Top 30+ JIPMAT 2026 Quant Questions PDF with Solutions

The required number is of the form K*LCM(13,15) - 2 where K is an integer.
So, the required number is the maximum four digit number of the form 195K-2 which is 9943

Let the maximum marks be ‘x’
Then, Pass marks = 25% of x + 10 = 35% of x - 10
⇒ 10% of x = 20
⇒ x = 200
Therefore, Pass marks = 25% of 200+10 = 50+10 = 60

Given, the side length of the triangle= $$3\sqrt{3}$$ cm.

Its altitude = $$\frac{\sqrt{3}}{2}\times3\sqrt{3}=\frac{9}{2}$$ cm.

Its inradius= $$\frac{1}{3}\times\frac{9}{2}=\ \frac{3}{2}\ \ $$ cm.

So, the circle inside the triangle has a radius of length 3/2 cm.

The diameter of the circle is the diagonal of the square= 3 cm.

.'. Side length of the square becomes$$\frac{3}{\sqrt{2}}$$ cm.

.'. Area of the square= $$\left(\frac{3}{\sqrt{2}}\right)^2$$= 9/2 sq. cms.

Let the amount of money be P.
Since it triples itself, the simple interest = 3P - P = 2P.
Now, 2P = P(30)r
r = 2/30

If sum becomes 6 times, them simple interest = 6P - P = 5P

Now, 5P = P(t)(2/30)
t = 75 years.

Let the cost price of the laptop be 100.
So, initial selling price of the laptop = 112
New cost price of the laptop = 110
New selling price of the laptop = (95/100)*112 = Rs 106.4
So, loss percentage = (110 - 106.4)/110 * 100% = 3.27%

Let the cost price of 1 kilo gram oil be Rs. 1000.
As, the trader sells 900 grams oil for Rs. 990, the profit percentage is (990-900)/900 = 10%

We cannot infer the answer to this question as we do not know the cost price or selling price of both the bicycles and hence cannot determine the overall percentage of profit.

LCM of fractions = $$\frac{\text{ LCM of Numerator}}{\text{HCF of Denominator}}$$

= $$\ \frac{\ LCM\ of\ 4,\ 18,\ 25,\ 21}{HCF\ of\ 3,\ 5,\ 3,\ 41}$$

= 6300 (LCM of 4, 18, 25, 21 is 6300 and HCF of 3, 5, 3, 41 is 1)

Total weight of boys = 45 * 40 = 1800 kg
Total weight of girls = 50*32 = 1600 kg
Total weight of the class = 1600 + 1800 = 3400 kg
Average weight of the class = $$\frac{total weight}{total students}$$ = $$\frac{3400}{45+50}$$ = 35.79 = 35.8 kg approx.
Hence, option B is the right answer.

Convert to positive exponents (reciprocals) and find their relative order. The order for the negative exponents will be the reverse of the order for the positive exponents.

Let $$X = \{2^{-1/3}, 3^{-1/2}, 5^{-1/4}, 6^{-1/6}\}$$ and $$Y = \{2^{1/3}, 3^{1/2}, 5^{1/4}, 6^{1/6}\}$$.

We compare $$Y$$ by raising each to the LCM of the denominators (3, 2, 4, 6), which is 12.

1. (A') $$2^{1/3}$$: $$(2^{1/3})^{12} = 2^4 = 16$$
2. (B') $$3^{1/2}$$: $$(3^{1/2})^{12} = 3^6 = 729$$
3. (C') $$5^{1/4}$$: $$(5^{1/4})^{12} = 5^3 = 125$$
4. (D') $$6^{1/6}$$: $$(6^{1/6})^{12} = 6^2 = 36$$

Comparison of Positive Values (Y): $$16 < 36 < 125 < 729$$

Order: $$\text{A'} < \text{D'} < \text{C'} < \text{B'}$$

Comparison of Original Negative Values (X): Since $$A = 1/A'$$, the increasing order for the negative exponents is the reverse of the positive order.

Reverse Order: $$\text{B} < \text{C} < \text{D} < \text{A}$$

Final Arrangement: (B), (C), (D), (A)

Hence, the correct answer is option B.

For the number to be divisible by 11, the difference in the sums of the alternating digits should be divisible by 11.

In this case, the difference is (2+1+9+7) - (a+8+2) = 9 - a
This should be divisible by 11. As 'a' is a natural number, it means that a=9

The figure will be as shown below.

Let the side of the square be ‘x’
It is known that AP = 4.5 cm
So AQ = (x-4.5) cm
The area of the triangle AQR will be
$$\frac{1}{2}*x*(x-4.5)$$
This can be equated to 50 $$cm^2$$
So we get,
$$x(x-4.5)=100$$
$$=>2x^2-9x-200=0$$
$$=>(2x-25)(x+8)=0$$
As $$x$$ cannot be negative it has to be 12.5 cm
Area of the square will be = 12.5*12.5 = 156.25 $$cm^2$$
Hence Option C.

