Quantitative Aptitude (QA) Test - 15

The total number of squares in the (6*6) board is 36. If we place a rook in one square, then the attacking square by the rook is (5 in that row, and 5 in that column) => 10 squares,

So we can't place another rook in that 10 attacking squares, and 1 square belongs to the other rook. Hence, the second rook can be placed (36-10-1) = 25 square.

Now, the number of ways that we can choose one square among the 36 squares is $$^{36}C_1$$. For each of the square, we can find 25 squares such that they do not attack each other.

Since the rooks are identical, which implies (first rook, second rook) = (second rook, first rook) (For example: $$\left(a_1,b_2\right)\ =\left(b_2,a_1\right)$$ are the same.

Hence, the total number of ways = $$\ \frac{\ ^{36}C_1\cdot25}{2}=\ \frac{\ 36\cdot25}{2}=450$$ ways

Let the numbers thrown on 4 dice be a,b,c,d
a+b+c+d=10 a,b,c,d lie between 2 and 5.
Only case valid are {2,2,2,4},{2,2,3,3}
Total no. of cases in{2,2,2,4}=4
Total No. of cases in{2,2,3,3}=6
Total cases=10 Now, x+y+z+w=10
Where x,y,z,w lie between 1 and 6.
Each of x,y,z,w has to be at least 1.
So the valid cases of the above equation is $$^9C_3$$
But it will include 4 cases 0f {7,1,1,1}
Total cases= $$^9C_3$$-4=80
Hence probability=10/80=1/8=0.125

From the given figure,

In triangle BDC Tan b =(DC/BC) -----(1)

in triangle ABX Tan b = (BX/8) -----(2)

By equating both the equations we get BX = 6.4 cm.

A = 5(C+D)
Let A,B,C be 21X, 30X and 2X.
21X = 5(2X + D)
D = 2.2X
B:D = 30X:2.2X = 150:11.
x-y = 139.

The amount of Aluminium in Alloy Q is 0%; this helps us figure out the ratio in which the two alloys were mixed:

Hence, the two alloys were mixed in the ratio of 5:3

Equating the amount of C in the alloys: let the amount of Cobalt in alloy Q be x, then

 

On solving, we get x = 44; hence, the Cobalt constitutes 44% of the alloy Q.

Thus, Option (B) is the correct choice.

The given curve is $$x^2 + y^2 - 4x +2y -20 = 0$$.
=> $$x^2 - 4x + 4 + y^2 + 2y + 1 -4 -1 -20 = 0$$
=> $$(x-2)^2 + (y+1)^2 = 5^2$$
This is the equation of a circle with radius 5 and center (2, -1)
If d is the distance between the center of the circle and the given point then the maximum and the minimum distance of the point from the given circle is d+r and d-r respectively.
$$d = \sqrt{(8-2)^2 + (7-(-1))^2} = \sqrt{36+64} = 10$$
Thus, the required ratio = (10+5):(10-5) = 15:5 = 3:1
Hence, option A is the correct answer.

Let common ratio for mangoes and apples be x
so mangoes will be 5x and apples will be 8x
He purchased mangoes and oranges in the ratio 4:7
so let common ratio here be y
Now mangoes will be 4y and oranges will be 7y ,
Now we can say 5x=4y
For fruits to be minimum
x will be 4 and y will be 5(smallest possible value)
so we get mangoes as 20
apples as 32
oranges will be 20*(7/4)=35
So minimum fruits will be 20+32+35=87

Money deposited in the bank = 90,000 - 23,000 - 18,000 - 21,000 - 13,000 = Rs 15,000

Money withdrawn from bank = $$15,000\left(1.08\right)^2$$ = Rs 17,496

Selling price of refrigerator and sofa set = (23,000+18,000)(1.12) = Rs 45,920

Selling price of TV and dining table = (21,000+13,000)(0.92) = Rs 31,280

Total amount at the end of two years = 31280 + 45920 + 17496 = Rs 94,696

% change = $$\ \frac{\ 94696-90000}{90000}\times\ 100$$ = 5.22%

Answer is option C.

Let the efficiency of $$A_1$$ be $$"e"$$; then each subsequent tank will have the efficiencies as $$\frac{e}{2},\frac{e}{4},\frac{e}{8},...\infty\ $$

The sum of these when they functio as inlets = $$\frac{e}{1-\frac{1}{2}}\ =2e$$

Let the tank capacity be $$T$$; the time taken to fill the tank is 4 hours. This can be represented as: $$\frac{T}{2e}=4;\ \therefore\ T\ =\ 8e$$

When the even-numbered pipes function as outlets, the sum of efficiencies ibecomes as follows: $$e-\frac{e}{2}+\frac{e}{4}-\frac{e}{8}+...\infty\ $$ 

We can group this as: $$\left(e+\frac{e}{4}+\frac{e}{16}+...\infty\ \right)-\left(\frac{e}{2}+\frac{e}{8}+\frac{e}{32}+...\infty\ \right)$$

$$\left(\frac{e}{1-\frac{1}{4}}\right)-\left(\frac{\frac{e}{2}}{1-\frac{1}{4}}\right)$$ = $$\frac{4e}{6}$$

Therefore, the time taken to fill the tank = $$\frac{8e}{\frac{4e}{6}}=12\ hrs$$

The shop keeper buys 60 mangoes each at a price of Rs 200 . Total money spent = RS 200*60 = Rs 12000.

