Quantitative Aptitude (QA) Test - 6

His hair grows 10 cm.
Per unit centimetre of hair growth, he loses 999 hairs.
So for 10 cm of hair growth he loses 9990 hairs.
Therefore his hair count is 10000-9990 = 10.

Let the number be $$(10X + Y)^{2} = AB36$$ for some A,B.
  We also know that the value of (10X+Y) must be in this range: $$32\le\ 10X+Y\le\ 99$$ to get a four-digit number. (31^2 is a 3-digit number, and 100^2 is a 5-digit number).

Y is either 4 or 6.

When Y = 6 => $$\left(10x+6\right)^2=AB36$$ => $$100x^2+120x+36=AB36$$

Therefore, $$\left(100x^2+120x\right)$$ has to be in the form of AB00, which implies $$\left(100x^2+120x\right)$$ must be divisible by 100.

We know that $$100x^2$$ is divisible by 100 => We need to check all the values of x such that 120x is divisible by 100.

The only value of x = 5, for which 120x is divisible by 100. Hence, (X, Y) = (5, 6)

When Y= 4 => $$\left(10x+4\right)^2=AB36$$ => $$100x^2+80x+16=AB36$$

Therefore, $$\left(100x^2+80x\right)$$ has to be in the form of AB20, which implies $$\left(100x^2+80x-20\right)$$ has to be in the form AB00, which implies $$\left(100x^2+80x-20\right)$$ must be divisible by 100.

We know that $$100x^2$$ is divisible by 100 => We need to check all the values of x such that (80x-20) is divisible by 100.

The above equation satisfies when the value of X is 4 and 9. Hence, (X, Y) = (4, 4) and (9, 4)

  We find that three pairs of (X, Y) satisfy the equation; (5,6); (4,4) and (9,4)

Let $$x,y,z$$ be the current ages of Amy,Bhanu and Carol.
$$\frac{y}{z} = \frac{x-6}{y-6}$$ .......(i)
Also $$z+6 = x- 6$$ ……..(ii)
By substituting the value of x-6 in (i)
We get $$y^2-6y = z^2+6z$$
$$y^2-z^2 = 6y+6z$$
$$(y+z)(y-z) = 6*(y+z)$$ (Using the property $$a^2-b^2=(a+b)(a-b)$$)
$$y-z =6$$
$$y =z+6$$
Therefore, the ages of Amy,Bhanu and Carol are z+12,z+6 and z
These will have two perfect squares only when z = 4.
Sum of their ages = $$3z+18 =30$$

The probability that he misses the target all four times is given by $${0.4}^4$$. Hence, the probability that he has hit the target is $$1 - {0.4}^4 = 0.9744.$$

Let us assume there are 'n' men that started the work together and each of person completed 1 unit of work in a day.

So total amount of work that is to done = 15 * n * 1 = 15n units

On the first day, 'n' men will work together so the amount of work the will be completed by the end of day1 = n units

On the second day, 5 men leave the job so 'n-5' men will work together and work done by them on 2nd day = n-5 units

There is no change in the number of men on the third day so the amount of work completed on 3rd day = n-5 units

On the fourth day, 5 men leave the job so 'n-10' men will work together and work done by them on 4th day = n-10 units

Hence total amount of work done in 25 days = n+(n-5)+(n-5)+(n-10)+(n-10)+(n-15)+(n-15) ... +(n-60) (Since people leave on every alternative day starting from 2nd day, hence 5 people will leave on 2nd day, 4th day, 6th day and so on)

$$\Rightarrow 15n = n + 2[(n-5)+(n-10)+...+(n-60)]$$
$$\Rightarrow 15n = n + 2\frac{12}{2} \times (n-5+n-60)$$
$$\Rightarrow 15n = n + 12(2n -65)$$
$$\Rightarrow n = 78$$

So total number of men that started the work = 78
Number of men that worked on day2 = 78 - 5 = 73 days

8 Chinese can be arranged in $$8!$$ ways

After arranging them in alternate places, the 4 Americans can be arranged in 9 gaps in  $$^9 P_4$$
Thus required no. of ways is $$8!*^9 P_4$$ = $$8!\times\frac{\ 9!}{5!}=9!\times8\times7\times6=336\times9!$$

The answer is option B.

