JIPMAT 2025 QA PYQs

Q1:jipmat 2025QA › Linear EquationsMediumQA · MCQ

The value of

999\frac{1}{7}+999\frac{2}{7}+999\frac{3}{7}+999\frac{4}{7}+999\frac{5}{7}+999\frac{6}{7}

is equal to:

A5997
B5979
C5994
D2997

Pick an option to attempt

The Setup: We need to evaluate the sum of six mixed fractions. Doing this the traditional way by converting to improper fractions is major NPC energy. Let's break it down using the property of mixed fractions:

A\frac{B}{C}=A+\frac{B}{C}

. Step 1: Separate the whole numbers from the fractions. Since

999

appears

6

times, we can group them up.

S=(999 \times 6)+(\frac{1}{7}+\frac{2}{7}+\frac{3}{7}+\frac{4}{7}+\frac{5}{7}+\frac{6}{7})

Step 2: Calculate the whole number part. Let's use girl math:

999

is practically

1000

.

999 \times 6=(1000-1) \times 6=6000-6=5994

Step 3: Sum up the fractions. The denominators are all matching vibes (they are all

7

).

\frac{1+2+3+4+5+6}{7}=\frac{21}{7}=3

Step 4: Add the two parts together for the ultimate W.

5994+3=5997

Final Answer: 5997

Q2:jipmat 2025QA › ProbabilityMediumQA · MCQ

A speaks truth in 75% cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, in narrating the same incident?

A5%
B15%
C35%
D45%

Pick an option to attempt

The Setup: We are looking for the exact probability that one person is telling the truth while the other is straight capping (lying). Step 1: Define the probabilities. Let

P(A)

be the probability A tells the truth.

P(A)=0.75

. So, probability A lies is

P(A')=0.25

. Let

P(B)

be the probability B tells the truth.

P(B)=0.80

. So, probability B lies is

P(B')=0.20

. Step 2: Find the contradiction scenarios. They contradict when (A tells truth AND B lies) OR (A lies AND B tells truth).

P(\text{Contradict})=(P(A) \times P(B'))+(P(A') \times P(B))

Step 3: Plug in the numbers and do the math.

P(\text{Contradict})=(0.75 \times 0.20)+(0.25 \times 0.80)

P(\text{Contradict})=0.15+0.20=0.35

Step 4: Convert to a percentage.

0.35 \times 100\%=35\%

. Final Answer: 35%

Q3:jipmat 2025QA › Quadratic EquationsMediumQA · MCQ

Amit and Alok attempted to solve a quadratic equation. Amit made a mistake in writing down the constant term and ended up with roots (4, 3). Alok made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of the equation are:

A-4, -3
B6, 1
C4, 3
D-6, -1

Pick an option to attempt

The Setup: We are reconstructing a standard quadratic equation

x^2+bx+c=0

. If you fumble the constant (

c

), your sum of roots (

-b

) is still safe. If you fumble the

x

coefficient (

b

), your product of roots (

c

) remains a W. Step 1: Analyze Amit's work. He messed up the constant term, meaning his sum of roots is straight facts. Roots:

4

and

3

. Sum of roots

=4+3=7

. Since sum of roots

=-b

, we get

b=-7

. Step 2: Analyze Alok's work. He messed up the

x

coefficient, meaning his product of roots is legit. Roots:

3

and

2

. Product of roots

=3 \times 2=6

. Since product of roots

=c

, we get

c=6

. Step 3: Build the correct equation and solve it like a boss. Equation:

x^2-7x+6=0

. Factorizing:

(x-6)(x-1)=0

. Correct roots are

6

and

1

. Final Answer: 6, 1

Q4:jipmat 2025QA › Permutation & CombinationEasyQA · MCQ

How many three-digit even numbers can be formed using the digits 1, 2, 3, 4 and 5, when repetition of digits is not allowed?

A36
B30
C24
D12

Pick an option to attempt

The Setup: We need to build a 3-digit number from the set

\{1, 2, 3, 4, 5\}

without any repeating digits. To be an even number, the last digit has to pass the vibe check (it must be divisible by 2). Step 1: Fix the units digit. The only even digits in our squad are

2

and

4

. That gives us

2

valid options for the final slot. Step 2: Fill the remaining slots. We have a 3-digit number (Hundreds, Tens, Units). We used

1

digit for the units place, leaving

4

digits available. For the Hundreds place, we have

4

options. For the Tens place, we have

3

options. Step 3: Multiply them together using the Fundamental Principle of Counting.

Total=4 \times 3 \times 2=24

Final Answer: 24

Q5:jipmat 2025QA › QuadrilateralsMediumQA · MCQ

From the four corners of a rectangular sheet of dimensions 25 cm x 20 cm, square of side 2 cm is cut off from four corners and a box is made. The volume of the box is:

A $828\text{ cm}^3$
B $672\text{ cm}^3$
C $500\text{ cm}^3$
D $1000\text{ cm}^3$

Pick an option to attempt

The Setup: Imagine we're Minecrafting an open box out of a 2D sheet. By cutting squares out of the corners and folding the flaps up, we create a 3D cuboid. The side length of the cut square automatically becomes the height of our new box. Step 1: Determine the dimensions of the box's base. The original length is

25\text{ cm}

. Cutting

2\text{ cm}

from *both* ends leaves a new length:

L=25-2(2)=25-4=21\text{ cm}

The original width is

20\text{ cm}

. Cutting

2\text{ cm}

from *both* ends leaves a new width:

W=20-2(2)=20-4=16\text{ cm}

Step 2: Identify the height. The flaps we fold up are exactly the length of the square's side, so

H=2\text{ cm}

. Step 3: Calculate the volume of the cuboid (

V=L \times W \times H

).

V=21 \times 16 \times 2

V=336 \times 2=672\text{ cm}^3

Final Answer: 672

Q6:jipmat 2025QA › Divisibility RulesMediumQA · MCQ

2^{122}+4^{62}+8^{42}+4^{64}+2^{130}

is divisible by which one of the following integers?

