Top 200 IPMAT 2026 Quant Questions PDF with Video Solutions

For the last 10 overs, if the required rate is max of 12, we can get the minimum run rate after 40 overs.

Let the run rate for the first 40 overs be x.

Then, $$40x+\left(10\times12\right)=\ 340\ $$

=> $$40x+120=\ 340\ $$

=>$$40x=\ 220\ $$

.'. x= 5.5 runs an over

Let the total time taken by all three of them together be X => One day work = $$\frac{1}{X\ }$$

So, time taken by Ram alone is 2X => One day work = $$\frac{1}{2X}$$  -------- (1)

time taken by Rahim alone is X+3.=> One day work = $$\frac{1}{X\ +\ 3}$$ --------(2)

time taken by Robert alone is X+3 => One day work = $$\frac{1}{X\ +\ 3}$$ ------ (3)

From (1) , (2), (3), the total work done in one day is $$\frac{1}{2X\ }\ +\ \frac{1}{X+3}\ +\ \frac{1}{X+3}$$ which is equal to one days of work when they work together.

So, $$\frac{2}{X\ +3}+\frac{1}{2X}=\frac{1}{X}$$. => X = 1

Let the volume of the tank be lcm(4,3) = 12 units

Rate of filling by pipe = 12/3 = 4 units/hours

Rate of emptying by drain = 12/4 = -3 units/hours

Effective rate of filling when 5 pipes and 3 drains are connected = 5(4)+4(-3) = 20-12 = 8 units/hours

Total time taken = 12/8 = $$1\ \frac{1}{2}$$ hours

$$\log{5^{100}}$$ = 100 * ( 1 - $$\log{2}$$ ) = 69.87. So, $$5^{100}$$ has 69+1=70 digits.

Given that $$(456)_n+\left(1277\right)_n=\left(1755\right)n$$

$$\therefore\ 4n^2+5n+6+n^3+2n^2+7n+7=n^3+7n^2+5n+5$$

$$\therefore\ n^2-7n-8=0$$

$$\therefore\left(n-8\right)\left(n+1\right)=0$$

$$\therefore n=8\ or\ -1$$

Since, base can not be negative, we get the base as 8.

We know that p+q = -b/a and pq=c/a.
Therefore: $$-p^2q^2= -c^2/a^2$$ and $$p^2-q^2=(p+q)(p-q)=(-b/a)*1$$.
Hence the eqn is :$$ x^2-(-b/a)x-c^2/a^2=0$$ => $$a^2x^2 + abx - c^2=0$$

Let the cost of each apple, orange and mango be ‘a’, ‘b’ and ‘c’ respectively.
2a+5b+7c = 200 ------ (1)
3a+5c = 6b+100 ------ (2)
Multiplying the first equation by 3 and the second equation by 4 and subtracting the 2nd equation from 1st gives us,
39b+c = 6a+200
Thus, the cost of 39 oranges and a mango exceeds the cost of 6 apples by Rs. 200.

A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.

(X+X+8) - (Y+7) = multiple of 11

2X-Y +1 = 0, 11

So, 2X + 1=Y or 2X - 10 = Y are the two possible equations.

2X+1 = Y $$\Rightarrow$$ The possible pairs of (x, y) are (1, 3),(2, 5), (3, 7), (4, 9)

2X - 10 = Y $$\Rightarrow$$ The possible pairs of (x, y) are (5, 0), (6, 2), (7, 4), (8, 6), (9, 8)

So, there are a total of 9 solutions.

Let the distance between the two stations be D.
Ratio of speeds = 42 : 32 = 21 : 16
So, distance covered by the faster train = (21/37) * D
Distance covered by the slower train = (16/37) * D
Difference between the distances travelled = (5/37) * D
This is equal to 120 km.
So, (5/37)*D = 120 * 37/5 => D = 24 * 37 km = 888 km
So, distance between the two stations = 888 km

Let the final speed of b be p and x be the total distance.

After meeting, the distance covered by A is $$\frac{4}{19}$$ of the total distance and distance covered by B is $$\frac{15}{19}$$ of the total distance. Assuming both meet at R, PR= $$\frac{4x}{19}$$, QR = $$\frac{15x}{19}$$

Speed after meet is in the ratio 12 : 15.

Now since both reach the same time, the time taken must be the same.

QR/12 = PR/p

=> $$\frac{15x/19}{12}$$ = $$\frac{4x/19}{p}$$

=> $$p = 3.2$$

=> Fraction of original speed = $$\frac{3.2}{15}$$ = $$\frac{16}{75}$$

$$\log_{\left(0.0625\right)}8+\log_3243\ -\ \log_{343}49$$

$$\log_{\left(2^{-4}\right)}2^3+\log_33^5\ -\ \log_{7^3}7^2$$

$$-\frac{3}{4}+5\ -\ \frac{2}{3}$$

$$-\frac{3}{4}+5\ -\ \frac{2}{3}=\frac{43}{12}$$ = $$3\ \frac{7}{12}$$

Let the volume of the tank be 6L and let the flow rates through pipes A, B, C be a, b, c (L/hr).

if A fills the tank in 2 hrs, flow rate = (6/2) = 3 L/hr

if B empties it in 6 hrs, flow rate = (6/6) = 1 L/hr

if C empties it in 3 hrs, flow rate = (6/3) = 2 L/hr

So A starts filling and B starts emptying simultaneously. (Until water level reaches half the height, C will not come into use)

Net filling rate = a-b = 3-1 = 2L/hr

Half of the tank(3 L) will be filled in 3/2 = 1.5 hrs.

After that point, C will also start emptying and the net filling rate = a-b-c = 3-1-2 = 0 L/hr

$$\therefore\ $$ The tank will never get filled to the brim.