Let the work of building the wall be equal to 280 units. Then the work done by Arijit, Bhole, Chetan and Diljit will be 2, 5, 7, and 8 units. We can obtain resultant work by taking the work of the rogue friend as negative when we look for cases.
For the wall to be completed in 90-100 days, one friend must be building it and the other one breaking it, such that the resultant work per day is between
$$\frac{280}{100}$$= 2.8 and $$\frac{280}{90}$$= 3.11 units.
Hence the difference between the rate of works has to be 3, which can be achieved by (8-5) or (5-2). This means that either Arijit or Bhole went rogue.
But three friends were working and the work was completed in 28 days, which means that 10 units of work were done per day. This can be achieved only if the rogue friend was working too.
This can be achieved in two ways: (8+7-5) or (7+5-2). We again see that either Arijit or Bhole went rogue.
Hence, we cannot determine which friend went rogue.

It is given that the reduction in the maximum distance is uniform. So it decreases equally every hour. Total reduction is 9-1 = 8 metres in 4 hours. Hence, it will decrease by 8/4 = 2 metres every hour. So after 1 hour it will be 0-2 = 7m

After 2 hours, it will be 7 - 2 = 5m

So required area is $$\pi * (9^2 - 5^2) = 176 $$sq m

Let $$t_1$$ be the time taken travelling 60km/hr and $$t_2$$ be the time spent travelling 45 km/hr.

As the distance covered is equal, $$60t_1=45t_2$$ or $$4t_1=3t_2$$ .....(I)

$$x=t_1+t_2$$

The time spent with the speed 50km/kr = 50x = $$50\left(t_1+t_2\right)$$

Total distance covered = $$60t_1+45t_2+50\left(t_1+t_2\right)$$

Putting $$t_1=\frac{3}{4}t_2$$ from (I)

Total distance covered = $$60\left(\frac{3}{4}t_2\right)+45t_2+50\left(\frac{7}{4}t_2\right)=177.5t_2$$

Equating with 710km we get $$177.5t_2=710$$ or $$t_2=4$$ hours.

Thus $$t_1=3$$

Total travel time = 2x = 2(3+4) = 14 hours

In rectangle ABCD, the point of intersection of the diagonals is F. AC = BD = 20 cm. Since the diagonals bisect each other, AF = FC = 10 cm = BF = FD.

$$AE^{2}+ EF^{2}= 10^{2}$$

AE=8cm.

Area of AFD = Area of ABF (Because the base is of the same length and they have the same height)
Similarly, Area of FCD = Area of BFC. Since Area of DFC = Area of ABF, we have:
Area of DFC = Area of BFC = Area of ABF = Area of AFD
Area of AFD = 1/2 * base * height = 1/2 * 10 * 8 = 40 $$cm^2$$
So, area of the rectangle = 160 $$cm^2$$

$$\frac{Time_{up}}{Time_{down}}=2$$

$$\frac{Speed_{up}}{Speed_{down}}=\frac{1}{2}$$

Let the speed of boat be b and the speed of the stream be b and s respectively.

Hence,

b + s = 2 (b - s)

b = 3s

New speed of stream = $$\frac{2}{3}s$$

Hence,

New upstream speed = $$3s+\frac{2}{3}s=\frac{11s}{3}$$

New downstream speed = $$3s-\frac{2}{3}s=\frac{7s}{3}$$

Hence ratio of speeds = 7/11

Ratio of time taken = 11/7.

Area of the shaded region = Side of the smaller square (HG) x half the side of the larger square (BP).
The diagonal of ABCD is equal to the diameter of the circle.
Thus, AC = 4cm
Also, $$AD^2 + DC^2 = 4^2$$
As AD = DC, $$2AD^2 = 16$$
=> AD = 2$$\sqrt2$$
Thus, BP = $$\sqrt2$$
In triangle $$EHO, EH^2 + OH^2 = EO^2$$
EH = 2OH AND EO = radius of the circle = 2cm
Thus, $$4OH^2 + OH^2 = 4$$
=> OH = 2/$$\sqrt5$$. Thus, HG = 4/$$\sqrt5$$.
Therefore the required area = BP x HG = $$\sqrt2$$ x 4$$/\sqrt5$$ = $$\frac{4\sqrt{10}}{5}$$

Let the market price and the cost price of the article be denoted by M and C respectively.

when the article is sold at p/2% discount, the profit was (p/2)%

$$M\left(1-\frac{p}{200}\right)=C\left(1+\frac{p}{200}\right)$$

=> $$\frac{M}{C}=\frac{\left(1+\frac{p}{200}\right)}{1-\frac{p}{200}}=\ \frac{\ 200+p}{200-p}$$