A3
B5
C7
D11

Pick an option to attempt

The Setup: This expression looks like a final boss battle, but it's really just base-2 math in disguise. We need to convert all terms to base

2

and factor out the smallest common power. Step 1: Convert bases

4

and

8

to base

2

.

4^{62}=(2^2)^{62}=2^{124}

8^{42}=(2^3)^{42}=2^{126}

4^{64}=(2^2)^{64}=2^{128}

Step 2: Rewrite the entire expression.

E=2^{122}+2^{124}+2^{126}+2^{128}+2^{130}

Step 3: Factor out the lowest power, which is

2^{122}

.

E=2^{122}(1+2^2+2^4+2^6+2^8)

E=2^{122}(1+4+16+64+256)

E=2^{122}(341)

Step 4: Check the divisibility of

341

among the given options. Let's test

11

using the alternating sum rule:

3-4+1=0

, which is completely divisible by

11

. Since

341=11 \times 31

, the whole expression is a multiple of

11

. Final Answer: 11

Q7:jipmat 2025QA › PolygonsMediumQA · MCQ

The cost of fencing of an equilateral triangular park and a square park is the same. If the area of the triangular park is

16\sqrt{3}\text{ m}^2

then the length of the diagonal of the square park is:

A $6\sqrt{3}\text{ m}$
B $8\sqrt{3}\text{ m}$
C $8\sqrt{2}\text{ m}$
D $6\sqrt{2}\text{ m}$

Pick an option to attempt

The Setup: The fencing costs are identical, meaning their perimeters perfectly match each other's energy (they are exactly equal). We need to find the side of the triangle, use it to get the perimeter, transfer that perimeter to the square, and finally find the square's diagonal. Step 1: Find the side length of the equilateral triangle (

a

). Area formula:

\frac{\sqrt{3}}{4}a^2=16\sqrt{3}

. Cancel out

\sqrt{3}

:

\frac{a^2}{4}=16

.

a^2=64 \implies a=8\text{ m}

. Step 2: Calculate the perimeter of the triangle.

P=3 \times 8=24\text{ m}

Step 3: Find the side of the square (

s

). Since perimeters are equal:

4s=24 \implies s=6\text{ m}

Step 4: Find the diagonal of the square using the formula

d=s\sqrt{2}

.

d=6\sqrt{2}\text{ m}

Final Answer:

6\sqrt{2}

Q8:jipmat 2025QA › Simple & Compound InterestMediumQA · MCQ

The difference between compound and simple interests on a certain sum of money at the interest rate of 10% per annum for

1\frac{1}{2}

years is Rs.183, when the interest is compounded semi-annually, then the sum of money is:

A₹22,000
B₹24,000
C₹26,000
D₹28,000

Pick an option to attempt

The Setup: Compound interest is basically the financial version of a snowball effect. Since it's compounded semi-annually, we need to adjust the rate (

R

) and time (

T

) to reflect half-year cycles before hitting the formula. Step 1: Adjust the variables for semi-annual compounding. Rate per half-year:

R=\frac{10\%}{2}=5\%

. Number of cycles in

1.5

years:

n=1.5 \times 2=3

cycles. Step 2: Use the standard formula for the difference between CI and SI for 3 compounding cycles.

\text{Diff}=P \times (\frac{R}{100})^2 \times (\frac{300+R}{100})

Step 3: Plug in our adjusted rate (

R=5

) and the given difference (Rs.183).

183=P \times (\frac{5}{100})^2 \times (\frac{300+5}{100})

183=P \times (\frac{1}{20})^2 \times (\frac{305}{100})

183=P \times \frac{1}{400} \times \frac{61}{20}

Step 4: Solve for the principal sum (

P

).

P=\frac{183 \times 400 \times 20}{61}

Since

183/61=3

:

P=3 \times 8000=24000

Final Answer: 24000

Q9:jipmat 2025QA › Linear EquationsMediumQA · MCQ

Which of the following statements is/are correct? A. If

2^x=3^y=6^{-z}

, then

\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0

B.

(243)^{0.16} \times (9)^{0.1}=0.3

C. If

3.105 \times 10^P=0.00239+0.000715

, then

P=-3

AB only
BB and C only
CA and B only
DA and C only

Pick an option to attempt

The Setup: We've got a classic 'two truths and a lie' situation. Let's audit each statement mathematically to see which ones are actually valid and which one is straight capping. Step 1: Fact-check Statement A. Let

2^x=3^y=6^{-z}=k

. This implies

2=k^{\frac{1}{x}}

,

3=k^{\frac{1}{y}}

, and

6=k^{-\frac{1}{z}}

. We know

2 \times 3=6

. Substituting the

k

terms:

k^{\frac{1}{x}} \times k^{\frac{1}{y}}=k^{-\frac{1}{z}}

k^{\frac{1}{x}+\frac{1}{y}}=k^{-\frac{1}{z}}

Equating the powers:

\frac{1}{x}+\frac{1}{y}=-\frac{1}{z} \implies \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0

. Statement A is a W (True). Step 2: Fact-check Statement B. Convert bases to powers of

3

:

243=3^5

and

9=3^2

.

(3^5)^{0.16} \times (3^2)^{0.1}=3^{0.8} \times 3^{0.2}

3^{0.8+0.2}=3^1=3

The statement claims it equals

0.3

. Total cap. Statement B is False. Step 3: Fact-check Statement C. Add the decimals on the right side:

0.00239+0.000715=0.003105

. Equation:

3.105 \times 10^P=0.003105

. To get from

3.105

to

0.003105

, we move the decimal left by

3

places, meaning multiplying by

10^{-3}

. So

P=-3

. Statement C is True. Final Answer: A and C only

Q10:jipmat 2025QA › AveragesMediumQA · MCQ

A car owner buys petrol at the rate Rs.17, Rs.19 and Rs.20 per litre, respectively for three consecutive years. Compute the average cost per litre, if he spends Rs.6,460 per year for the three consecutive years.