Let x, y and z be the number of coins of each denomination => x+4y+16z=100.
Max value of z=6. For z=6, max value of y=1.
Hence there are 2 cases y=0,1 for z=6. For z=5, there are 3 cases (y=5, x=0), (y=4, x=4), (y=3, x=8).
For z=4, there are 3 cases (y=9, x=0), (y=8, x=4) and (y=7, x=8).
Hence total of 8 cases

Let the capacity of tank X be lcm(90,120) = 360 units

Rate of filling done by pipe of type A = $$\frac{360}{90}=4$$ units/minute

Rate of emptying done be pipe of type B = $$\frac{-360}{120}=-3$$ units/minute

Total rate of filling when 3 pipes of type A and 2 pipes of type B are connected = 3(4)+2(-3)= 6units/minutes

The total amount of filling to be done  = (1- 30%)*3*capacity of X = 0.7*3*360 = 756 units

Time required  = $$\frac{756}{6}=126$$ minutes

The curve meets the Y axis when $$X=0$$.

Substituting X = 0 in equation $$X^3+Y^3-5X^2+4X-4Y=0$$, we get

$$Y^3-4Y=0$$

$$Y\left(Y^2-4\right)$$ = 0

$$Y\left(Y-2\right)\left(Y+2\right)\ =\ 0$$

Y = -2, 0, 2

The curves meets the X-axis when Y = 0

Substituting Y = 0 in equation $$X^3+Y^3-5X^2+4X-4Y=0$$, we get

$$X^3-5X^2+4X=0$$

$$X\left(X^2-5X+4\right)\ =\ 0$$

X(X-1)(X-4) = 0

X = 0, 1, 4

Curve intersects at 5 distinct points

Answer is option B.

The simplest way to solve this question is to assume how many ways there are so that the children get equal to or fewer than 7 toys.

We know that the maximum number of toys the three children can get is 7.

Let a, b and c be the three integers that represent the number of toys less than or equal to 7.

Hence, the three children would receive the toys in the following way:

$$7-a+7-b+7-c=16$$

a, b and c can be greater than or equal to 0 but less than 8.

Hence, we can say,

$$21-a-b-c=16$$

Or, $$a+b+c=5$$

Now, we know that n things can be distributed among r different persons in $$^{n+r-1}C_{r-1}$$

Hence, the number of ways of distributing 5 among 3 is $$^7C_2=21$$

Hence, there are 21 ways.

Let us put x = 6

We get $$f\left(18\right)+2f\left(8\right)\ =\ 6\cdot6\ =\ 36$$ $$\rightarrow 1$$

Now let us put x = $$\frac{8}{3}$$

We get  $$f\left(3\cdot\frac{8}{3}\right)+2f\left(\frac{48}{\frac{8}{3}}\right)=6\cdot\frac{8}{3}=16$$

f(8)+2f(18) = 16  $$\rightarrow 2$$

Equations 1 and 2 can be seen as linear equations in two variables.

Upon solving them, we get

f(18) = $$-\frac{4}{3}$$, f(8) = $$\frac{56}{3}$$

2f(18)+5f(8) = $$-\frac{8}{3}+\frac{280}{3}\ =\ \frac{272}{3}=90.67$$

A large sphere of radius R cm is melted to form N smaller spheres of radius r cm.

=> $$\dfrac{4}{3}\pi R^3=N\times\ \dfrac{4}{3}\pi r^3$$

$$=>R=N^{\frac{1}{3}}\times\ r$$

Now, the large sphere is painted blue at Rs. 36/cm².

=> Cost of painting the large sphere = $$36\times\ 4\pi R^2$$

And each smaller sphere is painted red at Rs. 1/cm²

=> Cost of painting all smaller spheres = $$N\times\ 1\times\ 4\pi r^2$$

The total painting cost remains the same for both cases.

=> $$36\times\ 4\pi R^2=N\times\ 1\times\ 4\pi r^2$$

Eliminating $$4\pi$$ both sides and then taking root-

=> $$6\times\ R=N^{\frac{1}{2}}\times r$$

Substituting the value of R from 1

$$6\times N^{\frac{1}{3}}\times\ r=N^{\frac{1}{2}}\times r$$

=> $$6=N^{\frac{1}{6}}$$

=> $$N=6^6=46656$$

Let percentage of growth in first year be $$5x\%$$

So percentage of loss in second year will be $$x\%$$

So, $$5000\left(1+\dfrac{5x}{100}\right)\left(1-\dfrac{x}{100}\right)=6750$$

or, $$\left(1+\dfrac{5x}{100}\right)\left(1-\dfrac{x}{100}\right)=\dfrac{6750}{5000}=1.35$$ ----->(1)

Let $$\dfrac{x}{100}=k$$

So, from equation (1),

$$\left(1+5k\right)\left(1-k\right)=1.35$$

or, $$4k-5k^2=0.35$$

or, $$100k^2-80k+7=0$$

or, $$\left(10k-1\right)\left(10k-7\right)=0$$

or, $$k=\dfrac{1}{10}$$ or $$\dfrac{7}{10}$$

or, $$x=10$$ or $$70$$

But, $$x=70$$ is rejected because in that case, after the first year the investment will become $$5000\left(1+5\times\ \frac{70}{100}\right)=22500$$ which is greater than $$10000$$

So, $$x=10$$

so, percentage loss in second year is $$10\ \%\ \ $$

The equation given is, $$x^2-\left(p+2\right)x+\left(2p+9\right)=0$$

The conditions we need to check for two negative roots will be, we already notice the coefficient of the $$x^2$$ term is positive, so taking that into account, we have:
1. $$D\ge0$$
2. The sum of the roots is negative
3. Product of roots is positive

1.  $$D\ge0$$:

$$\left(p+2\right)^2-4\left(2p+9\right)\ge0$$

$$\left(p^2+4+4p\right)-4\left(2p+9\right)\ge0$$

$$p^2-4p-32\ge0$$

$$\left(p+4\right)\left(p-8\right)\ge0$$

So we have either, $$p\le-4\ or\ p\ge8$$

2. $$p+2<0$$

We get: $$p<-2$$

So, the earlier case of $$p>8$$ is eliminated.

3. $$2p+9>0$$

We get: $$p>-4.5$$

The final condition is, $$p\in\ \left(-4.5,\ -4\right]$$

The only value that satisfies this condition is p=-4.

The man can finish the work in 14 days If he works 8 hours a day. So he requires a total of $$14\times\ 8=112$$ hours to complete the work.

Let the total work be 112 units, in that case his efficiency is 1 unit/hr.