Using Statement 1,
P+Q+R+S = 52*4 = 208
Q = 103 years
P+R+S = 208-103 = 105
The age of any of them can be 104, 1, 0
But using statement 2 we know that all the ages are natural numbers and are distinct. So the solution of 103, 104, 1, 0 is not valid. Hence Q is the oldest among them.

(A): $$2^{3^2}$$ = $$2^9=512$$

(B): $$\left(2^3\right)^2=2^6=64$$

(C): $$3^{3^2}=3^9=27^3=19683$$

(D): $$\left(3^2\right)^3=3^6=729$$

So, the correct order is: $$(C)>(D)>(A)>(B)$$

$$2^{m-1} + 2^{m+1} = 640$$
=> $$2^m / 2 + 2^m * 2 = 640$$
=> $$5 * 2^m = 1280$$
=> $$2^{m} = 256$$
=> m = 8

We know that A and B's speeds are in the ratio of 1:2. This means that B covers twice the distance covered by A. So when they meet for the first time A would have covered 10*$$\frac{1}{3}$$ and B would have covered 10*$$\frac{2}{3}$$. Now, B will reverse his direction while A will continue to travel in the same direction.

Since B is faster than A, he will reach his start point before A, so the distance he would have traveled is 10*$$\frac{2}{3}$$ again. In the same time, A would travel 10*$$\frac{1}{3}$$. So, the total distance covered by A is 10*$$\frac{1}{3}$$+10*$$\frac{1}{3}$$=10*$$\frac{2}{3}$$

Remaining distance=10*$$\frac{1}{3}$$

Of this distance A will travel 1/3rd while B will travel 2/3rd of it before they meet for the second time.

So, the total distance traveled by B is 10*$$\frac{2}{3}$$+10*$$\frac{2}{3}$$+10*$$\frac{1}{3}$$*$$\frac{2}{3}$$

=$$\frac{140}{9}$$

In this case, the time taken by A to reach other end is 10km/5kmph = 2hrs. B’s speed= 2*5= 10kmph. Hence, in two hours B would have run = 2*10kmph = 20km

Sum of the roots = $$\alpha\ +\beta\ +\alpha\ -\beta\ =2\alpha\ $$

Also, from the equation, sum of roots = -b/a = $$\beta\ $$

Hence, $$2\alpha=\beta\ $$

Product of roots = $$\left(\alpha\ +\beta\ \right)\left(\alpha\ -\beta\ \right)=\alpha\ ^2-\beta\ ^2$$

Also, from the equation, product of the roots = q

Hence q = $$\alpha\ ^2-\beta\ ^2$$

q = $$\alpha\ ^2-\left(2\alpha\ \right)^2=\alpha\ ^2-4\alpha\ ^2=-3\alpha\ ^2$$

The figure can be represented as :

The triangle formed is an equilateral triangle with the side of length 14cm

The height of an equilateral triangle= $$\frac{\sqrt{\ 3}}{2}\cdot a$$

=$$\frac{\sqrt{\ 3}}{2}\cdot 14$$

= $$7\sqrt{\ 3}$$

Height = 10+ $$7\sqrt{\ 3}$$+ 14 cm

= 24+$$7\sqrt{\ 3}$$

Option A is the correct answer.

Using Statement 1,we come to know there are 78 people in total in a queue.
Using statement 2,we come to know Rahul is standing at the 27th position from back and the no. of people between Raj and Rahul is 9. So Rahul is standing at 27th position from the back. Raj can be standing at either 37th or 17th position.We can't infer the exact position of Raj.

Statement 1 implies x = 14k + 10 which implies x = 7(2k+1) + 3
Statement 2 implies x = 28r + 24 which implies x = 7(4r+3) + 3
Hence, the question can be answered using either of the two statements alone.

2w + 9b = 7m

1w = 1.25m

Hence, 2.5m + 9b = 7m

2b = 1m

1b = 0.5m

In one day 1 man does $$\frac{1}{7}$$ work.

Hence, in one day, 1 woman does $$\frac{1.25}{7}$$ work

And, in one day, 1 child does $$\frac{1}{14}$$ work.

Considering the efficiency of each, in one day, total amount of work done = $$\frac{2}{7}\times\ 1.25\ +\ \frac{1.25}{7}\times\ 0.5+\frac{0.75}{14}=\frac{2.5}{7}+\frac{0.625}{7}+\frac{0.375}{7}=\frac{3.5}{7}=\frac{1}{2}$$

Therefore, total time needed is 2 days.

We know $$n(A\cup B) = n(A) + n(B) -n (A\cap B)$$

thus, $$n(N\cup C) = n(N) + n(C) - n(N\cap C)$$

$$n(N\cap C) = n(N) + n(C) - n(N\cup C)$$

thus the percent of people in both will decrease when $$n(N\cup C)$$ is maximum which is 100%

=> min $$n(N\cap C)$$ = 55% + 60% - 100% = 15%

Assuming the profit percentage is p.