A18.49
B18.58
C19.20
D21.66

Pick an option to attempt

The Setup: Gas prices are doing us dirty, and taking a simple arithmetic mean of the prices won't work since the money spent is constant, not the volume. We need to find the total money spent and divide it by the total liters of petrol bought to find the true weighted average. Step 1: Calculate the volume of petrol bought each year. Year 1:

\frac{6460}{17}=380

liters. Year 2:

\frac{6460}{19}=340

liters. Year 3:

\frac{6460}{20}=323

liters. Step 2: Calculate total volume and total cash dropped across all three years. Total liters

=380+340+323=1043

liters. Total spent

=6460 \times 3=19380

rupees. Step 3: Find the true average cost per liter.

\text{Average}=\frac{\text{Total Spent}}{\text{Total Liters}}=\frac{19380}{1043}

\text{Average} \approx 18.581

Final Answer: 18.58

Q11:jipmat 2025QA › Simple & Compound InterestMediumQA · MCQ

In 4 years, an amount of Rs.6,000 becomes Rs.8,000 at a certain rate of simple interest. In what time at the same simple interest rate, will an amount of Rs.525 become Rs.700?

A2 years
B3 years
C4 years
D5 years

Pick an option to attempt

The Setup: Simple interest is literally just basic scaling. We could find the exact rate (

R

), but that's NPC behavior. Instead, let's use ratio logic because the interest rate is the exact same vibe. Step 1: Calculate the interest generated in the first scenario. Amount becomes Rs.8000 from Rs.6000.

\text{Interest}=8000-6000=2000

. Step 2: Find the ratio of Interest to Principal (

\frac{I}{P}

) for the first scenario.

\frac{I}{P}=\frac{2000}{6000}=\frac{1}{3}

. This means in

4

years, the money grows by one-third of its original value. Step 3: Calculate the interest needed for the second scenario. Amount needs to become Rs.700 from Rs.525.

\text{Interest}=700-525=175

. Step 4: Find the

\frac{I}{P}

ratio for the second scenario to see if it matches the energy.

\frac{I}{P}=\frac{175}{525}=\frac{1}{3}

. Step 5: Since the growth ratio (

\frac{1}{3}

) is exactly the same, and the interest rate hasn't changed, the time it takes must also be exactly the same. No extra math required, it's a straight

4

years. Final Answer: 4

Q12:jipmat 2025QA › MiscellaneousMediumQA · MCQ

After adding 5 to both numerator and denominator of the following fractions, arrange them in ascending order of the magnitude of change in their values. A.

\frac{2}{3}

B.

\frac{3}{4}

C.

\frac{4}{5}

D.

\frac{5}{6}

AD < C < A < B
BD < A < B < C
CD < C < B < A
DA < B < C < D

Pick an option to attempt

The Setup: We are giving these fractions a +5 glow-up on both the top and bottom. We need to measure how much their value *changed* (New - Old), and then rank those changes from smallest to largest. Step 1: Create a general formula for the change. Let the fraction be

\frac{n}{d}

.

\text{Change}=\frac{n+5}{d+5}-\frac{n}{d}

\text{Change}=\frac{d(n+5)-n(d+5)}{d(d+5)}

\text{Change}=\frac{nd+5d-nd-5n}{d(d+5)}=\frac{5(d-n)}{d(d+5)}

Step 2: Notice the cheat code. For all our fractions (A, B, C, D), the difference between the denominator and numerator is exactly

1

(

3-2=1

,

4-3=1

, etc.). So

d-n=1

for all of them! The change formula simplifies to just

\frac{5}{d(d+5)}

. Step 3: Evaluate the magnitude of change for each based on its denominator (

d

). Since the numerator is always

5

, the fraction with the *largest* denominator will have the *smallest* change (inversely proportional vibes). Denominator of A (

d=3

)

\rightarrow

Smallest denominator = Largest change. Denominator of D (

d=6

)

\rightarrow

Largest denominator = Smallest change. Step 4: Rank them in ascending order (smallest change to largest change). Since the denominators are

6 > 5 > 4 > 3

, the changes are D < C < B < A. Final Answer: D < C < B < A

Q13:jipmat 2025QA › Profit & LossMediumQA · MCQ

Selling price of a glass is Rs.1,965 and loss percent is 25%. If its selling price is Rs.3,013, then what will be profit percent?

A13%
B10.40%
C15%
D20%

Pick an option to attempt

The Setup: They fumbled the bag on the first sale and took an L (loss). We need to reverse-engineer the original Cost Price (CP) and then calculate the stonks (profit) on the new Selling Price (SP). Step 1: Find the Cost Price. A

25\%

loss means the SP is

75\%

of the CP (

0.75 \times \text{CP}

).

1965=0.75 \times \text{CP}

\text{CP}=\frac{1965}{0.75}=\frac{1965}{3/4}

\text{CP}=1965 \times \frac{4}{3}=655 \times 4=2620

Step 2: Calculate the new profit. The new SP is

Rs.3013

.

\text{Profit}=\text{SP}-\text{CP}=3013-2620=393

Step 3: Convert that profit into a percentage.

\text{Profit \%}=\frac{\text{Profit}}{\text{CP}} \times 100

\text{Profit \%}=\frac{393}{2620} \times 100

Cancel a zero:

\frac{393}{262} \times 10

. Notice that

262 \times 1.5=393

.

\text{Profit \%}=1.5 \times 10=15\%

Final Answer: 15%

Q14:jipmat 2025QA › SolidsHardQA · MCQ

Consider the following statements, which of them is/are correct? A. If the height of cylinder is doubled, the area of curved surface is doubled. B. If the radius of a hemispherical solid is doubled, its total surface area becomes fourfold. C. If a hemisphere and cone have equal bases and equal heights, then the ratio of curved surface area is

\sqrt{2}:1

.

