Quantitative Aptitude (QA) Test - 8

As the numbers are in AP and the sum is more than 150, the middle number will be greater than 50.
The average of the three numbers will be the middle number.
The square of 50 is 2500.
So the middle number can be the prime just greater than 50.
53 is the next prime.
59 is the next prime, and three times the square of 59 is more than 8600.
SO only 53 is the only possible middle number.

$$23^{500}$$

=$$529^{250}$$

=$$(530-1)^{250}$$

=$$^{250}C_0(530)^{250}-^{250}C_1{530}^{249}+^{250}C_2{530}^{248}-...........+^{250}C_{248}(530)^{2}-^{250}C_{249}(530)^{1}+1$$

=1000m+(125*249*530*530)-(250*530)+1

=1000m+ 250*530{249*265-1}+1

=1000m+ 250*530*65984+1

=1000m+ 250*4{530*16496}+1

=1000m+ 1000*{n}+1

=1000{m+n}+1   {We can see that both m and n are integers}

We have to find the last three digits

Hence last three digits are 001.

It is given that $$5$$ litres of first container, $$6$$ litres of 2nd container and $$3$$ litres of 3rd container are taken.  Four vessels contain alcohol of $$80\%,60\%,40\%$$, and $$10\%$$ concentration, respectively.
Therefore,
In $$5$$ litres of the first container, the volume of alcohol => $$(5\times\ 0.8)=4L$$ is alcohol => $$(5-4)=1L$$ is water.
In $$6$$ litres of the second container, the volume of alcohol => $$(6\times\ 0.6)=3.6L$$ alcohol => $$(6-3.6)=2.4L$$ water
In $$3$$ litres of the third container, the volume of alcohol => $$(3\times\ 0.4)=1.2L$$ alcohol => $$(3-1.2)=1.8L$$ water
Hence, the volume of alcohol from the first three containers is $$(4+3.6+1.2)=8.8$$L, and the volume of water is $$(1+2.4+1.8)=5.2L$$
Let 10x litres of the fourth container be taken in the mixture to make the ratio of alcohol to water is $$1:4$$
Hence, In 10x litres of the fourth container, the volume of alcohol = $$(10x\times\ 0.1)=xL=>(10x-x)=9xL$$
Hence, the total volume of alcohol in the mixture is $$(8.8+x)L,$$ and the total volume of water in the mixture is $$(5.2+9x)L$$
Therefore,
$$\ \frac{\ 8.8+x}{5.2+9x}=\frac{1}{4}$$, which implies $$\ \frac{\ 8.8+x}{5.2+9x}=\frac{1}{4}\ =>\ 35.2+4x=9x+5.2=>\ 5x\ =30\ =>\ x\ =6\ L$$
Hence, the volume of the fourth container = ($$10x=10\times\ 6=60$$ L

Required probability = 1- both the balls are either red or blue
=$$1-\frac{^{3+5}C_2}{^{16}C_2} = 1-\frac{7*8}{15*16} = \frac{23}{30}$$

Let us assume that Amar takes ‘x’ days to complete the work alone and Akbar takes ‘y’ days to complete the work alone. Now we have been given that
$$\frac{1}{x} = \frac{1}{2}*(\frac{1}{40} + \frac{1}{y})$$
=> $$\frac{2}{x} = \frac{1}{40} + \frac{1}{y}$$
From, the second relation that we have been given we know that
$$\frac{1}{y} = \frac{1}{40} + \frac{1}{x})$$
Solving these two equations, we get
a = 20 and b = $$\frac{40}{3}$$
Total work done in 1 day by these two will be $$\frac{1}{20} + \frac{3}{40}$$ = $$\frac{5}{40}$$
Hence, in one day they can finish $$\frac{1}{8}$$ of the total work. Thus, the time taken by them to finish the entire work together will be 8 days.

If he tosses a head in the first attempt, the gain is 5. If he tosses it in the second attempt, the gain is 4. And so on.
Let G be the gain.

$$G = (1/2)*5 + (1/2)^2*4 + (1/2)^3*3 + (1/2)^4*2+(1/2)^5*1+(1/2)^6*0+(1/2)^7(-1)+….$$ $$G*(1/2) = (1/2)^2*5 + (1/2)^3*4 + (1/2)^4*3+(1/2)^5*2+(1/2)^6*1+(1/2)^7*0+….$$

$$-G/2 = -5/2 + (1/2)^2 + (1/2)^3 + (1/2)^4 + …$$
$$-G/2 = -5/2 + (1/4)*2 = -2 => G = 4.$$

Given, A can do 1/3rd of work in 10 days
⇒ A can do complete work alone in 30 days
B can do half of work in 15 days
⇒ B can do complete work in 30 days
Let total work be 30 units (LCM of 30,30)
Efficiency of A = 30/30 = 1 unit/day
Efficiency of B = 30/30 = 1 units/day
Then, Total work will be completed in 30/2 = 15 days