The man decreases his work hours by 20%, which means he works $$0.8\times\ 8=6.4$$ hours daily.

He needs to finish his work in 14 days only, so he has a total of $$14\times6.4=89.6\ $$ hours.

The total work however will remain unchanged at 112 units.

So his efficiency y in the second case should be $$\dfrac{112}{89.6\ }=1.25\ $$ units per day.

So percentage increase in efficiency is $$\dfrac{1.25-1}{1}\times\ 100=25\%$$

Let the efficiency of inlet pipes be a, and the outlet pipes be b.

10 inlet pipes fill the tank in 4 hours, thus the capacity of the tank will be = $$10\times4\times a=40a$$ litres

n outlet pipe empties half the tank in 3 hours; thus, the capacity of half the tank = $$n\times3\times b=3nb$$ litres. Thus, the capacity of the full tank is = $$6nb$$ litres.

=> $$40a=6nb$$ => $$a=\dfrac{3nb}{20}\rightarrow1$$

Now, (n+1) inlet pipes and $$\dfrac{n}{2}$$ outlet pipes are opened, and the tank is filled in 24 hours.

=> $$\left(\left(n+1\right)a-\left(\dfrac{n}{2}\right)b\right)\times24=6nb$$     (substituting the value of a from eq. 1)

=> $$\left(\left(n+1\right)\left(\dfrac{3nb}{20}\right)-\left(\dfrac{n}{2}\right)b\right)=\dfrac{nb}{4}$$

Cancelling the terms n and b from both LHS and RHS -

=> $$\left(n+1\right)\dfrac{3}{20}-\dfrac{1}{2}=\dfrac{1}{4}$$

=> $$\left(n+1\right)\dfrac{3}{20}=\dfrac{3}{4}$$

=> $$n+1=5$$ => $$n=4$$

Substituting the value of n in eq. 1 -

$$a=\dfrac{3\times4\times b}{20}$$

=> $$\dfrac{a}{b}=\dfrac{3}{5}$$

Alternate Explanation,

Let the volume of the tank be $$1$$ litre, and it takes 10 * 4 = 40 hours for the inlet pipe to fill the tank. Thus -

Efficiency of one inlet pipe = $$\dfrac{1}{40}$$ litre/hr

And n outlet pipes empty half the tank in 3 hours. Thus, 0.5 litres of the tank gets emptied by outlet pipes in n * 3 = 3n hours. Thus -

Efficiency of one outlet pipe = $$\dfrac{\dfrac{1}{2}}{3n}=\dfrac{1}{6n}$$ litre/hr

So, $$\left(\dfrac{\left(n+1\right)1}{40}-\dfrac{n}{2}\times\ \dfrac{1}{6n}\right)=\dfrac{1}{24}$$

or,$$\dfrac{n+1}{40}-\dfrac{1}{12}=\dfrac{1}{24}$$

or, $$\dfrac{n+1}{40}=\dfrac{3}{24}=\dfrac{1}{8}$$

So, $$n=4$$

So, efficency of inlet pipe to outlet pipe = $$\dfrac{\dfrac{1}{40}}{\dfrac{1}{6n}}=\dfrac{6n}{40}=\dfrac{6\times\ 4}{40}=\dfrac{3}{5}$$

Given, Density = Mass/Volume

So, Mass= Density*Volume

So, Mass of iron sphere: mass of copper cube = (Density of iron * Volume of iron sphere):(Density of copper * Volume of copper cube)

or, $$X\ =\ \dfrac{7000\times\ \dfrac{4}{3}\pi\ \times\ 21^3}{8000\times\ 10.5^3}$$

or, $$X\ =\ \dfrac{7000\times\ \dfrac{4}{3}\times\ \dfrac{22}{7}\times\ 21\times\ 21\ \times\ 21}{8000\times\ 10.5\times\ 10.5\times\ 10.5}$$

or, $$X\ =\ \dfrac{88}{3}$$

or, $$3X\ =\ 88$$

$$\log70=\log2+\log5+\log7=a\rightarrow1$$

$$\log28=2\log2+\log7=b\rightarrow2$$

$$\log245=\log5+2\log7=c\rightarrow3$$

Subtracting 1 from 3 -

$$\log7-\log2=c-a\rightarrow4$$

Subtracting 4 from 2 -

$$3\log2=b-c+a$$

$$\log2=\dfrac{(b-c+a)}{3}$$

Multiply eq. 1 by 2, and then subtracting eq. 3 from it -

$$2(\log2+\log5+\log7)-(\log5+2\log7)=2a-c$$

$$2\log2+\log5=2a-c$$

Putting the value of log2 -

$$\log5=\left(2a-c\ \right)-2\left(\dfrac{b-c+a}{3}\right)$$

$$\log5=\left(\dfrac{6a-3c-2b+2c-2a}{3}\right)$$

$$\log5=\dfrac{4a-2b-c}{3}$$

$$\dfrac{a+b+c}{3}=e$$ => $$a+b+c=3e$$ => $$b+c=3e-a\rightarrow1$$

Also, $$\dfrac{d+e}{2}=b$$ => $$d+e=2b\rightarrow2$$

And, $$\dfrac{e+f}{2}=c$$ => $$e+f=2c\rightarrow3$$

Adding eq. 2 and eq. 3 -

=> $$d+2e+f=2(b+c)$$        (substituting the value of (b+c) from eq. 1)

=> $$d+2e+f=6e-2a$$

=> $$d+f=4e-2a\rightarrow4$$

Now, the average of all 6 numbers is 11.