AB and C only
BA and C only
CA and B only
DA, B and C

Pick an option to attempt

The Setup: We are fact-checking 3D geometry claims. Let's run the formulas for each statement to see who is speaking straight facts and who is capping. Step 1: Check Statement A. Curved Surface Area (CSA) of a cylinder

=2\pi rh

. If

h

becomes

2h

, the new CSA

=2\pi r(2h)=2(2\pi rh)

. It perfectly doubles. Statement A is a W. Step 2: Check Statement B. Total Surface Area (TSA) of a hemisphere

=3\pi r^2

. If

r

becomes

2r

, the new TSA

=3\pi(2r)^2=3\pi(4r^2)=4(3\pi r^2)

. It scales by a factor of 4 (fourfold). Statement B is a W. Step 3: Check Statement C. Hemisphere CSA

=2\pi r^2

. For the cone, if it has the same base and height as the hemisphere, its radius is

r

and its height is also

r

(since a hemisphere's height equals its radius). Cone slant height (

l

):

l=\sqrt{r^2+h^2}=\sqrt{r^2+r^2}=\sqrt{2r^2}=r\sqrt{2}

. Cone CSA

=\pi rl=\pi r(r\sqrt{2})=\pi r^2\sqrt{2}

. Ratio of Hemisphere CSA to Cone CSA:

\frac{2\pi r^2}{\pi r^2\sqrt{2}}=\frac{2}{\sqrt{2}}=\frac{\sqrt{2}}{1}

The ratio is exactly

\sqrt{2}:1

. Statement C is a W. Step 4: Since all three are legit, option D is the one. Final Answer: A, B and C

Q15:jipmat 2025QA › IdentitiesHardQA · MCQ

\frac{(4.53-3.07)^2}{(3.07-2.15)(2.15-4.53)}+\frac{(3.07-2.15)^2}{(2.15-4.53)(4.53-3.07)}+\frac{(2.15-4.53)^2}{(4.53-3.07)(3.07-2.15)}

is simplified to

A0
B1
C2
D3

Pick an option to attempt

The Setup: Doing the decimals here would be a colossal waste of time. This is a classic algebraic identity in disguise. Let's sub in variables to reveal the true form. Step 1: Let

x=4.53-3.07

, let

y=3.07-2.15

, and let

z=2.15-4.53

. Notice that if you add them all up, everything cancels out perfectly:

x+y+z=(4.53-3.07)+(3.07-2.15)+(2.15-4.53)=0

. Step 2: Rewrite the absolute monstrosity of an expression using

x, y, z

.

E=\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}

Step 3: Find a common denominator (

xyz

) and add the fractions.

E=\frac{x^3}{xyz}+\frac{y^3}{xyz}+\frac{z^3}{xyz}=\frac{x^3+y^3+z^3}{xyz}

Step 4: Deploy the master identity. If

x+y+z=0

, then

x^3+y^3+z^3=3xyz

. Substitute

3xyz

into the numerator:

E=\frac{3xyz}{xyz}=3

Final Answer: 3

Q16:jipmat 2025QA › PercentagesMediumQA · MCQ

Match the LIST-I with LIST-II. LIST-I A. 60% of 264 B. 60% of 28% of 240 C. 85% of 485.5 D. 550% of 250 LIST-II I. 20% of 6875 II. 50% of 825.35 III. 15% of 1056 IV. 8.4% of 480 Choose the correct answer from the options given below:

AA-II, B-IV, C-III, D-I
BA-II, B-III, C-I, D-IV
CA-III, B-IV, C-I, D-II
DA-III, B-IV, C-II, D-I

Pick an option to attempt

Don't evaluate all eight - start with the entry that is quickest to compute and let elimination do the rest. D is the cleanest:

550\% \times 250 = \frac{550 \times 250}{100} = 55 \times 25 = 1375

. Scanning List-II, I gives

20\% \times 6875 = \frac{6875}{5} = 1375

. So D-I, which already eliminates Options 2 and 3. A is the next easy one:

0.6 \times 264 = 158.4

. In List-II, III gives

15\% \times 1056

- split it as

10\%=105.6

plus

5\%=52.8

, total

158.4

. So A-III, and that settles it as Option 4. The remaining two confirm it: B:

0.6 \times 0.28 \times 240 = 40.32

, matching IV:

0.084 \times 480 = 40.32

. C:

0.85 \times 485.5 = 412.675

, matching II:

0.5 \times 825.35 = 412.675

. Hence A-III, B-IV, C-II, D-I - Option 4.

Q17:jipmat 2025QA › IndicesHardQA · MCQ

\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}}}}

is equal to

A1
B2
C4
D $\sqrt{6}$

Pick an option to attempt

The Setup: This looks like a terrifying infinite radical, but it's actually a satisfying chain reaction. We work from right to left, using the difference of squares identity:

(a+b)(a-b)=a^2-b^2

. Step 1: Multiply the last two terms together. They are identical except for the sign in the middle. Let

X=\sqrt{2+\sqrt{3}}

. The last two terms are

\sqrt{2+\sqrt{X}}

and

\sqrt{2-\sqrt{X}}

.

\sqrt{2+\sqrt{X}} \times \sqrt{2-\sqrt{X}}=\sqrt{(2)^2-(\sqrt{X})^2}=\sqrt{4-X}

Substitute

X

back in:

\sqrt{4-(2+\sqrt{3})}=\sqrt{2-\sqrt{3}}

Step 2: Now our expression is shorter. Multiply this new result by the second term from the original equation. Let's drop the visual noise.

\sqrt{2+\sqrt{3}} \times (\sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2-\sqrt{3}})

Apply the difference of squares again! Let

Y=\sqrt{3}

.

\sqrt{2+\sqrt{Y}} \times \sqrt{2-\sqrt{Y}}=\sqrt{4-Y}=\sqrt{4-(2+\sqrt{3})}=\sqrt{2-\sqrt{3}}

Step 3: Final stage. Multiply this with the very first term.