Let the sides of the triangle be x,x,y.
Hence we have 2x + y = 43 and 2x > y ( sum of two sides > third side)
We will solve this by enumerating the values for x,y.
Since 2x = 43 - y, y can only take odd values for 2x to be even.
y = 1 => x = 21
y = 3 => x = 20
....
y = 19 => x = 12
y = 21 => x = 11
y = 23 => x = 10, but this violates the condition 2x > y.
Hence y can take odd values from 1 to 21. i.e. 11 values in total.
Hence 11 isosceles triangles can be formed.

Let the length of the race track be ‘d’ metres and speed of Raghu, Raghav and Raja be x, y and z respectively.

We have, $$\frac{x}{y} = \frac{d}{d-200}$$ and $$\frac{x}{z} = \frac{d}{d-350}$$

So, $$ \frac{y}{z} = \frac{d-200}{d-350}$$

Further, $$ \frac{y}{z} = \frac{d}{d-180}$$

So we have, $$ \frac{d-200}{d-350}= \frac{d}{d-180}$$

$$d*(d-350) = (d-200)*(d-180)$$

$$d^2 - 350d = d^2 - 180d - 200d + 200*180$$

30d = 200*180

d = 1200m

Thus the length of the track is 1200 m.

Drop 2 perpendiculars CE and CF to AB

BE = EF = CF = 4

Consider the triangle CEB. It is a right angled triangle with hypotenuse 8 and base equal to 4.
Hence, the length of the side CE is $$\sqrt{8^2 - 4^2} = \sqrt{48} = 4\sqrt{3}$$

In order to calculate the length of AC, consider the triangle AEC, the base equals 4+4 = 8 and the height is $$4\sqrt{3}$$.
Hence, the length of the hypotenuse AC equals $$\sqrt{48 + 64} = \sqrt{112} = 4\sqrt{7}$$

Let us say Captain America meets Batman after t seconds he starts from the starting point. Speeds of Captain America and batman are 5m/s and 2.5 m/s respectively.

t = $$\frac{1000}{5-2.5}$$ = 400 seconds

Distance travelled by Captain America when meets Batman for the first time = 2000m i.e., after 2 complete revolutions. So by the time he completes 10 revolutions, he will meet Batman 5 times.

So time for $$5^{th}$$ meet = 400 * 5 = 2000 seconds.

The fish is swimming along the sides of a rectangle formed by chords of the circle. Since the lengths of the sides are 300m and 400m, the length of the diagonal is 500m. The diagonal is the diameter of the circle. Therefore, the radius of the circle is 250m and the area is 62500$$\pi$$.

Let the cost price and selling price of Sita’s scarf be $$C_1$$ and $$S_1$$.

Similarly, Let the cost price and selling price of Gita’s scarf be $$C_2$$ and $$S_2$$.

Given,

$$\ \frac{\ S_1-C_1}{C_1}=\ \frac{\ S_2-C_2}{S_2}$$

$$ \frac{S_1}{C_1} + \frac{C_2}{S_2} = 2$$

Given,

$$C_2 = \frac {4C_1}{5}$$ and $$S_2 = \frac{5S_1}{4}$$

$$\frac{S_1}{C_1} + \frac{16C_1}{25S_1} = 2$$

$$25S_1^2 - 50S_1C_1 + 16C_1^2 = 0$$

$$(5S_1 - 2C_1)(5S_1 - 8C_1) = 0$$

Selling price should be greater than cost price. Therefore,

$$S_1 = \frac{8C_1}{5}$$

$$C_2 = \frac{4C_1}{5}$$ and $$S_2 = \frac{5S_1}{4} = (\frac{5}{4})(\frac{8C_1}{5}) =2C_1$$

Required profit % = $$\frac{S_2-C_2}{C_2} = \frac{2C_1-\frac{4C_1}{5}}{\frac{4C_1}{5}} * 100$$ = 150%

Answer is option A.

As can be seen from the pic, triangle AOC is equal to triangle ABC and so on. Hence, area of hexagon outside triangle = area of triangle AEC. Hence the area of the triangle is half the area of the hexagon.

Let the original price of the sweater be 100.

After a successive discount of 50 and 30, the price of the sweater will be 130(30 percent marked up price)

This is also the cost price for Kartik.

