Quantitative Aptitude (QA) Test - 9

It has been stated that 20% of X evaporates every hour; hence, after '$$n$$' hours, the amount of X remaining in the solution = $$7x\left(0.8\right)^n$$

Similarly, for Y, we identify that after '$$n$$' hours, the amount of Y remaining in the solution = $$2x\left(0.9\right)^n$$ 

Together, the ratio after  '$$n$$' hours = $$\frac{7x\left(0.8\right)^n}{2x\left(0.9\right)^n}=\frac{7}{2}\times\ \left(\frac{8}{9}\right)^n$$

We can extend the same to the solution in the beaker; the ratio after '$$n$$' hours =  $$\frac{4x\left(0.7\right)^n}{x\left(0.9\right)^n}=\frac{4}{1}\times\ \left(\frac{7}{9}\right)^n$$

We have been told that after 6 hours: $$\left\{\frac{7\left(8\right)^6}{2\left(9\right)^6}=k^5\frac{4\left(7\right)^6}{\left(9\right)^6}\right\}$$

On solving, $$\left\{\frac{8^5}{7^5}=k^5\right\}$$ OR k = 8:7

Option A is the correct choice. 

Let us first draw the fig.

Now PL =8cm , PQ=12cm 
In triangle PQL  CosP =$$\frac{8}{12}$$=$$\frac{2}{3}$$
⇒ PQ=$$\frac{3PL}{2}$$         (1)
Now in triangle PNR ; CosP =$$\frac{PN}{PR}=\frac{2}{3}$$
⇒PR = $$\frac{3PN}{2}$$       (2)
Dividing (1) and (2)

⇒$$\frac{PQ}{PR}=\frac{PL}{PN}$$
so we can say Triangles PQR and PLN are similar (SAS)
⇒$$\frac{\triangle\ PLN}{\triangle\ PQR}=\frac{4}{9}$$
⇒$$\frac{x}{x+40}=\frac{4}{9}$$
⇒ x =32
so area of PQR = 32+40
=72cm^2
Hence option (B) is the answer.

Draw perpendicular from A to BC as shown in the figure below -

So triangle ABD is 45-45-90 triangle,
AD = BD
Triangle ADC is 30 - 60 - 90 triangle,
Tan 30 = $$\frac{AD}{DC}$$
$$ DC = \sqrt{3}*AD = \sqrt{3}*BD $$
$$ BC = BD + DC = BD + \sqrt{3}*BD $$ = $$7 + 7\sqrt{3}$$

So BD = 7 and AD = 7
Area triangle ABC = $$\frac{1}{2} * AD * BC = \frac{7 * (7 + 7\sqrt{3})}{2} = 24.5*(1 + \sqrt{3})$$
Hence, option B is the right choice.

Let OA and OB be perpendicular to RS and PQ respectively and r be radius of the circle.
so we get RM=MS and PB=BQ =5cm
Now, we know that PM*MQ=MS*MR 
solving we get MR =2cm
Therefore OB =7-2=5cm
So radius of circle :$$\sqrt{\ 5^2+5^2}$$=$$5\sqrt{\ 2}$$cm.
Now if you observe in triangle POQ
OP=OQ=$$5\sqrt{\ 2}$$ cm and PQ=10cm
⇒ OPQ is a right triangle or PQ subtends and angle of 90 degrees at the center.
Area of quadrilateral POQR = 1/2*10*5 + 1/2*10*2 = 35
Now area of shaded region :
Area of sector POQ - area of quadrilateral POQR
=$$\frac{90}{360}\times\ \pi\ \times\ 50$$ -35
=12.5$$\ \pi$$-35
=4.25cm^2.

Total ways of choosing 2 cards from a pack of well shuffled cards are 52 cards = 52C2.
Our desired sample space consists of red non-face cards.
Total red cards = 26. Red face cards = 6
Hence desired set = 20
Thus, the required probability is
$$\frac{20C2}{52C2}$$ = $$\frac{20*19}{52*51}$$ = $$\frac{95}{663}$$

Whatever cost the company incurred till getting the 25% work done is already spent from their pockets. The only saving they can do it for the remaining 75% of the work. 

Let the efficiencies of P,Q,R be $$1$$,$$2$$ and $$3$$ and let their wages be ₹$$4y$$, ₹$$6y$$ and ₹$$11y$$ respectively. 

Now, to find the most economical way, we need to see the amount the manager spends on each of the workers to get a unit amount of work done. 

Let the total original work be $$100z$$. Work left = $$75z$$. 

Case 1: P is fired

Time taken by Q and R to complete the work = $$\frac{75z}{\left(5\right)}=15z$$

Cost = $$15z\left(6+11\right)\ =\ 255z$$

Case 2: Q is fired

Time taken by P and R to complete the work = $$\frac{75z}{\left(1+3\right)}=18.75z$$

Cost = $$18.75z\left(4+11\right)\ =\ 281.25z$$

Case 3: R is fired

Time taken by P and Q to complete the work = $$\frac{75z}{\left(1+2\right)}=25z$$

Cost = $$25z\left(4+6\right)\ =\ 250z$$

Thus, in the case when R is fired, the cost incurred will be the least.

