Topic - Quant Mixed Bag

Solved ? - Yes

Source - Selected questions/solutions from Nitin Gupta Sir (AlphaNumeric) ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - Selected questions/solutions from Nitin Gupta Sir (AlphaNumeric) ]]>

(a) 114

(b) 117

(c) 129

(d) 136

(e) 130 ]]>

Subsets with 0 element = 1.

Subsets with 1 element = 12.

Subsets with 2 element = 10C2 = 45.

Subsets with 3 element = 8C3 = 56.

Subsets with 4 element = 6C4 = 15

So total subsets = 1 + 12 + 45 + 56 + 15 = 129.

Method 2

Let f(n) be number of subsets of {1,2, .....n} which contain no more than one out of any three consecutive natural numbers.

then f(n) = f(n-1) + f(n-3)

Logic is to include f(n-1) subsets and add n to all the subsets for f(n-3) to get all possible subsets for n

with f(1) =2, f(2) = 3 , f(3) =4.

solving we get f(12)=129.

Hence, choice (c) is the right answer.

]]>(a) can not be determined

(b) Rs 318

(c) Rs 375

(d) Rs 384

(e) none of the foregoing ]]>

If the profit %ages are different, then it must be clear that the ratio of apples : Oranges in the 60Rs. sale and the 72Rs. sale must be the same.

Hence, we can safely say that 1.2 KG of oranges = 20 Oranges and 1KG of apples = 12 apples.

SP of 20 Oranges and 12 Apples = 72 => SP of 5 Oranges and 3 apples = 18

CP of 1.2KG of Oranges and 1KG of Apples = 60 => CP of 6KG Oranges and 6KG apples = 300

Hence, choice (b) is the correct option

]]>(a) 20

(b) can not be determined

(c) 16

(d) greater than 20

(e) none of the foregoing ]]>

Let M be the feet of perpendicular from C to PQ. Now, by alternate segment theorem we have

=> < CQM = < CPB and < CPM = < CQD. Thus, right triangles CQM and CPB are similar and also CPM and CQD. Thus, CQ/CP = CM/CB and CP/CQ = CM/CD. Thus, CB * CD = CM^2 = 16.

Method 2

Assume the vertices of A, B, C, D as (0, b), (a, b), (a, 0) and (0,0).

Since, AP = AQ, let AQ = m. Thus P = (0, b-m) and Q = (m, b).

Now x = 0 is tangent to the circle at P.

The eq. of circle is x^2 + (y - b+m)^2 + kx = 0, here k is variable. But, it's also tangent at Q = (m, b).

Thus we get k = -2m.

Now, the circle x^2 + (y-b+m)^2 -2mx passes through (a, 0) => a^2 + (b-m)^2 = 2am.

Equation of PQ is x-y = b-m.

Also |(a-m+b)/v2| = 4.

Squaring and using the relation we get 2ab = 32 => ab = 16.

Hence, choice (c) is the correct option

(a) 0 < A < = 1/2

(b) 1/6 < = A < = 1/4

(c) 0 < A < = 1/3

(d) 0 < A < 2/3

(e) 1/6 < = A < = 1/3 ]]>

The area formed is of a triangle with vertices (p/3, p/3), (p/2, 0) and (p, p).

Thus, the area is p^2/6 sq. units.

Case 2: 1 < = p < = 3

The figures formed is a quadrilateral (p/3 ,p/3),(p/2,0), ((p+2)/3 , (4-p)/3) and (1,1).

Thus, the area formed is 1/3 - 1/6*(p-2)^2

Case 3: 3 < = p < = 4

The area formed is the image of case 1. Area = (4-p)^2/6

Thus, min(A) = 0 at p = 4 and max(A) = 1/3 at p = 2

Hence, option (c) is the right answer ]]>

(a) 1.80

(b) 2.0

(c) 1.85

(d) 1.95

(e) None of the foregoing ]]>

x - y < z ----------------(1)

z - x < y ----------------(2)

y - z < x ----------------(3)

squaring and adding all this three equation we get:

(x-y)^2 + (z-x)^2 + (y-z)^2 < x^2 + y^2 + z^2

=> 2 * (x^2 + y^2 + z^2) - 2 * (xy + yz + zx) < x^2 + y^2 + z^2

=> x^2 + y^2 + z^2 < 2 * (xy + yz + zx)

=>(x^2 + y^2 + z^2)/(xy + yz + xz) < 2

Hence, choice (b) is the right option ]]>

(a) x^2 + 4z^2 is composite

(b) x^2 + y^2 + z^2 is non-prime

(c) x^2 + 4z^2 is prime

(d) both (a) and (b)

(e) none of the foregoing ]]>

= (xy +√3(xz - y^2) -3yz)/(y^2 - 3z^2) is rational

=> y^2 = xz.