Selling price for Kartik = 110(10 percent marked on the price)

Thus, loss percent = $$\frac{20}{130}\times\ 100=15.4\%$$

Thus, the correct option is C.

$$x^2,y^2,z^2$$ are in AP
Add xy+yz+zx in each term
$$x^2+xy+yz+zx,y^2+xy+yz+zx,z^2+xy+yz+zx$$ are in AP
(x+z)(x+y),(y+z)(x+y),(z+x)(z+y) are in AP
Divide (x+y)(y+z)(z+x) from each term
1/(y+z) , 1/(x+z), 1/(x+y) are in AP
y+z,x+z,x+y are in HP

Let the total no. of months for which the account was held be n months.

Now,

The principle($$P_1$$) deposited in the first month is held in the account for a duration of $$\frac{n}{12}$$ years.

The principle($$P_2$$) deposited in the second month is held in the account for a duration of $$\frac{n-1}{12}$$ years.

The principle($$P_3$$) deposited in the third month is held in the account for a duration of $$\frac{n-2}{12}$$ years.

Similarly, we can say that the principle($$P_n$$) deposited in the nth month is held in the account for a duration of $$\frac{1}{12}$$ years.

Therefore, the total interest earned will be equal to

$$\frac{P_1\times\ r\times\ n}{1200}+\frac{P_2\times\ r\times\left(n-1\right)}{1200}+\frac{P_3\times\ r\times\left(n-2\right)}{1200}+.....+\frac{P_n\times\ r\times\ 1}{1200}$$

Since the principle deposited every month is equal so $$P_1=P_2=P_3=....=P_n=600$$

So, Total Interest earned becomes

$$\frac{600\times\ r}{1200}\left\{\left(\left(n\right)+\left(n-1\right)+\left(n-2\right)+.....+1\right)\right\}$$

$$=\frac{600\times\ 12}{1200}\left\{\frac{n\left(n+1\right)}{2}\right\}=\frac{12n^2+12n}{4}$$

Therefore, total maturity value = $$\left(600\times\ n\right)+\left(\frac{12n^2+12n}{4}\right)=16200$$

or, $$n^2+201n-5400=0$$

or, $$\left(n+225\right)\left(n-24\right)=0$$

As no. of months cannot be negative so n=24 months

Let $$\log _X 5 = P, \log _X 7 = Q$$ and $$\log _X 11 = R$$.

So, $$1/A = P+Q; 1/B = P+R; 1/C = Q+R$$

$$\log _{385} X = 1/ (\log _X 385)=1/(P+Q+R)$$

$$P+Q+R = (AB+BC+AC)/(2ABC)$$

So, $$\log _{385} X = 2ABC/(AB+BC+AC)$$

 In alloy P, if the total weight of P is 11 gm, then the weight of gold is 4gm, and the weight of silver is 7 gm. In alloy Q, if the total weight of Q is 12 gm, then the weight of gold is 7 gm, and the weight of silver is 5 gm.
L.C.M of (11,12) = 132
Therefore, if the weight of the alloy P is 132 gm => (48 gm gold and 84 gm silver). If the weight of the alloy Q is 132 gm => (77 gm gold and 55 gm silver).
It is given that quantities of alloys from P and Q melted are in the ratio of 4:3, which implies If the quantity of alloy P that is melted = (4*132) gm, then the quantity of alloy Q that is melted =(3*132) gm.

In (4*132) gm of alloy P, the quantity of gold is (4*48) = 192 gm, and the quantity of silver is (4*84) = 336 gm.
In (3*132) gm of alloy Q, the quantity of gold is (77*3) = 231 gm, and the quantity of silver is (55*3) = 165 gm.
Hence, the total quantity of silver in alloy R is (336+165) = 501 gm, and the total quantity of gold in alloy R is (192+231) = 423 gm.

Hence, the ratio of silver to gold in alloy R is 501:423 = 167:141
The correct option is D

As A is positive, $$A^{1/7} > A^{1/11}$$ implies that $$A^4>1$$ or $$A>1$$ So, $$A^2 < A^3$$

Let ‘t’ be the price of 1 banana and 3 apples. Let us assume that the price of 1 mango is Rs. m.
For Krishna, 10m + t = 212
For Ravi, 6m + 3t = 228
Solving for m and t, m = 17, t = 42
Hence the amount paid by Ramesh = 9m + 5t = 9*17 + 5*42 = Rs. 363
Therefore, option B is the correct answer.

y = $$\frac{1}{4x+\frac{9}{x}+15}$$

Since x>0 , y will be max when demoninator is minimum.

$$4x+\frac{9}{x}$$ is minimum when $$4x=\frac{9}{x}=>x=\frac{3}{2}$$

$$\therefore\ y_{\max}=\frac{1}{6+6+15}=\frac{1}{27}$$