\sqrt{2+\sqrt{3}} \times \sqrt{2-\sqrt{3}}

\sqrt{(2)^2-(\sqrt{3})^2}=\sqrt{4-3}=\sqrt{1}=1

Final Answer: 1

Q18:jipmat 2025QA › HCF & LCMEasyQA · MCQ

The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec, respectively. They all change simultaneously at 08:20:00 hours, when will they again change simultaneously?

A9:27:12 hours
B8:27:12 hours
C7:27:13 hours
D11:00:12 hours

Pick an option to attempt

The Setup: To find when cyclic events sync up perfectly IRL, we need the Least Common Multiple (LCM) of their intervals. Step 1: Find the prime factorization of each number to build the LCM.

48=16 \times 3=2^4 \times 3^1

72=8 \times 9=2^3 \times 3^2

108=4 \times 27=2^2 \times 3^3

Step 2: Calculate the LCM by taking the highest power of each prime factor present.

\text{LCM}=2^4 \times 3^3=16 \times 27

\text{LCM}=432\text{ seconds}

Step 3: Convert the seconds into minutes and seconds.

432 \div 60=7

with a remainder of

12

. So, it takes

7\text{ minutes and }12\text{ seconds}

for them to sync up again. Step 4: Add this time gap to the original starting time. Starting time: 08:20:00. Adding

0\text{ hours, }7\text{ mins, }12\text{ secs}

gives us 08:27:12. Final Answer: 8:27:12 hours

Q19:jipmat 2025QA › IdentitiesMediumQA · MCQ

If

x^3+y^3=468

and

x+y=12

, then value of

x^4+y^4

will be

A3620
B2036
C3025
D3026

Pick an option to attempt

The Setup: We are stepping up the algebraic ladder here. We need to find

xy

, use it to find

x^2+y^2

, and then square that to finally unlock

x^4+y^4

. It's a three-phase boss fight. Step 1: Find

xy

using the sum of cubes identity.

x^3+y^3=(x+y)((x+y)^2-3xy)

468=12(12^2-3xy)

468=12(144-3xy)

Divide by 12:

39=144-3xy \implies 3xy=105 \implies xy=35

Step 2: Find

x^2+y^2

.

x^2+y^2=(x+y)^2-2xy

x^2+y^2=(12)^2-2(35)

x^2+y^2=144-70=74

Step 3: Find

x^4+y^4

by squaring the squares.

(x^2+y^2)^2=x^4+y^4+2x^2y^2

We can rewrite this as

x^4+y^4=(x^2+y^2)^2-2(xy)^2

. Substitute the values we've farmed:

x^4+y^4=(74)^2-2(35)^2

74^2=(70+4)^2=4900+560+16=5476

.

2(35)^2=2(1225)=2450

.

x^4+y^4=5476-2450=3026

Final Answer: 3026

Q20:jipmat 2025QA › Ratio, Proportion & VariationMediumQA · MCQ

Rs.11,550 has to be divided between A, B and C such that A gets 4/5 of what B gets and B gets 2/3 of what C gets. How much more does C get in comparison to A (in Rs.)?

A7,200
B1,800
C1,170
D2,450

Pick an option to attempt

The Setup: We're splitting the bill, but it's nested. We need to standardize their shares into a single continuous ratio

A:B:C

to figure out who secured what bag. Step 1: Write out the individual ratios.

A=\frac{4}{5}B \implies A:B=4:5

.

B=\frac{2}{3}C \implies B:C=2:3

. Step 2: Combine them into one super-ratio. B is the middleman, so we make B's value equal in both ratios by multiplying. Multiply

A:B

by

2 \rightarrow 8:10

. Multiply

B:C

by

5 \rightarrow 10:15

. Now they link up perfectly:

A:B:C=8:10:15

. Step 3: Find the value of one 'part' of the ratio. Total parts

=8+10+15=33\text{ parts}

.

33\text{ parts}=11550

.

1\text{ part}=\frac{11550}{33}=350

. Step 4: Find the difference between C's bag and A's bag. C's parts

=15

. A's parts

=8

. Difference

=15-8=7\text{ parts}

. Value of difference

=7 \times 350=2450

. Final Answer: 2450

Q21:jipmat 2025QA › PercentagesEasyQA · MCQ

Three numbers A, B and C are such that A is 40% less than B, and C is 40% of the sum of A and B. The difference between A and B is what percentage of C?

A60.50%
B64%
C62.50%
D60%

Pick an option to attempt

The Setup: This is a classic percentage chain. The easiest way to not fumble the bag is to anchor one of the variables to

100

and let the rest fall into place. Step 1: Let B be the anchor. We set

B=100

. A is

40\%

less than B. So,

A=100-40=60

. Step 2: Find the value of C based on A and B. C is

40\%

of the sum of A and B.

\text{Sum}=60+100=160

.

C=0.40 \times 160=64

. Step 3: Find the difference between A and B.

\text{Difference}=B-A=100-60=40

. Step 4: Calculate what percentage this difference is of C.

\text{Percentage}=\frac{\text{Difference}}{C} \times 100

\text{Percentage}=\frac{40}{64} \times 100

Simplify the fraction by dividing by 8:

\frac{5}{8}

. Since

\frac{1}{8}=12.5\%

, then

\frac{5}{8}=5 \times 12.5\%=62.5\%

. Final Answer: 62.50%

Q22:jipmat 2025QA › PolygonsEasyQA · MCQ

Arrange the following in descending order based on their perimeters: A. A square with an area of 36 sq. cm. B. An equilateral triangle with a side of 9 cm. C. A rectangle with 10 cm as length and 40 sq. cm as area. D. A circle with radius of 4 cm.

