

So, selling price of article A = ₹ $$(x+200)\times\ 0.9=0.9x+180$$

Selling price of article B = ₹ $$1.25x$$

So, total selling price = ₹($$1.25x+0.9x+180$$) = ₹($$2.15x+180$$)

Also, overall profit = 4%

So we can say, ($$2.15x+180$$) = $$1.04\left(x+x+200\right)$$

or, $$2.15x+180=2.08x+208$$

or, $$2.15x-2.08x=208-180$$

or, $$0.07x=28$$

or, $$x=\dfrac{28}{0.07}=400$$

So, cost price of article B = ₹400

Let efficiency of work done by a man be 1 unit/day.

So, total units of work = $$40\times\ 48\times\ 1=1920$$ units

Now, if 64 men worked for x days, so amount of work done = $$64\left(x\right)\left(1\right)=64x$$ units

Now, after x days, total number of men = $$64+32=96$$

So, work done by these 96 men in $$16\ \dfrac{2}{3}$$ days = $$96\times\ 1\times\ 16\ \dfrac{2}{3}$$ units = $$96\times\ 1\times\ \dfrac{50}{3}=1600$$

So, we can say,

$$1920=64x+1600$$

or, $$1920-1600=64x$$

or, $$64x=320$$

or, $$x=\dfrac{320}{64}=5$$

So, value of $$x$$ is 5.

Since, $$a+b=2c$$

or, $$b-c=c-a$$

So, $$\dfrac{a}{a-c}+\dfrac{b}{b-c}$$

=$$\dfrac{a}{a-c}+\dfrac{b}{c-a}$$

=$$\dfrac{a}{a-c}-\dfrac{b}{a-c}$$

=$$\dfrac{a-b}{a-c}$$

Substituting $$a+b=2c$$,

=$$\dfrac{a-b}{a-\dfrac{\left(a+b\right)}{2}}$$

=$$\dfrac{a-b}{\dfrac{2a-a-b}{2}}$$

=$$\dfrac{a-b}{\dfrac{a-b}{2}}$$

=$$\dfrac{2\left(a-b\right)}{\left(a-b\right)}$$

=$$2$$

According to question,

$$x$$ is the least number which when divided by 8, 12, 22, 28, 35 leaves a remainder of 5.

Now instead of dividing $$x$$ individually by 8, 12, 22, 28, 35, we can divide by the LCM of 8,12,20,28,35.

Now, $$8=2^3$$

$$12=2^2\times\ 3$$

$$20=2^2\times5$$

$$28=2^2\times\ 7$$

$$35=5\times\ 7$$

So, LCM of 8,12,20,28,35 = $$2^3\times\ 3\times\ 5\times\ 7$$ = $$840$$

So, I can write, $$x=840k+5$$, where $$k$$ is any positive integer.

So, for the least value of $$x$$, $$k=1$$

So, $$x=840\times\ 1+5$$

or, $$x=845$$

So, sum of digits = $$8+4+5=17$$