The cost price of the rest of the 40 mangoes is Rs 8000.

He decides to earn an overall profit of 5 percent on the total sale = (5/100)*12000 = Rs600

So he must sell the mangoes at Rs 12600 .

20 of the mangoes were sold at a price of Rs 140 = Rs 2800.

He must sell the rest of the 40 mangoes at a price of 12600 - 2800 = Rs 9800.

1) If he marks up the price by 25 percent and to meet his target he must provide a discount of x percent .

8000*(1.25)*((100-x)/100) = 9800.

100-x/100 = 98/100 , x = 2%.

2) If he marks up the price by 50 percent to meet his requirement he must provide a discount of y percent,

8000(1.5)((100-y)/100) = 9800.

y = 18.333 

[y-x] = [16.333] = 16

$$5X < 3Z^{2} +28$$. This equation suggests nothing about the relation between X and Z. For example, (2,1) and (1,2) both satisfy the above inequality for (X,Z)

Let the capacity of the tank be 60 litres.

Rate of A  = 3 litres per hour

Rate of B = 4 litres per hour

Rate of C = 2 litres per hour

Rate of L1 = -1 litre per hour

Rate of L2 = -1 litre per hour.

First one-fourth of the tank is filled by A, B and C in $$\ \frac{15}{9}$$hours

Second one fourth is filled by A, B, C and L1 in $$\ \frac{15}{8}$$hours

Final 30 litres is filled by A, B, C, L1 and L2 in $$\ \frac{30}{7}$$hours

$$\therefore\ $$ Total time taken = $$\frac{1315}{168}$$ hours. = $$7\frac{139}{168}$$ hours

Let a, b, c and d be the percentages of households buying milk from one vendor, 2 vendors, 3 vendors and all 4 vendors respectively.
a+b+c+d = 100 .... (1)
a + 2b+ 3c + 4d = 310 .... (2)
eqn (2) minus 3 times eqn (1) gives
-2a-b +d = 10
So, the minimum value of d = 10% when a and b are zero.

Let the side of the outermost triangle be a.

Hence, the inradius = $$\frac{a}{2\sqrt{3}}$$

This value is equal to 1.

Hence, the value of a = $$2\sqrt{3\ }$$

For the right ranged triangle to be of maximum area, it needs to be isosceles as well.

Area = 1/2 x 2r x r = $$r^2$$. [r is the radius of the circle]

Area = a^2/12 = 1

Area of the remaining shaded region = Area of the outermost triangle - ARea of the bigger circle - 3*ARea of the smaller circle.

The smaller circle can be considered to be the incircle of an equilateral triangle of height 1. Hence, radius = 1/3

Hence, remaining shaded region = $$\left[\frac{\sqrt{\ 3}}{4}*2\sqrt{\ 3}*2\sqrt{\ 3}-\pi\ -3\pi\ \left(\frac{1}{3}\right)^2\right]$$ = $$3\sqrt{\ 3}-\pi\ -\frac{\pi}{3}=3\sqrt{\ 3}-\frac{4\pi}{3}$$

Hence, total shaded area = $$3\sqrt{\ 3}-\frac{4\pi}{3}+1$$ = 2.005

The distance travelled by each person is the same i.e. 8 km.
Now the ratio of speed is given, so the speeds can be assumed to be s, 2s, 3s and 8s.
Time = $$\frac{Speed}{Dis\tan ce}$$  So the respective times will be $$\frac{8}{s},\frac{8}{2s},\frac{8}{3s},\frac{8}{8s}$$ 
A ratio remains the same if each term is multiplied by a constant term. So, to remove fraction and s, we multiply each term by 3s
So the ratio of time = 24:12:8:3

Hence, the times can be assumed as 24x, 12x, 8x, 3x

The difference between the time taken by B and C is 20 minutes.
If B and C could reach Q at same time => 12x-8x = 20 min => 4x=20   => x = 5 minutes
A’s reaching time = 8 a.m. + 24*5= 10 a.m.
D’s time = 9 a.m. + 3*5=15 =9:15 a.m.
Required difference = 45 minutes

Let the percentage increase and decrease be r%

As per the question; $$504000\left(1+\frac{r}{100}\right)\left(1-\frac{r}{100}\right)=496125$$

or, $$1-\left(\frac{r}{100}\right)^2=\frac{496125}{504000}=\frac{63}{64}$$

or, $$\left(\frac{r}{100}\right)^2=1-\frac{63}{64}$$

or, $$\frac{r}{100}=\frac{1}{8}$$

r= 12.5%

Investment at the end of the fifth year = $$496125\left(1+\frac{r}{100}\right)=558141$$

$$\log_2 (9^{y-1} + 7) = 2 + \log_2 (3^{y-1} + 1)$$
=> $$\log_2 (9^{y-1} + 7) = \log_2 4 + \log_2 (3^{y-1} + 1)$$
=> $$\log_2 (9^{y-1} + 7) = \log_2 4*(3^{y-1} + 1)$$
=> $$(9^{y-1} + 7) = 4*(3^{y-1} + 1)$$
=> $$(3^{y-1})^2 + 7 = 4*3^{y-1} + 4$$
Let $$3^{y-1} = t$$
=> $$t^2 + 7 = 4t + 4$$
Solving this equation for t, we get t = 1 or 3
=> $$3^{y-1}$$ = $$3^0$$ or $$3^1$$
=> y = 1 or y = 2
So, ‘y’ is a natural number