AC > A > B > D
BD > A > B > C
CC > B > A > D
DC > B > D > A

Pick an option to attempt

The Setup: We need to calculate the perimeter (or circumference) for each shape to see who is really built different, then rank them from largest to smallest. Step 1: Calculate A (Square).

\text{Area}=s^2=36 \implies s=6\text{ cm}

.

\text{Perimeter}=4s=4 \times 6=24\text{ cm}

. Step 2: Calculate B (Equilateral Triangle). Side is

9\text{ cm}

.

\text{Perimeter}=3s=3 \times 9=27\text{ cm}

. Step 3: Calculate C (Rectangle).

\text{Area}=L \times W \implies 40=10 \times W \implies W=4\text{ cm}

.

\text{Perimeter}=2(L+W)=2(10+4)=2(14)=28\text{ cm}

. Step 4: Calculate D (Circle). Radius

r=4\text{ cm}

.

\text{Circumference}=2\pi r=2 \times 3.1415 \times 4 \approx 25.13\text{ cm}

. Step 5: Rank them in descending order (biggest first).

C(28) > B(27) > D(25.13) > A(24)

. Final Answer: C > B > D > A

Q23:jipmat 2025QA › SolidsMediumQA · MCQ

A solid brass sphere of radius 21 cm is converted into a right circular cylindrical rod of length 28 cm. The ratio of total surface areas of the rod to sphere is :

A3:1
B7:6
C7:3
D3:7

Pick an option to attempt

The Setup: Melting one 3D shape into another means their volumes are identical. We'll use the volume equality to find the missing radius of the cylinder, then compare their Total Surface Areas (TSA). Step 1: Equate the volumes to find the cylinder's radius (

R

). Sphere Volume = Cylinder Volume

\frac{4}{3}\pi r^3=\pi R^2h

Plug in the given specs:

r=21

and

h=28

.

\frac{4}{3} \times (21)^3=R^2 \times 28

\frac{4}{3} \times 9261=R^2 \times 28

4 \times 3087=R^2 \times 28

Divide by 28:

R^2=\frac{4 \times 3087}{28}=\frac{3087}{7}=441

R=\sqrt{441}=21\text{ cm}

So, the cylinder's radius matches the sphere's radius! Step 2: Calculate the Total Surface Area (TSA) of the cylinder (rod).

\text{TSA}_{\text{rod}}=2\pi R(R+h)

\text{TSA}_{\text{rod}}=2\pi(21)(21+28)=2\pi(21)(49)

Step 3: Calculate the Total Surface Area (TSA) of the sphere.

\text{TSA}_{\text{sphere}}=4\pi r^2=4\pi(21)^2

Step 4: Find the ratio.

\text{Ratio}=\frac{2\pi(21)(49)}{4\pi(21)(21)}

Cancel the

\pi

and one

21

:

\text{Ratio}=\frac{2 \times 49}{4 \times 21}=\frac{98}{84}

Divide top and bottom by 14:

\text{Ratio}=\frac{7}{6}

Final Answer: 7:6

Q24:jipmat 2025QA › Mixture & AlligationMediumQA · MCQ

The ratios of copper to zinc in alloys A and B are 3:4 and 5:9, respectively. If A and B are taken in ratio 2:3 and melted to form new alloy C, then the ratio of copper to zinc in C is:

A8:13
B3:5
C9:11
D27:43

Pick an option to attempt

The Setup: We are mixing two different alloys like mixing tracks. To avoid terrifying fractions, we need to pick weights for A and B that are easily divisible by their internal ratio sums. Step 1: Analyze the parts. Alloy A parts:

3+4=7

. Alloy B parts:

5+9=14

. The LCM of 7 and 14 is 14. We want to mix them in a

2:3

ratio. Let's make the multiplier

14

to keep things clean. Weight of A we use

=2 \times 14=28\text{ units}

. Weight of B we use

=3 \times 14=42\text{ units}

. Step 2: Break down Alloy A into Copper and Zinc. Copper in A

=\frac{3}{7} \times 28=12

. Zinc in A

=\frac{4}{7} \times 28=16

. Step 3: Break down Alloy B into Copper and Zinc. Copper in B

=\frac{5}{14} \times 42=15

. Zinc in B

=\frac{9}{14} \times 42=27

. Step 4: Combine them for the ultimate Alloy C. Total Copper

=12+15=27

. Total Zinc

=16+27=43

. The final ratio of Copper to Zinc in C is exactly

27:43

. Final Answer: 27:43

Q25:jipmat 2025QA › AveragesMediumQA · MCQ

Let x be median of the data 13, 8, 15, 14, 17, 9, 14, 16, 13, 17, 14, 15, 16, 15, 14. If 8 is replaced by 18, then the median of data is y, then the sum of x and y is equal to:

A30
B27
C28
D29

Pick an option to attempt

The Setup: The median is literally just the middle child of a sorted dataset. We need to sort the raw data to find the first median (

x

), swap a number, re-sort, find the second median (

y

), and add them. Step 1: Count the data points and sort the initial list. There are

15

numbers. The median will be the

(\frac{15+1}{2})^{\text{th}}

term, which is the

8^{\text{th}}

term. Sorted data:

8, 9, 13, 13, 14, 14, 14, \textbf{14}, 15, 15, 15, 16, 16, 17, 17

. The

8^{\text{th}}

term is

14

. So,

x=14

. Step 2: Execute the swap. Replace the

8

with an

18

. The

18

will move to the far right of the sorted list, shifting everything below it down a spot. New sorted data:

9, 13, 13, 14, 14, 14, 14, \textbf{15}, 15, 15, 16, 16, 17, 17, 18

. Step 3: Find the new median (

y

). The

8^{\text{th}}

term is now

15

. So,

y=15

. Step 4: Calculate the final sum.

x+y=14+15=29

Final Answer: 29

Q26:jipmat 2025QA › TrigonometryMediumQA · MCQ

Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from two ships are 30 deg and 45 deg respectively. If the lighthouse is 100 m high, the distance between two ships is approximately:

A173 m
B200 m
C273 m
D300 m

Pick an option to attempt

The Setup: We are dealing with two right-angled triangles sharing the same vertical height (the lighthouse). The ships are on *opposite* sides, meaning the total distance between them is the sum of their individual horizontal distances to the base of the lighthouse. Step 1: Calculate distance for the first ship (

d_1

). Using the tangent ratio (opposite/adjacent):

\tan(30^\circ)=\frac{100}{d_1}

\frac{1}{\sqrt{3}}=\frac{100}{d_1} \implies d_1=100\sqrt{3}

Since

\sqrt{3} \approx 1.732

:

d_1=100(1.732)=173.2\text{ m}

Step 2: Calculate distance for the second ship (

d_2

).