The correct answer is C.

Let s be the normal speed of the train and t hours be the normal time taken by the train to cover the distance from Ludhiana to Jalandhar.
The new speed = s + 2/3s = 5/3 s and new time = t - 36/60 = t - 3/5
Since, the distance is equal in both the cases; speed is inversely proportional to time
=> $$\frac{s}{5s/3} = \frac{t - 3/5}{t} $$
thus, 3t = 5t - 5
t = 1.5 hr

To obtain a ratio of 1:1 in the resultant solution, at least one of the mixtures in any of the test tubes needs to have a silver to nitric acid ratio greater than 1. This is, however, not plausible since $$p:q$$ shall always be less than 1, irrespective of the combination. Hence, the correct answer is zero. 

Initially let the number of t-shirts of Pepe, Levis and Adidas with Harry be 5x, x and 4x respectively
Let the number of gifted t-shirts of Pepe and Adidas be y and z.
Now, number of t-shirts of Pepe left with Harry becomes 5x-y
Number of t-shirts of Levis left with Harry becomes x
Number of t-shirts of Adidas left with Harry becomes 4x-z

x, y and z should be whole numbers as the number of t-shirts cannot be negative or fractions

So, $$\frac{\left(5x-y\right)}{x}=\frac{5}{3}$$

$$15x-3y=5x$$

For y to be a whole number, x should be a multiple of 3

Similarly, $$\frac{\left(4x-z\right)}{x}=\frac{7}{3}$$

$$12x-3z=7x$$

$$z=\frac{5}{3}x$$

For z to be a whole number, x should be a multiple of 3

So, from above two conditions x should be a multiple of 3

Total number of t-shirts initially with Harry$$=5x+x+4x=10x$$

To minimize the total number of t-shirts initially with Harry the minimum possible value of x is 3.

Therefore $$10x=30$$

Alternate Method:

Initially let the number of t-shirts of Pepe, Levis and Adidas with Harry be 5x, x and 4x respectively.

Let the final number of t-shirts of Pepe, Levis and Adidas with Harry be 5y, 3y and 7y respectively.

Now, x=3y

Therefore, Total no. of t-shirts initially with Harry =5x+x+4x=10x = 30y

So, minimum no. of t-shirts initially with Harry will be 30.

Consider A and B are always together
Total number of ways of arranging the persons on 10 chairs = 2*9!
Suppose A,B,C are always together with A and B together
Total number of ways = 4*8! (There are 4 possiblities- CAB, CBA, ABC, BAC)
Required number = 2*9!- 4*8! = 2*8!(9-2) = 14*8!

Let's consider the radius to be 5 units and the depth of the cone to be 12 units. The addition of the base plate effectively reduces the depth to two-thirds of the original. Hence, we obtain an inverted frustum with a depth of 8 units. The radius of the base plate can be calculated using the reduction in depth since a proportional change will apply to the radius i.e. the radius of the base plate is one-third the radius of the cone (5/3 units). Using the above, we can calculate the volume of the inverted frustum as follows: 

$$\frac{\pi}{3}h\left(R^2+R.r+r^2\right)\ =\frac{\pi}{3}\left(8\right)\left(5^2+5\times\ \frac{5}{3}+\frac{5^2}{3^2}\right)$$

On solving we obtain the volume as $$\left(\frac{26}{27}\right)100\pi\ \ unit^2$$.

The volume of milk that is actually supposed to be sold = $$\frac{\pi}{3}\left(5\right)^2\left(12\right)=100\pi\ units^2$$

$$\therefore\ \%profit=\frac{\left\{100\pi\ -\left(\frac{26}{27}\right)100\pi\ \right\}}{\left(\frac{26}{27}\right)100\pi\ }\times\ 100\approx3.85\%\ $$ 

Let the initial amount of grain be 100kgs and price be Rs 100/kg.

Let us say that 2a kgs of rice is unburnt. Hence, 100-2a kg was burnt, a kg was sold at 150%*100 = Rs 150/kg and remaining a kg were sold at 1.3*Rs 100= Rs 130/kg.

Hence, we are told that loss from 100-2a kg = Profit from a kg sold at Rs 150

Hence, 100*(100-2a)=50*a

10000-200a=50a

a=40kg

Hence, 100-2*40 = 20 kgs was burnt, 40kgs was sold at Rs 150 and 40 kgs was sold at Rs 130.

Overall profit % = $$\frac{20*0+150*40+130*40}{100*100} - 1$$

= 1.12-1

=12%

It is given that three installments are in A.P, and the second installment is 40% less than the first.
Let the three installments be 3a+d, 3a , 3a-d
$$3a\ =\frac{3}{5}\left(3a+d\right)$$
d = 2a
Three installments are 5a, 3a, a
$$5a(1.3)^2+3a(1.3)+a=54735(1.3)^3$$
$$13.35a=54735(1.3)^3$$
$$a=4100(1.3)^3$$
a = 9007.7
Average amount paid per installment = 3a = Rs 27023.1. Rounded down the value is 27023.

The answer is 27023.