=> $$\dfrac{a+b+c+d+e+f}{6}=11$$

=> $$(a+e)+(b+c)+(d+f)=66$$     (substituting the value of (b+c) and (d+f) from eq. 1 and eq. 4)

=> $$(a+e)+(3e-a)+(4e-2a)=66$$

=> $$8e-2a=66$$        (And we are given the value of a = 3)

=> $$8e=72$$ => $$e=9$$

Let the cost price of Peanut Butter be A and the cost price of Jam be B.
So, A+B = 100 (first equation)
He sells Peanut Butter at a loss of 20%. So, the selling price is 0.8A
He sells Jam at a profit of 25%. So, the selling price is 1.25B
He makes an overall profit of Rs. 7. So, the overall selling price is Rs. 107

So, 0.8A + 1.25B = 107 or 3.2A + 5B = 428
5A+5B = 500 (5 times the first equation)
Therefore, 1.8A = 72
Or, A = 40

Hence, the cost price of a packet of Peanut Butter is Rs. 40

Let us first calculate the number of man hours needed to complte the work.
Total man-hours = 50 x 10 x 45.
Hence, 50 x 10 x 45 = 40 x t x 30, where t is the number of hours per day
Solving, we get t = 150/8 = 18.75 hours.
Hence, additional hours = 18.75 - 10 = 8.75 = 8 hrs 45 mins

Let us assume that the volume of tank is 300 cc(LCM of 20,25,30 )
A fills the tank at the rate of 15 cc/min
B fills the tank at the rate of 10cc/min
C empties the tank at the rate of 12 cc/ min.
Time taken by A and B when C is open =$$\frac{300}{15+10-12}$$ = $$\frac{300}{13}$$
Time taken by A and B when C is closed = $$\frac{300}{15+10}$$ = 12 min
required ratio = $$\frac{\frac{300}{13}}{12}$$ = 25:13

Let the entire work consist of 100 units.

Hence, in 1 day, A does 100/20 = 5 units, B does 100/25 = 4 units and C does 100/40 = 2.5 units.

On the first day, the total amount of work done = 5+4 = 9 units

On the second day, the total amount of work done = 4+2.5 = 6.5 units

On the third day, the total amount of work done = 5+2.5 = 7.5 units

On the fourth day, the total amount of work done = 5+4+2.5 = 11.5 units

Hence, in 4 days, total units of work done = 34.5 units.

In the next 4 days, in a similar way, total units of work done = 34.5 units.

If we again add the work for 4 days, the number of units crosses 100, hence we will take one day at a time.

On the 9th day amount of work done = 5+4 = 9 units.

Total = 78 units.

On the 10th day amount of work done = 4+2.5 = 6.5 units.

Total = 84.5 units.

On the 11th day amount of work done = 5+2.5 = 7.5 units.

Total = 92 units.

Remaining units = 8

On the 12th day, all 3 work together and contribute to 11.5 units.

Hence, part of 12th day needed = 8/11.5 = 16/23

Hence, total number of days = $$11\frac{16}{23}$$

Let assume the marked price to be Rs.300M
Cost price of watch shop owner = 2/3rd of 300M = 200M
Minimum profit = 30% of 200M= 60M

Minimum selling price =260M

Thus the maximum discount he can provide = 300M-260M = 40M
=> Discount % = $$\frac{40M}{300M}\times 100$$% = 13.33%

C is the correct answer.

Assume a man does x unit of work in a day and woman does y unit of work in a day and the total work is 56 units. (LCM of 7,8)

(4x+6y)8 = 56

4x+6y = 7

=> 2x+3y = 3.5

(5x+5y)7 = 56

5x+5y = 8

x+y= 1.6

Hence, x =1.3 and y = 0.3

Now, x+5y = 1.3+0.3*5 = 2.8

Hence, the number of days = $$\ \frac{\ 56}{2.8}$$ =20

Let the marked price of the pen be x. Hence, he must be selling each pen for .75x.
Thu, the amount earned by selling 100 pens will be .75x*100 = 75x
We have been given that profit is equal to the selling price of 20 pens. Hence, profit = 20*.75 = 15x.
Thus, the cost price of 100 pens must be 60x
If he had not offered any discount then, he would have earned 100x. Hence, the profit would have been $$\frac{40x}{60x}*100$$ = 66.66 %

If p = $$y^2+\frac{1}{y^2}$$, then $$p^2=y^4+\frac{1}{y^4}+2$$

K = $$\frac{p^2-1}{p+1}=p-1$$
So, $$K=y^2+\frac{1}{y^2} -1=(y-\frac{1}{y})^2+1$$
So, the least possible value of K is 1 for any real value of y.
Hence, K doesn't take the value of 0.5

Let f (x) = min (7 + min(y, - 3), y - 6)
Case I: y < - 3
Therefore, f (x) = min (7 + y, y - 6)
Since, y < - 3, therefore f (x) = y - 6 < - 9
Therefore, f (x) < - 9
Case II: y > - 3
Therefore, f (x) = min (4, y - 6)
Sub Case I: y - 6 < 4
Or, y < 10
For - 3 < y < 10, f (x) = y - 6
Therefore, - 9 < f (x) < 4
Sub Case II: y - 6 > 4
Therefore, f (x) = 4
Therefore, maximum possible value of f (x) = 4.

We have $$f(x) = ax^2+bx+c$$

and, $$f(9)=9f(1)$$, therefore,

$$9^2a+9b+c =9(a+b+c)$$

$$81a+c=9a+9c$$

Which gives $$c=9a$$

Product of roots $$=\ \dfrac{c}{a}$$

Given that $$f(x)$$ has equal roots, say $$\alpha$$ and $$\alpha$$

Product of roots $$=\ \alpha^2 = \ \dfrac{c}{a} = 9$$

$$\alpha\ = \pm3$$

The only possible root among the options provided is $$3$$. Option C is the correct answer.

Let the price of bat be ‘a’, ball be ‘b’ and glove be ‘c’. So we have
2a + 3b + 7c = 950
7a + 6b + 2c = 1750
Adding both the equations, we get
9a + 9b + 9c = 2700
a + b + c = 300
Hence, the price of 1 bat, 1 ball and 1 glove is 300.

The discriminant of the given quadratic equation is equal to $$4*[(q^2-pr)^2-(r^2-pq)*(p^2-qr)]$$
Which equals $$4q*[p^3+q^3+r^3-3pqr]$$
As $$p+q+r=0,p^3+q^3+r^3-3pqr=0$$
Hence, the discriminant of the given quadratic equation equals 0.
So the roots are real and equal

Let the number of rows be 'a' and let the number of columns be 'b'
So, the number of boys is '8a' and the number of girls is '10b'
The total number of spots in the rectangle is a*b and there are 16 empty spots in the rectangle. So, 8a+10b+16 = ab
Or, ab - 8a - 10b +80 = 96
Or, (a-10)*(b-8)=96

Note that a-10 and b-8 are both integers. So, the number of solutions for this equation is equal to the number of divisors of 96 which equals 12. We can equate every divisor of 96 to 'a-10' and find the respective values of 'a' and 'b'.

