Best GMAT Quant Course 2026

$$\frac{\left(P+36\right)}{P}=1\ +\frac{36}{P}$$

We are told that the above expression is an integer, so P must be factor of 36.

Therefore, possible values of P are 1, 2, 3, 4, 6, 9, 12, 18 and 36

Sum = 1+2+3+4+6+9+12+18+36 = 91

Let the cost price of each mango be $x. Therefore, the cost price of 8 mangoes is $8x, which equals the selling price of 6 mangoes.

Hence, the selling price of 6 mangoes is $8x, which implies the selling price of 1 mango is (8x/6) = $(4x/3)

Hence, the profit percentage is {(4x/3 -x)/x}*100% = 33.33% of the cost price.

The correct option is D

Let the Length of race be d 

Speed of A = 3 m/s and B started 15 seconds after A 

A must be ahead by 3*15 = 45 meters, when B started.

When C started, A will be ahead by 3*20 = 60 meters and B will be ahead by 6*5 = 30 meters 

Let t be the time taken by C to cover the d distance 

Speed of c = 9 = d/t       -Eqn 1

Speed of a = 3= (d-60)/(t+20)     -Eqn 2

Substituting value d from Eqn 1 in Eqn 2, we get -

3 =(9t-60)/t+20)

3t+60 = 9t-60

6t= 120 

t = 20 

d = 9t 

d = 180m

B had already covered a distance of 30 meters when C started.

Time taken by B = (180-30)/6 = 25 seconds 

Hence, B will finish 5 seconds after C

Let scores of ram be 16x, 23x, 18x, 19x, 20x, 21x, and 24x

Total score = 16x +23x + 18x + 19x + 20x + 21x + 24x = 141x 

This is 47% of the total marks 

maximum marks = (100/47)*141x = 300x

maximum marks per test = 300x/7

42% marks per test = (42/100)*300x/7 = 18x

Hence we can conclude that he scored more than 42% marks in 5 tests- 19x, 20x, 21x, 23x, and 24x

We are replacing 12 litres of the 60 litres water solution with wine.

We are replacing 1/5th of the water solution with wine every replacement.

Hence water will reduce by 1/5th and remain 4/5th after every replacement.

We can say that after n replacements 

Quantity of water = $$60\times\left(\frac{4}{5}\right)^n$$

Quantity of water should be less than 35%

Quantity of water should be less than $$\frac{35}{100}\times60\ =21$$

$$60\times\left(\frac{4}{5}\right)^n$$ $$<21$$

$$\left(\frac{4}{5}\right)^n<\frac{21}{60}$$

$$\left(\frac{4}{5}\right)^n<\frac{7}{20}$$

min. n = 5

Let a, b and m be the prices of each apple, banana and mango, respectively.

Writing the given data in the form of equations, we get

4a + 3b + 10m = 96 and 2a + 6b + 7m = 95

We need to find the value of 10a + 3b + 23m, for which we need to write this expression as a linear combination of the given equations.

10a + 3b + 23m = x(4a+3b+10m) + y(2a+6b+7m)

Equating the coefficients of a and b, we get 10 = 4x + 2y and 3 = 3x + 6y

Multiplying the equation 4x + 2y = 10 with 3 and subtracting 3x + 6y = 3 we get,

12x + 6y - 3x - 6y = 30 - 3 => 9x = 27 => x = 3; substituting this in any of the above two equations, we get y = -1.

Therefore, 10a + 3b + 23m = 3(4a + 3b + 10m) -1(2a + 6b + 7m) = 3(96)-95 = 193.

We have five different observations.

In order to have the lowest value be as large as possible, we need the other values to be as low as possible.

Since 25 is the median, it will be the middle term of the five terms when arranged in ascending order.

We are given that the median is 25, so our numbers are _, _, 25, _, _.

Inserting the lowest available values for the top two gives us _, _, 25, 26, 27.

Let the lowest number be x and the second lowest number be x+1.
x + (x+1) + 25 + 26 + 27 = 5*19
2x + 79 = 95
2x = 16
x = 8

The probability of a surprise test each day is $$\frac{1}{2}$$, and the probability of no surprise test is also $$\frac{1}{2}$$.

The number of ways to select on which $$2$$ days out of $$6$$, there will be a surprise test is $$^6C_2=15$$

Hence, the probability that the surprise test is on exactly 2 days in a 6-day period is

$$^6C_2\times\ \left(\frac{1}{2}\right)^{^{6-2}}\times\ \left(\frac{1}{2}\right)^{^2}$$

$$=15\times\ \left(\frac{1}{2}\right)^{^6}\ =\ \frac{15}{64}$$

The correct option is D

For 6 to be the largest value among four cards, the other three cards must be from (1 to 5).

The number of ways to select these three cards is $$^5C_3$$.

The number of ways to choose four cards among nine cards is $$^9C_4$$.

Hence, the probability of getting six as the largest value = $$\ \frac{\ ^5C_3}{^9C_4}=\frac{10}{126}=\frac{5}{63}$$

The correct option is D