For the inequality to hold good, an odd number of terms should be less than zero. If x >= 0, all the terms are greater than 0. So, the inequality does not hold true. For x = -2, -6…, -98, an odd number of terms are less than 0. So, the product of those terms is also less than 0 and the inequality holds good.
These numbers form an Arithmetic Progression.
-2 + (n-1)*(-4) = -98 => n-1 = 24 => n = 25
So, the number of integers for which the inequality holds good is 25.
Option d) is the correct answer.

Let us assume that Ram started from point ‘x’ and Lakhan started from point ‘y’.
Let the speed of Ram and Lakhan be R kmph and L kmph.

On first meeting, Since they met at 3 pm Ram ran for 4 hours while Lakhan ran for 1 hour.Hence they must have covered distance 4R and L before the 1st meeting.

$$\Rightarrow \frac{xo}{oy} = \frac{4R}{L}$$   …(1)

On second meeting, Let us assume they met for the second time 'n' hours after 1st meeting Hence they must have covered distance nR and nL kms respectively between 1st and second meeting.
$$\Rightarrow \frac{2oy}{2ox} = \frac{nR}{nL}$$
$$\Rightarrow \frac{ox}{oy} = \frac{L}{R}$$ … (2)

From equation (1) and (2)
$$\Rightarrow (\frac{L}{R})^2 = \frac{4}{1}$$
$$\Rightarrow \frac{L}{R} = 2 : 1$$ (Answer)

Let the numbers be a(larger) and b(smaller) respectively.
$$a^2 + b^2 = 25a$$
$$a^2 - b^2 = 7a$$
=> $$2a^2 = 32a$$
=> a = 0(Ignore this case as a is a positive integer) or a = 16
=> $$b^2 = 400 - 256 = 144$$
=> b = -12 or +12
As b is a positive integer, b = 12 is the answer.

$$\log_{10}x+\log_{10}(x-7)+\log_{10}5-\log_{10}3=2$$

$$\rightarrow\log_{10}(x(x-7)*5/3)=\log_{10}100$$
$$\rightarrow 5x(x-7)=300$$
$$\rightarrow x^2-7x-60=0$$
$$\rightarrow x=12 \textrm{ or} -5$$.
However, since log is not defined for negative values, x cannot be -5. Hence, the sum of values =12.

Let ABC be the triangle and we are rotating it around AC. The length of AB is 12 cms.

As seen from the above figure, the solid formed is a double cone, with base radius = $$6 \sqrt3$$ cm and height = 6 cm.
So volume $$= 2( 1/3 \pi r^2 h)$$ = $$2*1/3 * \pi *(36*3) * 6$$ = $$432\pi$$

$$\frac{7}{15} + \frac{77}{15^2} + \frac{777}{15^3} + …$$
$$ =\frac{7}{15}[1 + \frac{11}{15} +\frac{111}{15^2} + …]$$
$$ =\frac{7}{15}[1+(\frac{1}{15} + \frac{10}{15}) + (1/15^2 +10/15^2 + 100/15^2) +(1/15^3 +10/15^3 + 100/15^3+1000/15^3)…]$$

Rearranging the terms we get

$$ =\frac{7}{15}[(1 + \frac{1}{15} + 1/15^2 +1/15^3......)+ (10/15 +10/15^2 + 10/15^3 +.......) + (100/15^2+100/15^3+..................)…]$$ 
$$ =\frac{7}{15}[\frac{1}{{1} - \frac{1}{15}} + (\frac{10}{15})*(\frac{1}{{1} - \frac{1}{15}}) + \frac{100}{15^2}*(\frac{1}{{1} - \frac{1}{15}})+…$$
$$ =\frac{7}{15}*15/14*[1+10/15+100/15^2 +…] = 3/2$$

Let the total work to be done be 60 units. So Ram does 4 units of work in a day and Shyam does 3 units of work in a day. Hence when they work together, they will be able to do 7 units in 1 day. Hence in 5 days, they will do 35 units of work. Thus, the remaining 25 units will have to be done by Shyam. The time taken by Shyam to do the remaining 25 units of work = 25/3 = 8.33 days.
Hence the total time taken = 5 + 8.33 = 13.33 days.
Hence the work will be completed on 14th day.

One of X and Y is even as X+Y is prime.
As the only even prime is 2 and Y<X, Y = 2.
So, X-2, X and X+2 are prime.
This is true only if X=5.
Hence, number of divisors of XY+2 = 12 is 6

Let the number of people in each bullock cart is x, and the number of bullock carts be y.

Hence, the total number of people = (x*y)=xy. It is given that 20 bullock carts broke down, so each of the other bullock carts had to carry 2 extra people.

We have, xy = (y-20)(x+2)

=> xy = xy+2y-20x -40 => 2y-20x = 40

=> y-10x = 20 .... Eq (1)

Similarly, on the return journey, 30 more bullock carts broke down, so each of the remaining bullock carts had to carry 6 more people than when they started out in the morning.

Therefore,

xy= (y-50)(x+6)

=> xy = xy +6y-50x-300  => 6y-50x = 300

=> 3y-25x = 150

Solving both the equations, we get x=18 and y=200 and xy=3600.