For each case, the number of boys will be 8a and the number of girls will be 10b

Let (h,k) be the point of intersection for the curves.
$$k = h^4+2h^3-40 =2h^3+2h^2+23$$
$$h^4-2h^2-63 = 0$$
$$(h^2-1)^2 = 64$$
$$h^2-1 =+8$$ or $$-8$$
$$h^2 =9$$ (square of a number can’t be -7)
$$h = +3$$ or $$-3$$
Therefore, it has two points of intersection in the given range.

$$X *(100-a) = 7200$$ ...(1)
$$(X-20)*(100+a) = 6600$$ ...(2)

By adding equation (1) and (2)

$$200X - 2000 -20a = 13800$$
$$10X - a = 790$$ ... (3)

Substituting value from equation (3) to equation (1)

$$X *(100-(10X-790)) = 7200$$

$$X *(890 - 10X) = 7200$$

$$X^2 - 89X + 720 = 0$$

$$(X - 9)(X - 80) = 0$$

X can't be 9 as he sold X - 20 shirts. So X = 80 and a = 10.

Let the amount with B = x

=>Amount with A, C and D will be x + 80, x-50 and x+120+x-50 = 2x + 70 respectively.

Given x + x +80+x-50+2x+70 = 500

=> x = 80

Amount with A= 160

Amount with B= 80

Amount with C= 30

Amount with D=230

Since Amounts with C and D are not multiples of 20, they should have atleast one Rs 10 note

Least number of ten rupee notes = 2

Let the number of 5 rupee notes be x, 10 rupee notes be y and 20 rupee notes be z.
We know that,
x + y + z = 100 -----------------------(1)
5x + 10y + 20z = 1500--------------(2)
5x + 20y + 10z = 1100-------------(3)

Multiplying equation (1) by 5 and substituting it in equations (2) and (3), we get,
5y + 15z = 1000-------(4)
15y + 5z = 600 ----(5)

(4)*3 - (5) => 40z = 2400
z = 60
Therefore, the number of 20 rupee notes is 60.
Therefore, option D is the right answer.

Time at which Ajay and Vijay meet is 40*(11-5)/(80-40)+11 = 17:00 pm. Assuming that Alay started at 't', (17-t)*10 = (t-5)*40 or t =7.4 which is 7:24 am
Alternatively,
Vijay travelled for 6 hours before Ajay started, so he would have covered 6*40 = 240 km. Now Relative speed between them is 40, so they will meet after 240/40 hours. i.e. at 5 pm. and by then the total distance traveled by them would have been 80*6 = 480 km. Alay travels at 50 km/hr, so to travel 480 he would require 480/50 = 9 hours 36 minutes. Hence, he would start at 7:24 am

Total acid = 0.4a+0.6(a+1)+0.8(a+2) = 1.8a+2.2
$$\frac{1.8a+2.2}{3a+3} = 0.6444$$
a = 2

Let the efficiency of each person be 1 unit of work per day. So, number of units of work to be done in total = 100 units
Work done on the first day = 1 unit
Work done on the second day = 2 units
Work done on the third day = 3 units and so on
Let the work be completed in ‘n’ days
Work completed in ‘n’ days = 1 + 2 + 3 +…+ n = n(n+1)/2 units
So, n(n+1)/2 = 100 => n(n+1) = 200
Solving this, we get n = 13.65 days
So, the work gets completed on the 14th day

Let the efficiency of A = a, B = b, C = c

Let the initial work be 6 units.

Then a = 6units/2days = 3 units/day

b = 6units/3days = 2 units/day

c = 6units/6days = 1 unit/day

Together their efficiency is 3+2+1 = 6 units/day

The work given by the boss would have taken them 4 days if they worked together,

So work = 6units/day * 4 days = 24 units

Now, A&B; together work for 2 days, A&C; together work for 3 days and rest is completed by B alone.

Let the time required by B to complete the remaining work be 'n' days.

So, the equation is: (a+b)*2 + (a+c)*3 +b*n = 24

=> 5*2 + 4*3 + 2*n = 24

=> n = 1 day

Let the total work be $$8x$$ units.

Thus, the total amount of work done by a man in 1 day = $$\dfrac{8x}{8} = x$$ units

When multiple men work together, the average amount of work done by a man in one day

= $$x \times (1 - \dfrac{20}{100}) = x \times (1 - \dfrac{1}{5}) = x \times \dfrac{4}{5} = 0.8x $$

Number of men needed to finish the job in 1 day = $$\dfrac{8x}{0.8x} = 10$$ men.

In 90 mins, 90 tons of water enter the ship.
So, the workers should empty 60 tons of water in 90 minutes.

One worker would empty 3 tons of water every 90 minutes.

Number of workers needed is 60/3 = 20

Assuming the distance is d, based on the question let the distance travelled by Atul be d-30.
The distance travelled by Bala = d+30.
Since the time taken to meet is the same, assuming the speeds are 5x and 2x, we get:

(d-30)/2x = (d+30)/5x

Now, the ratio of distance = (d-30)/(d+30)=2/5

=> d=70km

Let x be the initial volume of the solution. Hence, 5x/8 is the amount of milk and 3x/8 is the amount of water. 25% of x=2x/8 water is added.
Hence, the solution has 5x/8 milk and 5x/8 water now. Hence, the solution has equal parts of milk and water.
Removing 100 liters of solution and adding water is the same as removing 50 liters of milk and adding 50 liters of water.
So,
(5x/8 -50)/(5x/8+50) = 3/7. So, x = 200

Since, the distance between Ahmedabad and Surat is 250 km. So, at 12:00 both of them together cover a distance of 250 KMs.

For Ambika, from 9:00 am to 12:00 noon, distance travelled= Speed$$\times$$time= 50$$\times$$3= 150 KMs.

Bhanu starts at 10:00 am and covers the remaining distance in 2 hours.

