Topic - Quant Mixed Bag

Solved ? : Yes

Source : Learningroots forum ]]>

Topic - Quant Mixed Bag

Solved ? : Yes

Source : Learningroots forum ]]>

a. Scheme A, Rs. 600

b. Scheme B, Rs. 600

c. Scheme B, Rs. 800

d. Scheme A, Rs. 800 ]]>

a. 72

b. 100

c. 144

d. 45 ]]>

Similarly, for a 9 sided figure, each internal angle will be equal to (9 - 2) * 180/9=140 degrees.

So, external angle = 40 degrees and so, angle at each point of the start will be 180 - 2 * 40 = 100.

(2-1) * 1! + (3-1) * 2! + (4-1) * 3! + ... + (8-1) * 7!

2! - 1! + 3! - 2! + 4! - 3!... 8! - 7!

8! -1 = 40319 is the answer. ]]>

a. 6

b. 7

c. 8

d. 9 ]]>

3x(4y + 7) + 5(4y + 7) = 81

(3x + 5)(4y + 7) = 81

Using the AM > = GM property,

(3x + 5 + 4y + 7)/2 > = sqrt((3x + 5)(4y + 7))

3x + 4y + 12 > = 18

3x + 4y > = 6

So minimum value is 6. ]]>

a. 1216

b. 2716

c. 216

d. 1716 ]]>

We are falling short by 6 in this case and this 6 has to be taken from one or many of the 8 digits in question.

a + b + c + d + e + f + g + h = 6

13c7 solutions = 13 * 12 * 11 * 10 * 9 * 8/720 = 1716 cases.

Also, you have to understand that none of the digits can be 0 as then the sum of digits won't exceed 63 and so, we are saved of additional steps. ]]>

a. 5(2/3) days

b. 5(1/3) days

c. 7(1/3) days

d. None of these ]]>

The concept to understand here is that R is taking 13 days to finish off a piece of work that M+S are doing in 12 days. So, considering that total work is 156 units, R will do 12 units of work per day which is his efficiency and M+S will do 13 units of work per day which is their combined efficiency. So, if one day's work is considered a total of 25 units will be done out of which R has contributed 12 and so, his contribution will be 12/25.

]]>(a) 3

(b) 7

(c) 5

(d) 11 ]]>

100a+10b+c-(100b+10a+c)=90

90a-90b=90

a-b=1

a can take values from 1 to 9 and b can take corresponding values from 0 to 8. So total of 9 cases. ]]>

a. 1800

b. 1891

c. 300

d. 330 ]]>

We have to eliminate the cases which have repetitions and then check the cases that have all unequal integral solutions.

Total solutions are 62c2 = 61 * 31 = 1891

When 2 people get similar number of balls - 2a + c = 60, a will vary from 0 to 30 i.e. 31 cases one among which will be 20 + 20 + 20. Ignoring that and ordering the remaining among a, b, c we get 30 * 3 = 90 solutions. Adding the one 20 + 20 + 20 case we get 91 cases in total with at least two terms being similar to each other.

1800 cases will be equally distributed among 6 cases which are a > b > c, a > c > b, b > a > c, b > c > a, c > a > b and c > b > a. So, 1/6th of 1800 satisfies our required condition of a > b > c and so the answer is 300. ]]>

(a^2 + x^2)/2 > = sqrt(a^2 * x^2)

(a^2 + x^2) > = 2ax.

Similarly,

(b^2 + y^2) > = 2by

(c^2 + z^2) > = 2cz

Adding all three,

2 > = 2(ax+by+cz)

So, max value of ax + by + cz = 1.

Alternatively, you can check for a = b = c = x = y = z = 1/sqrt(3) and solve.

]]>Equating the volume of the cone with 30pi, we get the radius of the cone to be 3 units. If we look at the combination properly, we can see that we have two intersecting chords of 3+3 on one side and 9+x on the other. So, 3 * 3 = 9 * x and so, x=1. So, diameter of the sphere is 9+1=10. So, radius is 5 units. So volume is 4pi(5^3)/3 = 500pi/3.

]]>For any equation, a + b + c + … + r terms = n

Number of non-negative integral solutions will be (n + r – 1) C (n – 1)

Here it will be 27c25 which will be 351. ]]>

(a) 4 : 1 : 4 : 3

(b) 1 : 2 : 2 : 1

(c) 2 : 3 : 6 : 1

(d) 5 : 2 : 7 : 3 ]]>