Hence, $$\frac{M}{C}=\frac{\left(200+p\right)}{\left(200-p\right)}$$

Now when the article is sold at a p% discount, the loss was (p/4)%

$$M\left(1-\frac{p}{100}\right)=C\left(1-\frac{p}{400}\right)$$

or, $$\frac{M}{C}=\frac{\left(400-p\right)}{\left(400-4p\right)}$$

Therefore, $$\frac{\left(200+p\right)}{\left(200-p\right)}=\frac{\left(400-p\right)}{\left(400-4p\right)}$$

=> $$\ \ \ \left(200+p\right)\left(400-4p\right)=\ \ \left(400-p\right)\left(200-p\right)$$

=>$$80000-400p-4p^2=80000-600p+p^2$$

=>$$5p^2=200p$$

or, 5p(p-40)=0

Since p is not equal to 0 so p= 40

Therefore, $$\frac{M}{C}=\frac{\left(200+40\right)}{\left(200-40\right)}=\frac{3}{2}$$

Now let the market price and cost price be 300 and 200 respectively.

$$SP=300\left(1-\frac{10}{100}\right)=270$$

So profit = 70

Therefore, profit% = $$\frac{70}{200}\times\ 100=35\%$$

The ratio of the areas can be calculated by using the ratio of the sides. 

$$\frac{Area\ \left(\triangle\ ABC\right)}{Area\ \left(\triangle\ RAQ\right)}=\frac{\frac{1}{2}\times\ 3\times\ 3\times\ \sin\ \theta\ }{\frac{1}{2}\times\ 1\times\ \left(3+1\right)\times\ \sin\ \left(180-\theta\right)}\ =\ \frac{9}{4}\ $$

Similarly, $$\frac{Area\ \left(\triangle\ ABC\right)}{Area\ \left(\triangle\ QCP\right)}=\frac{\frac{1}{2}\times\ 3\times\ 3\times\ \sin\ \alpha\ \ }{\frac{1}{2}\times\ 1\times\ \left(3+2\right)\times\ \sin\ \left(180-\alpha\ \right)}\ =\ \frac{9}{5}\ $$

Similarly, $$\frac{Area\ \left(\triangle\ ABC\right)}{Area\ \left(\triangle\ PBR\right)}=\frac{\frac{1}{2}\times\ 3\times\ 3\times\ \sin\ \beta\ \ \ }{\frac{1}{2}\times\ 2\times\ \left(3+1\right)\times\ \sin\ \left(180-\beta\ \ \right)}\ =\ \frac{9}{8}\ $$

Hence, let us consider the area of $$\triangle\ ABC\ =\ 9x\ $$; then, $$\triangle\ RAQ\ =\ 4x,\ \ \triangle\ QCP\ =\ 5x\ ,\ \triangle\ PBR\ =\ 8x$$ 

$$\therefore\ Area\ of\ \triangle\ PQR=9x+4x+5x+8x=26x$$

$$\frac{Area\left(\triangle\ ABC\right)}{Area\left(\triangle\ PQR\right)}=\frac{9x}{26x}=\frac{9}{26}$$

Hence, Option D is the correct choice

Profits will be shared after Vansh’s salary is accounted for. At the year end, Vansh’s salary will be 12 X 25000 = Rs 3 lacs.
So, effective profit to be shared is 5-3=2 lacs.
Now, their profits will be in the ratio 10 X 12: 20 X 9 = 2:3
So, Vansh’s share will be 2/5 X 2 = Rs 80000
And Amrit’s share will be 3/5 X 2 = Rs 1.2 lacs
Vansh also earns a salary of Rs 3 lac. So, Vansh’s total share is Rs 3.8 lacs while that of Amrit is Rs 1.2 lacs. So, the ratio is 38:12 or 19:6.