The two trains are $$150$$ km apart. Thus, time for the collision = $$\frac{150}{70+80}$$ = $$1$$ hour

The distance travelled by the bird before the two trains collide will be the distance that the bird can travel in $$1$$ hour = $$1\times 100 = 100$$ km.

Hence, option D is the answer.

Total No. of Divisors of $$a^x$$x$$b^y$$x$$c^z$$ = (x+1)(y+1)(z+1), where a,b,c are prime numbers

Here $$n\times\ \left(n+5\right)$$ has 4 divisors that is possible if n=1 or if n and (n+5) are prime (i.e n=2).

If n=1, $$n^2\ +\ 5n$$ = 6 (4 divisors = 1,2,3 & 6)

If n=2, $$n^2\ +\ 5n$$ = 14 ( 4 divisors = 1,2,7 &14)

So, total number of such natural numbers is 2 (n=1 & n=2)

In triangle PQR

Area of triangle PQR = $$\sqrt{s. (s - a)(s - b)(s - c)}$$                     (Here s = $$\frac{28 + 26 +30}{2}$$ = 42)

$$\Rightarrow$$ Area = $$\sqrt{42.(16).(14).(12)} = 336$$ 

We also know that Area of triangle PQR = $$\frac{1}{2} PQ. UR $$

$$\Rightarrow$$  336 = $$\frac{1}{2} * 28 * UR$$

$$\Rightarrow$$  UR = 24cm 

In right-angled $$\triangle$$ QUR

QU = $$\sqrt{QR^2 - RU^2}$$ = $$\sqrt{26^2 - 24^2}$$ = 10 cm i.e. PU = PQ - QU = 28 - 10 = 18 cm.

Now in triangle QUR using angle bisector theorem

$$\Rightarrow$$ $$\frac{QU}{QR} = \frac{UY}{YR}$$

$$\Rightarrow$$ $$\frac{10}{26} = \frac{UY}{24 - UY}$$ 

$$\Rightarrow$$ $$UY = \frac{20}{3} cm$$

Similarly in triangle PUR using angle bisector theorem

$$\Rightarrow$$ $$\frac{PU}{PR} = \frac{UX}{XR}$$

$$\Rightarrow$$ $$\frac{18}{30} = \frac{UX}{24 - UX}$$

$$\Rightarrow$$ UX = 9cm 

Hence XY = UX - UY = $$9 - \frac{20}{3} = \frac{7}{3}$$ cm

$$x^2-lx+3 = x^2+kx-4$$
x = 7/(k+l)

Since x is an integer k+l can take values 7,-7, 1, -1

When k+l = 7, x = 1
l = 4 and k = 3

These values satisfy k+l = 7

When k+l = -7, x = -1
l = -4, k = -3

These values satisfy all the conditions.

Two more cases remain when k+l = 1 and x = 7
And when k+l = -1 and x = -7

We can substitute the value of x in the equations $$x^2-lx+3 = 0$$ and $$x^2+kx-4 = 0$$ and see that k and l are not integers in both the cases.

Hence, the number of possible solutions is equal to 2

We have the series:

0.8, 0.98, 0.998, 0.9998, 0.99998....

0.8, 0.9+0.08, 0.99+0.008, 0.999+0.0008, 0.9999+0.00008 and so on.

So there are two series, 0.8, 0.08, 0.008... and 0.9, 0.09, 0.009....

Keep in mind that the 0.9, 0.09... series starts from the second term. So the number of terms will be one less.

We can combine the two series, by shifting 0.9 from second term to first, 0.99 from third term to second, and so on:

0.8+0.9, 0.08+0.99, 0.008+0.999... till tenth term, and then subtract 0.9999999999, as it is present in the 11th term

=> 1.7, 1.07, 1.007, 1.0007....., 1.0000000007 - 0.9999999999

0.9999999999 can be written as $$1-0.1^{10}$$.

We can separate each term into 1 and $$\frac{7}{10^n}$$.

=$$1+\frac{7}{10},\ 1+\frac{7}{100},.....,1+\frac{7}{10^{10}}-1+0.1^{10}$$

We get the sum as:

=$$10+\frac{\frac{7}{10}\left(1-\frac{1}{10^n}\right)}{1-\frac{1}{10}}-1+\frac{1}{10^{10}}$$

=$$9+\frac{7}{9}\left(1-\frac{1}{10^{10}}\right)+\frac{1}{10^{10}}$$

=$$\frac{88}{9}+\frac{2}{9\times\ 10^{10}}$$