Quantitative Aptitude (QA) Test - 18

⇒ BH = 11.25BH=11.25cm. ...(2)

From equation (1) and (2) we can say that BG = 7.5+11.25 = 18.75.

We can see that both the triangles △BGC and △DGC are of same height.

This is because their bases lie on the same line and have a common 3rd vertex. Hence, the perpendicular dropped from that vertex will be the same for both triangles.

Hence, the ratio of the area of the triangles will be same as the ratio of base lengths.Therefore

Let f(x) = |x+3|+|x-3|+|x-6|+|x-5|+|x+5|

f(x) is a linear equation in x at all times. It will be a union of straight lines with inflection points at -3, 3, 6, 5 and -5. Hence, the minimum value of the expression occurs at one of these inflection points. We will calculate the value of f(x) at each of these points and then find out the least possible value of f(x)

f(-3) = 25

f(3) = 19

f(6) = 24

f(5) = 21

f(-5) = 31

19 is the least value.

Number of non-negative integral points on x + 2y = 20 ⇒ (20,0), (18,1),......,(2,9) and (0,10) ⇒ 11 points

Intersection point of 2x + y = 16 and x + 2y = 20 is (4,8)

So, all the points among the above 11 points that have x coefficient less than or equal to 4 are removed.

⇒ 3 points are removed.

Hence, required number of points = 11 - 3 = 8

3x + 4y + z = a --> (1)

2x + 6y + 4z = b --> (2)

x - y - 2z = c --> (3)

Eliminating z from (1) and (2) , (1)x4 - (2) : 12x + 16y + 4z - (2x + 6y + 4z) = 4a - b

10x + 10 y = 4a - b

⇒ x + y = (4a - b)/10 - (4)

Eliminating z from (1) and (3), (1)x2 + 3 = 6x + 8y + 2z + (x - y - 2z) = 2a + c

⇒ x + y = (2a + c)/7 - (5)

Equating equations (4) and (5), we get,

(4a - b)/10 = (2a + c)/7

28a - 7b = 20a + 10c

Thus, 8a = 7b + 10c

thus only one value of x i.e. 1/√2 can satisfy the equation.

Let us take the two roots to be A and B. A and B would be values of 3^x  that satisfy the equation.

However, we need to find the product of 3^x+2 that satisfy the equation. So, we must find the value of (A+2)*(B+2) = AB + 2*(A+B) + 4.

From the quadratic equation t^2-18t+27=0, we know that A+B = 18 and AB=27

So, the required product will be 27+36+4 = 67

So the answer is Option C

The LCM of 3 and 11 is 33. Thus, the number of numbers we find in the first 33 numbers will be the number of numbers that are there in every consecutive set 33 numbers.

Multiples of 3 less than 33 that have odd remainders when divided by 11 : 3,9,12,18,27

In the next set of 33 numbers, the numbers that satisfy the conditions are : 36,42,45,51,60 ie 33+3,33+9,33+12,33+18,33+27 respectively.

Below 1000, there will be  which means there will be 30 such sets.

From 33x30 to 1000, ie from 990 to 1000, there are only 2 more numbers that satisfy the conditions ie 993 and 999.

To find sum of all the numbers, first we find the sum of the first series of numbers ie all numbers below 33 that satisfy the condition.

S₁ =3+9+12+18+27=69

The sum of the second series of numbers S₂ = (3+33)+(9+33)+(12+33)+(18+33)+(27+33) = 69 + (5 * 33) = 69+165

Similarly, the sum of the third series of numbers S₃ =69+(165×2)

Thus, sum of the n^th series of numbers S_n =69+[165×(n−1)]

Thus, total sum of the series till the 30^th set of numbers S=(69×30)+[165×(1+2+3.....29)]

We should remember that this series does not take into consideration the last 2 numbers.

Therefore the actual sum S' = 73845 + 993 + 999 = 75837