We can rationalise the terms in the denominator in the following manner:

First term = $$\frac{1\left(2\sqrt{1}-1\sqrt{2}\right)}{\left(2\sqrt{1}+1\sqrt{2}\right)\left(2\sqrt{1}-1\sqrt{2}\right)}=\frac{\left(2\sqrt{1}-1\sqrt{2}\right)}{2}=\left(1-\frac{1}{\sqrt{2}}\right)$$

Second term = $$\frac{1\left(3\sqrt{2}-2\sqrt{3}\right)}{\left(3\sqrt{2}+2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\right)}=\frac{\left(3\sqrt{2}-2\sqrt{3}\right)}{6}=\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)$$

Third term = $$\frac{1\left(4\sqrt{3}-3\sqrt{4}\right)}{\left(4\sqrt{3}+3\sqrt{4}\right)\left(4\sqrt{3}-3\sqrt{4}\right)}=\frac{\left(4\sqrt{3}-3\sqrt{4}\right)}{12}=\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)$$

Fourth term = $$\frac{1\left(5\sqrt{4}-4\sqrt{5}\right)}{\left(5\sqrt{4}+4\sqrt{5}\right)\left(5\sqrt{4}-4\sqrt{5}\right)}=\frac{\left(5\sqrt{4}-4\sqrt{5}\right)}{20}=\left(\frac{1}{\sqrt{4}}-\frac{1}{\sqrt{5}}\right)$$

Hence, for $$S_{24}$$ we can rewrite the expressions as: $$1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}...\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}=\left(1-\frac{1}{5}\right)=\frac{4}{5}$$

$$\therefore\ 5\times\ S_{24}=4$$

Take option 1: The amounts invested in each scheme will be 20000, 60000 and 20000, respectively.

The returns obtained in BD, FD and MF will be 0.06*20000 = 1200, 0.1*60000 = 6000 and 0.08*20000 = 1600 respectively. Thus, the total return from option A is = Rs. 8800

Similarly, returns obtained from options B, C and D will be Rs. 7500, Rs. 7400 and Rs. 7800, respectively.

Thus, option A provides maximum returns.

We have to find out $$\sum_{i = 1}^{10} f(i)$$

$$f(1) = \frac{1}{1*3}$$ ; $$f(2) = \frac{1}{1*3}$$ ; $$f(3) = \frac{1}{3*5}$$; $$f(4) = \frac{1}{3*5}$$... 

We can see that each even term is same as previous odd term.

We know that $$f(1) = \frac{1}{1*3} = \frac{1}{2}\times ({1} - \frac{1}{3})$$

$$f(3) = \frac{1}{3*5} = \frac{1}{2}\times (\frac{1}{3} - \frac{1}{5})$$

therefore, $$\sum_{i = 1}^{10} f(i) = (f(1) + f(2) + ... +f(10))$$

$$\Rightarrow$$ $$2(f(1) + f(3) + f(5) + f(7) +f(9))$$

$$\Rightarrow$$ $$(1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + .... + \frac{1}{9} - \frac{1}{11})$$

$$\Rightarrow$$ $$(1 - \frac{1}{11})$$

$$\Rightarrow$$ $$\frac{10}{11}$$

Hence option A is the correct answer. 

let A's age be 11x, B's age = 7x
let C's age be 13y, D's age is 9y
It is given, average age of A, B, C, D is 34.5 years. Therefore,
11x + 7x + 13y + 9y = 34.5*4
18x + 22y = 138
9x + 11y = 69
The only possible solution is x = 4 and y = 3
The ages of A, B, C and D in 2020 are 44, 28, 39 and 27 years, respectively.

In 2020, three members joined, and the average became 27 years.
Let the sum of the ages of three new members be 'S.'
138 + S = 27*7
S = 51
It is given that 51 is the sum of three distinct prime numbers greater than 10
The only possibility is 11, 17 and 23.
Maximum difference = 44 - 11 = 33

Answer is option D.

Let $$\frac{y+5}{y^2+5y+25} = k$$

So, $$ky^2+5ky+25k-y-5=0$$

For the above quadriatic equation to have real roots, its discriminant should be greater than or equal to zero.

$$(5k-1)^2 \geq 4*k*5*(5k-1)$$
$$25k^2-10k+1\ge20k\left(5k-1\right)$$
$$25k^2-10k+1\ge100k^2-20k$$
$$75k^2-10k-1\le0$$

solving we get
$$(15k+1)(5k-1)\leq 0 $$
So $$-1/15 \leq k \leq 1/5$$

So, the difference in the maximum and minimum value is 4/15

27*4 = 108; 26*5 = 130; 24*7 = 168. Hence, 224 is the answer.

At the points where the two curves intersect,

$$3X^3+4X^2-2X-2 = 4X^2+X-2$$
$$X^3=X$$
So, X=0 or -1 or 1
As, we are only looking for points where X >= 0, the number of such points is 2