Remaining distance= (250-150) Kms = 100 Kms.

Therefore Speed of Bhanu= $$\ \frac{\ Dis\tan ce}{Time}=\ \ \frac{\ 100}{2}=\ 50\ Kmph$$

Distance between them at 8:15 AM is 7.5 KM. So, time taken to meet is 7.5/(35-30) = 1.5 hrs. Distance travelled is 1.5*35 = 52.5 Km

A, B and C complete one circle in 300 s, 240 s and 200 s respectively.
So, the minimum time when they meet at the starting point is LCM (200,240,300) = 1200 secs = 20 minutes

The LCM of 20, 40 and 80 is the time when all three of them meet for the first time at the starting point. So, the answer is 80. Prior to that time, two of them meet at points other than the starting point.

$$p^q = q^p$$
So, $$p^{9p} = (9p)^p$$
So, $$p^9 = 9p$$
So, $$p^8 = 9$$ =>$$p = 9^{1/8}$$

21 divides the number n.
So, the smallest divisor of n is either 2 or 3 (as 3 divides 21)
So, the largest divisor of n is either 21*2 = 42 or 21*3 = 63
Hence, n = 84 (21*2 = 42) and n = 189 (21*3=63)
So, number of values of n is 2.

As the three digit number has exactly three factors, it is the square of a prime number. Let the prime number be 'a'

So, $$pqr=a^2$$ and the three factors are $$1,a,a^2$$
Now, the six digit number $$pqrpqr = 1001*pqr = 7*11*13*a^2$$

The number of factors of this number will depend on the value of the prime number a. If a is one of the two numbers 11 or 13, the number of factors equals 2*2*4 = 16
If a is different from the three numbers, the number of factors equals 2*2*2*3=24

Hence, the number of factors can be either 16 or 24 and so none of the first three options are definitely true.

The number of ways in which 1350 can be resolved into two factors.

Prime factorization of 1350

$$1350=2\times3^3\times5^2$$

Total number of factors = (1+1)*(3+1)*(2+1) = 24

The number of ways in which this can be resolved into two factors = Total factors/2 = 24/2 = 12

1*1350

2*675

3*450

5*270

6*225

9*150

10*135

15*90

18*75

25*54

30*45

50*27

$$(ABC)_{10}=(PQR)_7$$

Since ABC in base 10 and the value of A is greater than 2, the minimum value of ABC is 300.

Now, 300 in base 7 is represented as 606.

The maximum value of ABC can be 342, which corresponds to 666 in base 7

Now, since the values of P and R are equal,

The possible values are 606, 616, 626, 636, 646, 656, 666

Thus, the total possible values are 7.

$$154_B$$ is a factor of $$451_B$$

$$B^2+5B+4 = (B+1)(B+4)$$

$$451_B= 4*B^2+5B+1 = (B+1)(4B+1)$$

$$(B+1)(B+4)*t = (B+1)(4B+1)$$

t = $$\frac{\left(4B+1\right)}{\left(B+4\right)}$$

t = $$\frac{\left(4B+16\ -\ 15\right)}{\left(B+4\right)}$$

t = $$4-\frac{15}{B+4}\ $$

$$\frac{15}{B+4}\ $$ should be an integer.

This is possible only when B = 11

Let the two numbers be 7m and 7n where m and n are coprime.
So we have
7m + 7n = 1540
=> m + n = 1540/7 = 220
So we have to write 220 as the sum of two co-prime numbers.
Total numbers below 220 which are also co-prime to 220 are given by the euler no. of 220 = $$220*\frac{1}{2}*\frac{4}{5}*\frac{10}{11}$$ = 80.
For each of these numbers, there will be a corresponding number such that the sum is 220. For example, 3 + 217, 7 + 213, 13 + 207 and so on.
Since the question asks for ordered pairs so both a and b can take 80 values each. Hence the required answer is 80.

A, B and C can complete the task in $$\frac{60}{37}$$ days i.e. in a day they’ll complete $$\frac{37}{60}$$ of the work i.e.
$$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{37}{60}$$.......(1) Similarly, $$\frac{1}{A}+\frac{1}{B}=\frac{9}{20}$$........(2)
From these 2 equations, we get C = $$6$$.
Also given B and C can complete the task in $$\frac{30}{11}$$ days therefore, $$\frac{1}{B}+\frac{1}{C}$$=$$\frac{11}{30}$$......(3)
From 1 and 3 we get A = $$4$$. Putting the value of A and C in 3 we get B = 5.
In 1st 3 days, the work completed is $$\frac{37}{60}$$.
On 4th day, A will complete another $$\frac{1}{4}$$ of the work. Therefore, the work remaining is $$1$$-$$\frac{37}{60}$$-$$\frac{1}{4}$$= $$\frac{2}{15}$$.
On 5th day B will complete the task by $$\frac{2}{15}$$*5 = $$\frac{2}{3}$$rd of the day. Therefore, time required to complete the task = 4.66 days
Therefore, our answer is option ‘c’.

A can complete the project in 10 days.
Therefore, in 1 day he is completing $$\frac{1}{10}$$th of the project and hence in 2 days he is completing $$\frac{2}{10}$$th = $$\frac{1}{5}$$th of the project. Therefore, $$\frac{4}{5}$$th of the project is still remaining. His efficiency now decreases by 50%. Hence, now he can complete $$\frac{1}{20}$$th of the project in 1 day. He works for 6 days completing $$\frac{6}{20}$$th of the remaining project.
The amount of project remaining is = $$\frac{4}{5}$$-$$\frac{6}{20}$$ = $$\frac{1}{2}$$ of the project is remaining.
If he is to complete the project in next 2 days he should complete $$\frac{1}{4}$$ of the work in 1 day i.e. $$\frac{1}{20}$$+efficiency of assistant = $$\frac{1}{4}$$
Therefore, efficiency of assistant = $$\frac{1}{5}$$.
Therefore, efficiency of assistant : efficiency of A = $$\frac{1}{5}$$ : $$\frac{1}{10}$$ i.e. $$\frac{2}{1}$$.
Therefore, our answer is option ‘E’.