\tan(45^\circ)=\frac{100}{d_2}

1=\frac{100}{d_2} \implies d_2=100\text{ m}

Step 3: Add them together for the total distance.

\text{Total Distance}=d_1+d_2=173.2+100=273.2\text{ m}

Rounding to the nearest whole number gives us

273

. Final Answer: 273

Q27:jipmat 2025QA › Profit & LossMediumQA · MCQ

Cost price of article A is Rs.200 more than the cost price of article B. Article A was sold at 10% loss and article B was sold at 25% profit. If the overall profit earned after selling both the articles is 4%, then what is the cost price of article B?

A₹300
B₹400
C₹450
D₹500

Pick an option to attempt

The Setup: This is a balancing act of stonks and Ls. We can set up an algebraic equation representing the net profit across both items to zero in on B's cost price. Let the cost price of B be

x

. That makes the cost price of A equal to

x+200

. Step 1: Express the profit/loss on each individual item. Profit on B

=+25\%

of

x=0.25x

. Loss on A

=-10\%

of

(x+200)=-0.10(x+200)=-0.10x-20

. Step 2: Combine them to find the net monetary profit.

\text{Net Profit}=0.25x-0.10x-20=0.15x-20

Step 3: Set up the overall percentage equation. The total cost price of everything is

x+(x+200)=2x+200

. The problem states the overall profit is

4\%

on the total cost. Overall Profit

=0.04(2x+200)=0.08x+8

. Step 4: Equate the two profit expressions and solve for

x

.

0.15x-20=0.08x+8

Subtract

0.08x

from both sides:

0.07x-20=8

0.07x=28

x=\frac{28}{0.07}=\frac{2800}{7}=400

So, the cost price of article B is Rs.400. Final Answer: 400

Q28:jipmat 2025QA › Time & WorkMediumQA · MCQ

40 men can complete a work in 48 days. 64 men started and did the same work for x days. After x days, 32 men increased, so, the remaining work is completed in

16\frac{2}{3}

days, then the value of x is

A5
B8
C10
D6

Pick an option to attempt

The Setup: We are measuring this task in 'man-days' (the amount of grinding one guy does in one day). The total work is a fixed pool of points, and we just need to track who did how much. Step 1: Calculate the total work pool.

\text{Total Work}=40\text{ men} \times 48\text{ days}=1920\text{ man-days}

Step 2: Track the grinding phases. Phase 1:

64\text{ men}

work for

x\text{ days}

. Work done in Phase 1

=64x

. Phase 2: The squad increases by

32

men (

64+32=96\text{ men}

). They grind for

16\frac{2}{3}

days. Convert

16\frac{2}{3}

to an improper fraction:

\frac{50}{3}

. Work done in Phase 2

=96 \times \frac{50}{3}

.

96 \div 3=32

, so Phase 2 work

=32 \times 50=1600\text{ man-days}

. Step 3: Set up the master equation. Total work must equal the sum of both phases.

64x+1600=1920

64x=1920-1600

64x=320

x=\frac{320}{64}=5

Final Answer: 5

Q29:jipmat 2025QA › Linear EquationsMediumQA · MCQ

If

a+b=2c

, then the value of

\frac{a}{a-c}+\frac{b}{b-c}

is

A1/2
B1
C2
D3

Pick an option to attempt

The Setup: You could do full algebra to solve this, but picking smart numbers is a massive brain play because if the expression holds true for variables, it holds true for any real numbers that satisfy the condition. Step 1: Choose values for

a

,

b

, and

c

that satisfy

a+b=2c

. Let's make sure the denominators don't become zero (

a \neq c

,

b \neq c

). Let

a=3

and

b=1

.

3+1=4

, so

2c=4 \implies c=2

. Step 2: Plug these values straight into the expression.

E=\frac{a}{a-c}+\frac{b}{b-c}

E=\frac{3}{3-2}+\frac{1}{1-2}

Step 3: Evaluate.

E=\frac{3}{1}+\frac{1}{-1}

E=3-1=2

*Bonus Algebra Flex:* If you rearrange

a+b=2c

, you get

b-c=c-a

. Substitute

b-c

with

-(a-c)

. The expression becomes

\frac{a}{a-c}-\frac{b}{a-c}=\frac{a-b}{a-c}

. Since

b=2c-a

, then

a-b=a-(2c-a)=2a-2c=2(a-c)

. So the fraction simplifies perfectly to

\frac{2(a-c)}{a-c}=2

. Final Answer: 2

Q30:jipmat 2025QA › HCF & LCMMediumQA · MCQ

Let x be the least number which when divided by 8, 12, 20, 28, 35 leaves a remainder of 5 in each case, then the sum of digits of x is:

A17
B14
C11
D15

Pick an option to attempt

The Setup: This is an LCM boss battle. The number we're looking for (

x

) is going to be exactly

5

steps larger than the Least Common Multiple (LCM) of all those divisors. Step 1: Find the prime factorization of each divisor.

8=2^3

12=2^2 \times 3

20=2^2 \times 5

28=2^2 \times 7

35=5 \times 7

Step 2: Calculate the LCM by grabbing the highest power of each prime number present in the squads. Highest power of

2

is

2^3=8

. Highest power of

3

is

3^1=3

. Highest power of

5

is

5^1=5

. Highest power of

7

is

7^1=7

.