Then cost price of the car = $$\ \dfrac{\ 18}{1+\dfrac{\ p}{100}}$$

Similarly, the cost price of bike = $$\ \dfrac{\ 2}{1-\dfrac{\ p}{100}}$$

Also, $$\ \dfrac{\ 2}{1-\dfrac{\ p}{100}}$$ = $$\ \ \dfrac{\ 1}{6}\times\ \dfrac{\ 18}{1+\dfrac{\ p}{100}}$$

=> $$\ \dfrac{\ 3}{1+\dfrac{\ p}{100}}=\dfrac{\ 2}{1-\dfrac{\ p}{100}}$$

=>  p =20

We know that $$Time=\dfrac{Dist}{Speed}$$

Let Atul's speed be A, Bala's speed be B, Chirag's speed be C and Dheeraj's speed be D.

Atul beats Bala by 10m. This means that by the time Atul covers 100m, Bala only covers 90. Therefore -

-> $$\dfrac{100}{A}=\dfrac{90}{B}\rightarrow1$$

Atul beats Chirag by 19m. This means that by the time Atul covers 100m, Chirag only covers 81m. Therefore -

-> $$\dfrac{100}{A}=\dfrac{81}{C}\rightarrow2$$

Equating eq. 1 and eq. 2 -

$$\dfrac{90}{B}=\dfrac{81}{C}$$

$$\dfrac{90}{B}\times\dfrac{100}{90}=\dfrac{81}{C}\times\dfrac{100}{90}$$

$$\dfrac{100}{B}=\dfrac{90}{C}\rightarrow3$$

Bala beats Dheeraj by 23.5m. This means that by the time Bala covers 100m, Dheeraj only covers 76.5m. Therefore -

-> $$\dfrac{100}{B}=\dfrac{76.5}{D}\rightarrow4$$

Equating eq. 3 and eq. 4 -

$$\dfrac{90}{C}=\dfrac{76.5}{D}$$

$$\dfrac{90}{C}\times\dfrac{100}{90}=\dfrac{76.5}{D}\times\dfrac{100}{90}$$

$$\dfrac{100}{C}=\dfrac{85}{D}$$

This means that by the time Chirag covers 100m, Dheeraj covers only 85m. Therefore, Chirag beat Dheeraj by 15m.

$$\alpha\ $$ and $$\beta\ $$ are root of the equation $$4x^2+7x-13=0$$

Thus $$\alpha\ +\beta\ =\ -\frac{7}{4}$$ and $$\alpha\ \beta\ \ =\ -\frac{13}{4}$$

$$\frac{1}{\alpha\ +\beta\ }=\ -\frac{4}{7}$$ and $$\frac{1}{\alpha\ \beta\ }=\ -\frac{4}{13}$$

Quadratic eqn with roots $$-\frac{4}{7}\ \&-\frac{4}{13}$$ can be written as $$a\left(x+\frac{4}{7}\right)\left(x+\frac{4}{13}\right)$$

On expanding the equation becomes $$a\left(x^2+\frac{80x}{91}+\frac{16}{91}\right)$$

taking a=91, we get the equation to be $$91x^2+80x+16$$

2*3.14*r = 4*l = 3*s = k (assume) So, r = k/6.28; l = k/4, s = k/3. Area of circle = $$k^{2}$$ / 12.56. Area of square = $$k^{2}$$/16. Area of triangle = $$k^{2}$$/20.79

After factorising we can see that the equation has become (x-9)(x-9)(x-8) =0.

Hence, option C is the correct answer.

Using Pythagoras theorem, the other side can be deduced to be 14 metres, if 50 is the diagonal and 48 is one of the sides. So, the area of the rectangle is 14 x 48 = 672

If the cost of cutting the grass is 24/ sq mtr, the total cost = 24 x 672 = 16,128 Rs.

It is given that $$a+b=c$$
=>$$a+b-c=0$$
Squaring on both sides we get,
$$(a+b-c)^2=0$$
=>$$ a^2+b^2+c^2+2ab-2bc-2ca=0$$
Hence Option B.

It is given that $$sin^2\theta - cos^2\theta = \frac{3}{4}$$ … (1)
We know that $$sin^2\theta + cos^2\theta = 1$$ … (2)

On multiplying both the equations , we get,
$$(sin^2\theta - cos^2\theta)*(sin^2\theta + cos^2\theta) = \frac{3}{4} * 1$$
$$sin^4\theta - cos^4\theta = \frac{3}{4}$$
Therefore, option D is the right answer.

There will be two freinds who get their own bag.

Selecting 2 out of B, C, D, E, F can be done in $$^5C_2$$ ways.

The remaining four should not get their own bags. Using derangements formula for n=4.

$$n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\right)$$

$$4!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\right)$$ = 9

Total number of ways will be 9*$$^5C_2$$ = 90 ways.

$$a^2+b^2=c^2$$

By $$m^{th}$$ power theorem,

$$\frac{\left(a^2+b^2\right)}{2}\ge\ \ \left(\frac{a+b}{2}\right)^2$$

$$=>\ \ \frac{\left(a+b\right)^2}{4}\le\ \frac{c^2}{2}$$

$$=>\ \ \left(a+b\right)^2\le\ 2c^2$$

$$=>\ \ \frac{\left(a+b\right)}{c}^{ }\le\ \sqrt{\ 2}$$