we are given that the sum of lengths of sides is 99. Let the lengths of the sides be a, b and c and semi perimeter be s=> a + b + c = 99
All side lengths are integers and also sum of lengths of any two sides is greater than the third side.
a + b > c => a + b + c > 2c => c < s => Length of any side is less than semi-perimeter.
=> Each side is less than 99/2 => 49.5. As all the sides are integers, each side must be less than or equal to 49.
So, each side can be equal to 49-x, 49-y, 49-z.
=> (49-x) + (49-y) + (49-z) = 99
=> x + y + z = 48 where x can vary from 0 to 48.
Number of solutions = $$^{48+3-1}C_{3-1}$$ = $$^{50}C_2$$ = 1225
Now, we have to remove the equilateral and isosceles triangles from these 1225 triangles.
Number of equilateral triangles = 1 (with each of the sides equal to 33)
None of the sides must be greater than 49. Keeping this in mind the isosceles triangles can be (25, 25, 49), (26, 26, 47),.......,(49, 49, 1). There are 25 such cases. But we are counting the (33, 33, 33) case again => 24 isosceles and 1 equilateral triangles.
The 1225 triangles we got are ordered triplets, but we need unordered triplets as the question does not define the sides for which the lengths are being found.
1 equilateral triangle of the form (x, x, x) => 1 triangle
24 Isosceles triangles of the form (x, x, y) => 3*24 = 72 triangles.
Total Scalene triangles = 1225 - 1 - 72 = 1152 triangles
1152 scalene triangles are of the form (x, y, z) => $$\frac{1152}{3!}$$ = 192 (unordered)
=> 192 scalene triangles can be formed with perimeter 99 units.
Shortcut formula:
Number of triangles that can be formed with perimeter p
• $$[\frac{p^2}{48}]$$ if p is even ($$[x]$$is the nearest integer)
• $$[\frac{(p+3)^2}{48}]$$ if p is odd
Number of scalene triangles that can be formed with perimeter p
• $$[\frac{(p-6)^2}{48}]$$ if p is even
• $$[\frac{(p-3)^2}{48}]$$ if p is odd

Let the cost of 1 pen be a, cost of 1 notebook be b and the cost of 1 eraser be c.
So we have,
3a + 2b + 4c = 80
5a + 6b + 4c = 176
Adding the two equations we get
8a + 8b + 8c = 256
=> a + b + c = 32
Hence the cost of 1 pen, 1 notebook and 1 eraser is 32.

In 3 hours, the bus travelled 120 km.
Relative speed of the car with respect to bus = 60 - 40 = 20kmph
Time taken for the car to meet bus = 120/20 = 6 hours
Distance travelled by the bike in 6 hours = 100 * 6 = 600 km

The figure is as shown in the diagram.

We have to find the area of triangle AGF.

In triangle AOC and AO1E,

Angle OCA = Angle O1EA = 90

Angle CAO = Angle EAO1

Triangle AOC and AO1E are similar.

AO/AO1 = 2/4

AO/(AO+6) = ½

AO = 6

Since A is an external point,

AB = AC and AD = AE

AD-AB = AE-AC

BD = EC.

Since F is an external point,

BF = FH = FD = y

F is a mid- point of BD

Similarly GC = GE = GH

G is a mid-point of CE.

GC = GE = y

Direct tangent of the circle = BD = BF+FD = 2y

2y = $$\sqrt{d^2-(R-r)^2}$$

2y = $$\sqrt{6^2-(4-2)^2}$$

2y = 4$$\sqrt{2}$$

Area of triangle AFG = ½ *FG* AH = ½ * (FH+HG)(AO+OH) = ½ (y+y)(6+2) = $$16\sqrt{2} = 8\sqrt{8}$$

5200 mangoes were left in the whole farm.

Had he plucked 20 more mangoes from each tree, 4400 mangoes would have left in the farm.

5200-4400=800

Number of trees = 800/20=40 trees

Total number of mangoes in the farm initilally= 240*40=9600

x mangoes from each tree

9600-40x= 5200

x= 4400/40= 110

Option D

Let Ram current age be r and Shyam's be s

$$\frac{r+10}{s+10}=\frac{6}{7}$$

$$7r+10\ =\ 6s$$ .....(I)

Their age differece will always remain same thus s-r = 5 ....(II)

Putting eqn value of s from eqn(II) to eqn (I)

$$7r+10\ =\ 6(r+5)$$

$$7r+10\ =\ 6r+30$$

On solving we get r =20

Given that we are unknown to the ages of arjun and atul .

Let them be x and y .

Given (x +1/y)(y +1/x) = 49/6

xy +1/xy = 37/6

Solving this we get xy = 6 .

Now we know that none of them are 1 year .

Hence we can have x , y to be 2 and 3 in any order .

The average of the first 4 numbers is (4N+6)/4 = N+1.5 The average of all the 6 numbers is (6N+15)/6 = N+2.5

Let Kavya score be K and Nikhila score be N.
Given K=N+19
Also given K=(5/9)(K+N)
9K=5(2K-19)
9K=10K-95
K=95
N=95-19
N=76
Hence, option D is the correct answer.

Let the incomes of A and B be 19x, 25x and their expenditures be 3y, 4y respectively.
So,
19x -3y = 234 … (1)
25x - 4y = 234 … (2)
On solving we have y = 1404
Now expenditure of A =3y = 3(1404) =4212

Let $$u = \dfrac{1}{p-q}$$ and $$v = \dfrac{1}{p+q}$$

Now, replacing, we get:

$$15u+18v=5$$ .... Eq 1

$$10u+36v=6$$ ..... Eq 2

Multiplying Eq 1 by 2,

We get:

$$30u+36v=10$$... Eq 3

Subtracting Eq 2 from Eq 3:

$$20u = 4$$

$$u = \dfrac{4}{20} = \dfrac{1}{5}$$

Now substitute $$u = \dfrac{1}{5}$$ back into Equation 1 to find $$v$$:

$$15\left(\dfrac{1}{5}\right) + 18v = 5$$

$$3 + 18v = 5$$

$$18v = 2$$

$$v = \dfrac{2}{18} = \dfrac{1}{9}$$

Substituting u and v back into the initial equations:

$$\dfrac{1}{p-q} = \dfrac{1}{5} \implies p - q = 5 \quad \text{--- (Equation 4)}$$

$$\dfrac{1}{p+q} = \dfrac{1}{9} \implies p + q = 9 \quad \text{--- (Equation 5)}$$

Adding Eq 4 and 5 to find p:

$$(p - q) + (p + q) = 5 + 9$$

$$p = 7$$

When X=1, number of solutions is 27

When X=2, number of solutions is 25

When X=3, number of solutions is 23 and so on.