\text{LCM}=8 \times 3 \times 5 \times 7

\text{LCM}=24 \times 35=840

Step 3: Form the number

x

. Since it leaves a remainder of

5

, we just add

5

to the LCM.

x=840+5=845

Step 4: Find the sum of the digits of

x

.

\text{Sum}=8+4+5=17

Final Answer: 17

Q31:jipmat 2025QA › TrigonometryHardQA · MCQ

If

\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = 3

, then value of

\sin^4\theta - \cos^4\theta

is

A $\frac{1}{5}$
B $\frac{2}{5}$
C $\frac{3}{5}$
D $\frac{4}{5}$

Pick an option to attempt

The Setup: This looks like a trig nightmare, but we can easily finesse it by isolating a single trig ratio (like

\tan\theta

) first, then substituting it back into a simplified version of the target expression. Step 1: Cross-multiply the initial equation to find the relationship between sine and cosine.

\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{3}{1}

\sin\theta + \cos\theta = 3(\sin\theta - \cos\theta)

\sin\theta + \cos\theta = 3\sin\theta - 3\cos\theta

Group the terms together:

4\cos\theta = 2\sin\theta

Divide by

2\cos\theta

to get tangent:

\frac{\sin\theta}{\cos\theta} = \frac{4}{2} \implies \tan\theta = 2

Step 2: Build a right-angled triangle. Since

\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{2}{1}

, we can find the hypotenuse using Pythagoras.

\text{Hypotenuse}^2 = 2^2 + 1^2 = 5 \implies \text{Hypotenuse} = \sqrt{5}

This means

\sin\theta = \frac{2}{\sqrt{5}}

and

\cos\theta = \frac{1}{\sqrt{5}}

. Step 3: Simplify the target expression using the difference of squares identity:

a^2 - b^2 = (a - b)(a + b)

.

\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta)

Since

\sin^2\theta + \cos^2\theta = 1

(the most basic trig fact), the expression simply becomes:

\sin^2\theta - \cos^2\theta

Step 4: Plug in our values from the triangle.

\sin^2\theta = (\frac{2}{\sqrt{5}})^2 = \frac{4}{5}

\cos^2\theta = (\frac{1}{\sqrt{5}})^2 = \frac{1}{5}

\text{Final Value} = \frac{4}{5} - \frac{1}{5} = \frac{3}{5}

Final Answer: 3/5

Q32:jipmat 2025QA › ProbabilityMediumQA · MCQ

Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children is:

A $\frac{1}{9}$
B $\frac{2}{9}$
C $\frac{1}{12}$
D $\frac{10}{21}$

Pick an option to attempt

The Setup: We are building a squad of 4 people from a total pool of 9 (

3\text{M} + 2\text{W} + 4\text{C}

). To find the probability, we need to divide the number of 'winning' combinations (exactly 2 kids and 2 non-kids) by the total possible combinations. Step 1: Calculate the total number of ways to pick any 4 people out of 9. This is a standard combination formula

{}^nC_r = \frac{n!}{r!(n-r)!}

.

{}^9C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126

So, there are

126

possible squads. Step 2: Calculate the favorable outcomes. We explicitly need *exactly*

2

children. That means the other

2

slots must be filled by adults (men or women). Total children =

4

. Ways to pick 2 kids =

{}^4C_2 = \frac{4 \times 3}{2 \times 1} = 6

. Total adults =

3 + 2 = 5

. Ways to pick 2 adults =

{}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10

. Multiply them together to get the total favorable combos:

6 \times 10 = 60\text{ ways}

Step 3: Calculate the final probability.

P = \frac{\text{Favorable}}{\text{Total}} = \frac{60}{126}

Divide top and bottom by

6

:

P = \frac{10}{21}

Final Answer: 10/21

Q33:jipmat 2025QA › Ratio, Proportion & VariationMediumQA · MCQ

P's income is Rs.140 more than Q's income and R's income is Rs.80 more than S's. If the ratio of P's and R's incomes is 2:3 and the ratio of Q's and S's incomes is 1:2, then the incomes of P, Q, R and S are, respectively:

A₹260, ₹120, ₹320 and ₹240
B₹300, ₹160, ₹600 and ₹520
C₹400, ₹260, ₹600 and ₹520
D₹320, ₹180, ₹480 and ₹360

Pick an option to attempt

The Setup: Setting up a massive system of linear equations here is total NPC behavior. Since they literally give us all four incomes in the options, the smartest strategy is to vibe check the choices against the constraints until only one survives. Step 1: Check Constraint 1:

P - Q = 140

and

R - S = 80

. (A)

260 - 120 = 140

(Pass).

320 - 240 = 80

(Pass). (B)

300 - 160 = 140

(Pass).

600 - 520 = 80

(Pass). (C)

400 - 260 = 140

(Pass).

600 - 520 = 80

(Pass). (D)

320 - 180 = 140

(Pass).

480 - 360 = 120

(Fail!

120 \neq 80

). Eliminate D. Step 2: Check Constraint 2: Ratio of

P:R

must be

2:3

. (A)

P=260, R=320 \implies \frac{260}{320} = \frac{13}{16}

(Fail! Not

2:3

). Eliminate A. (B)

P=300, R=600 \implies \frac{300}{600} = \frac{1}{2}

(Fail! Not

2:3

). Eliminate B. (C)

P=400, R=600 \implies \frac{400}{600} = \frac{4}{6} = \frac{2}{3}

(Pass!). Step 3: Just to be absolutely sure, check the final constraint for Option C: Ratio of

Q:S

must be

1:2

.

Q=260, S=520 \implies \frac{260}{520} = \frac{1}{2}

. It's a perfect match. Option C secures the W. Final Answer: Rs.400, Rs.260, Rs.600 and Rs.520

JIPMAT 2025 — QA

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