Let y = xr, z = xr^2

=> x^2 + 4z^2 = x^2 * (1+4r^4)

= x^2 * (2r^2 + 2r+1)(2r^2-2r+1).

If x = 1 then r ≠ 1;

Thus, x^2 + 4z^2 can be written as product of 2 natural numbers > 1 which are x(2r^2 + 2r+1) and x(2r^2 - 2r + 1).

Similarly, x^2 + y^2 + z^2 = x(r^2 + r + 1) * x(r^2 - r + 1) and can be written as product of 2 natural numbers > 1.

Hence, option (d) is the right answer

(a) 9

(b) 11

(c) 21

(d) not be uniquely determined

(e) none of the foregoing ]]>

In the 1st group no. of points won (by Asians) = x(x-1)/2

In the 2nd group no. of points won (by Europeans) = (x+9)( x+8 )/2

Let k be the points Asians gain in the 3rd group. Therefore those won by Europeans = x(x+9) - k

Using the given conditions we have

9*[x(x-1)/2 + k] = (x+9)( x+8 )/2 + x(x+9) - k

This gives 3x^2 - 22x + 10k - 36 = 0

=> Factorize it to get the following: (3x+5)(x-9) = -10k - 9

Since RHS is negative, LHS too should be negative. Hence, x lies between 0 and 9. On checking, we find that the only integer solutions of the above equation occur when either x=8 and k=2 or when x=6 and k=6.

The maximum no. of matches an Asian team can win if x = 6 and k = 6 is

5 (won from other 5 Asian teams in 1st Group) + 6 (If the same Asian team is the one which wins all the Asian-European matches in 3rd Group) = 11

and if x = 8 and k = 2 it is 7 + 2 = 9

So the maximum no. of points an Asian team can have is 11. ]]>

(a) 13

(b) 14

(c) 15

(d) 16

(e) None of the foregoing . ]]>

Indeed, if 10a + b is (where a, b are integers 1–9), then 7 divides 10a + b and 7 divides 10b + a.

Thus, 7 divides{ 3(10a + b) − (10b + a)} = 29a − 7b = a + 7(4a − b),

so that 7 divides a, and similarly 7 divides b, so we’d better have a = b = 7.

There are 16 3-digit numbers. First consider the 12 palindromic ones (ones where the hundreds and units digits are the same): 161, 252, 343, 434, 525, 595, 616, 686, 707, 777, 868, and 959.

Now consider the general case: suppose 100a + 10b + c is, where a, b, c are integers 1–9. Then 7 divides 100a + 10b + c and 7 divides 100c + 10b + a, so 7 divides (100a+10b+c)−(100c+10b+a) = 99(a−c), and so 7 divides a−c. In order for this not to result in a palindromic integer, we must have a − c = +/- 7 and, moreover, both 100a + 10b + a and 100c + 10b + c must be palindromic integers.

Consulting our list above, we find 4 more integers: 168, 259, 861, and 952.

So K = 1 + 12 + 4 = 17

Hence, choice (e) is the right answer

(a) 1/2*(a^3 + b^3 + c^3)

(b) (a^2 + b^2 + c^2)^2/(a+b+c)

(c) a^3 + b^3 + c^3

(d) 3*(a^3 + b^3 + c^3)

(e) (a^4 + b^4 + c^4)/(a + b +c) ]]>

Otherwise, put a = 1 and b = 0 the equation (a^5+b^5+c^5)/((a^2+b^2+c^2)(a^3+b^3+c^3)) = 5/6 reduces to (c+1)^3*(c^2 - 3c + 1) = 0 => c^2 - 3c + 1 = 0 as a+b+c ≠ 0.

We are asked to find ab(a+b) + bc(b+c) + ca(c+a) - 3abc = c(1+c)

Looking at the choices (a) becomes 1/2 * (1+c^3).

Multiplying both sides of c^2 - 3c + 1 = 0 by c + 1 we get c(1+c) = 1/2*(1+c^3)

Hence, option (a) is the right answer

]]>Also , area of the triangle with side (2,2,a) = (a * 2 * 2)/(4 * R) = (a/4) * root(4 * 2^2 - a^2) {Where R is the circumradius}

=> R = 4/root(16 - a^2) ------------(1)

Also from the second triangle with sides(3,3,b) we get

R = 9/root(36-b^2) ------------------(2)

Equating (1) and (2) we get,(36-b^2)/(16-a^2) = 81/16

Using (A) we have a/b = 9/4 ]]>

(a) 9/4

(b) 7/3

(c) 7/2

(d) 3/2

(e) None of the foregoing ]]>

Triangles SUW & SRT are similar, 2/(3+RW) = 3/(2+TU)

Hence UV = 8/3

Hence, choice (e) is the right option ]]>

(a) 5%

(b) 10%

(c) 15%

(d) can not be determined

(e) none of the foregoing ]]>