So, total number of solutions is 1+3+5+. . .+27 = 196

Let the two numbers be A and B.
So, A-B = 2 and 3A+2B = 86
Multiplying the first equation by two and adding it to the second equation, we get

2A-2B + 3A + 2B = 4 + 86 = 90
So, 5A = 90 or A=18
Hence, B=16 and their sum equals 18+16 = 34

Let the fixed enrollment fee, book insurance fee and book issuance fee be x, y and z.

When the student borrows 4 books, he pays Rs. 554.

The equation will be x + 4y + 4z = 554......     ........Eq 1

Similarly, when he borrows 8 books, he pays Rs. 1094.

The equation will be x + 8y + 8z = 1094.................Eq 2

We need to find the amount when the student issues 6 books.  i.e. x + 6y + 6z

Adding Eq 1 & 2

we get,

2x + 12y + 12z = 1648

Dividing both side with 2,

we get x + 6y + 6z = 824

Hence, the answer is 824.

$$\log_x4+\log_{16}x=\dfrac{3}{2}$$

$$\dfrac{1}{\log_4x}+\dfrac{\log_4x}{2}=\dfrac{3}{2}$$

Let $$log_4x=a$$

$$\dfrac{1}{a}+\dfrac{a}{2}=\dfrac{3}{2}$$

$$a^2-3a+2=0$$

$$(a-2)(a-1)=0$$

$$a=1$$ or $$a=2$$

$$log_4x=1$$ or $$log_4x=2$$

$$x=4$$ or $$x=16$$

Sum of all values of x $$=4+16=20$$

Let the numbers be xy and ab

Thus, xy = 10x+y and ab = 10a+b

It is given that sum of 10x+y+10a+b = 60

10(x+a) + y+b = 60

We see that y+b has to a multiple of 10, so y+b can be 10,20 (but 20 is not possible)

Thus, y+b = 10 such that both numbers are prime.

Only possible when numbers are 3 and 7.

So, numbers are 10x+3 and 10a+7

Thus. 10(x+a)+10 = 60

x+a = 5

So, two possibilities:

= (1,4) and (2,3)

So (1,4) is not possible as the difference between numbers will be more than 10.

So, (33,27) are the possible numbers.

33*27 = 891

Option B is the correct answer.

$$\frac{x}{(x^2 + x-56)} = \frac{x}{(x - 7)(x + 8)}$$ This is greater than 0 only if $$x(x-7)(x+8) > 0$$

From the graph we can see that, -8 < x < 0 and x > 7. Option a)

Given $$ x^{2} - 10x + 21 < 0$$

Factorizing this, we get $$\left(x-7\right)\left(x-3\right)<0$$

Now we draw the number line for this with critical points 3 and 7.

So these inequalities hold true for $$x\in\ \left(3,7\right)$$

Hence, the minimum integral value is 4.

Hence, the correct answer is option B.

$$A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$$

$$A^2 = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$$

$$A^2 = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix}$$

$$A^4=\begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix}$$

$$A^4 = \begin{bmatrix} 1 & 4a \\ 0 & 1 \end{bmatrix}$$
.
.
.
.

$$A^n = \begin{bmatrix} 1 & na \\ 0 & 1 \end{bmatrix}$$

But it is given that $$A^n = \begin{bmatrix} 1 & 30 \\ 0 & 1 \end{bmatrix}$$

Thus, we can say that $$na=30$$.

Now, we are provided that $$a$$ and $$n$$ are Natural Numbers. Thus, the possible values of $$a$$ and $$n$$ can be -
(a,n) = (1,30) -> a + n = 1 + 30 = 31 (maximum)
(a,n) = (2,15) -> a + n = 2 + 15 = 17
(a,n) = (3,10) -> a + n = 3 + 10 = 13
(a,n) = (5,6) -> a + n = 5 + 6 = 11 (minimum)

Thus, the difference between the maximum and the minimum value of $$a+n$$ will be $$31-11=20$$.

$$A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$$, $$B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$$, and $$C = \begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix}$$

$$A+B+C = \begin{bmatrix} \beta+1 & 1 \\ 2 & \alpha+2 \end{bmatrix}$$

$$(A+B+C)^2 = \begin{bmatrix} \beta+1 & 1 \\ 2 & \alpha+2 \end{bmatrix}\begin{bmatrix} \beta+1 & 1 \\ 2 & \alpha+2 \end{bmatrix}$$

$$(A+B+C)^2 = \begin{bmatrix} \beta^2+2\beta+3 & \beta+\alpha+3 \\ 2\beta+2\alpha+6 & \alpha^2+4\alpha+6 \end{bmatrix}$$

We are given that $$\left(A+B+C\right)^2=11\begin{bmatrix} 1 & 1 \\ 2 & 6 \end{bmatrix}$$

$$11\begin{bmatrix} 1 & 1 \\ 2 & 6 \end{bmatrix}= \begin{bmatrix} \beta^2+2\beta+3 & \beta+\alpha+3 \\ 2\beta+2\alpha+6 & \alpha^2+4\alpha+6 \end{bmatrix}$$

$$\beta^2+2\beta+3=11$$ => $$\beta^2+2\beta-8=0$$ => $$\beta=-4$$ or $$\beta=2$$.

$$\alpha^2+4\alpha+6=66$$ => $$\alpha^2+4\alpha-60=0$$ => $$\alpha=-10$$ or $$\alpha=6$$.

$$\beta+\alpha+3=11$$ => $$\beta+\alpha=8$$. Thus, the values which satisfy the above equation are -

$$\alpha=6$$ and $$\beta=2$$

Therefore, $$\alpha-\beta=6-2=4$$