Quantitative Aptitude (QA) Test - 14

Now we have a sequence of numbers with 1st term as 2
Now average of sequence with number 2 =2
Now as per given condition let x be the 2nd term of the sequence
Therefore
$$\frac{\left(2+x\right)}{2}=3$$
x=4
so second term will be 4
Now $$\frac{\left(2+4+y\right)}{3}=4$$
y=6
so we can say the sequence is as follows:
2,4,6,8,10,...2n
Now sum of first 20 terms will be
2(1+2+3+4+5+....+20)
=2(20)(21)/2
=(20)(21)
=420
Therefore option C is the correct answer.

If n+1 is a prime number, n! is not divisible by (n+1) by Wilson's theorem.
Number of prime numbers less than 50 is 15.
The only composite number for which n! is not divisible by (n + 1) is if n+1 = 4.
So, total is 15+1=16
Hence, option D is the correct answer.

If no machine was faulty the sum of weights of “1 pill from machine 1, 2 pills from machine 2, … and 10 pills from machine 10” would be 5*(1+2+…+10) = 275.
But the weight is 6 grams less. That is, machine 3 produces 3 gram pills and hence 6 grams is less in weight.

If there are 2 red, 2 green and 2 orange balls, it implies that two red and one green ball are picked from the first jar and two orange and one green ball are picked from the second jar.
The probability of picking 2 red and 1 green ball from the first jar is $$\frac{^3C_2\times5}{^8C_3} = \frac{15}{56}$$
The probability of picking 2 orange and 1 green ball from the second jar is $$\frac{^2C_2\times4}{^6C_3} = \frac{1}{5}$$
Hence, the required probability is $$\frac{15}{56} \times \frac{1}{5} = \frac{3}{56}$$

Based on the given info, $$M\ \alpha\ \ \left(A\right)\times\ \left(D\right)\times\ \left(S\right)$$
We can remove the proportinality: $$M\ \ =\ k\times\ \left(A\right)\times\ \left(D\right)\times\ \left(S\right)$$ where k is a constant
It is given that 230 = k (0.5)(1.5)(20); therefore, $$k\ =\ \frac{230}{15}$$
For finding the value of M in the new system:
M = (230/15)(0.75)(2)(30)
M = 690
Hence, Option (D) is the correct choice

The path can be written as a string of U - up and R - right commands. Hence to get from one end to the other, Suresh has to walk up 7 times and right 9 times. Thus the string is of 7 Us and 9 Rs. This can be arranged in $$\frac{16!}{7!9!} = ^{16} C _7 $$ ways.

Let W be the amount of work that needs to be completed. Let the efficiencies of M and N be 'm' and 'n'. Then, based on the information provided in the question, we obtain: 

$$\frac{W}{m+n}+p=\frac{W}{m}$$ ; $$p=\left(\frac{W}{m}-\frac{W}{m+n}\right)\ ;\ p\ =\frac{W\left(n\right)}{m\left(m+n\right)}$$ .....(1)

Here, $$\left(\frac{W}{m+n}\right)$$ is the time taken to complete the work when M and N work together, while $$\left(\frac{W}{m}\right)$$ is the time taken to complete the work when M works alone. 

Similarly, $$\frac{W}{m+n}+q=\frac{W}{n}$$ ; $$q=\left(\frac{W}{n}-\frac{W}{m+n}\right)\ ;\ q\ =\frac{W\left(m\right)}{m\left(m+n\right)}$$ ......(2)

Here, $$\left(\frac{W}{m+n}\right)$$ is the time taken to complete the work when M and N work together, while $$\left(\frac{W}{n}\right)$$ is the time taken to complete the work when N works alone. 

We are tasked with finding $$\left(\frac{m}{n}\right)$$; hence, let us re-write the above equations in terms of this value. We can re-write (1) as: $$\frac{n}{m}=\frac{p\left(m+n\right)}{W}$$......(3)

Similarly, we can re-write (2) as: $$\frac{m}{n}=\frac{q\left(m+n\right)}{W}$$......(4)

Dividing (3) by (4), we obtain: $$\frac{n^2}{m^2}=\frac{p}{q}\ \ OR\ \frac{m}{n}=\sqrt{\frac{p}{q}}$$

Hence, the required ratio is $$\sqrt{q}:\sqrt{p}$$. Option B is the correct choice. 

Total number of points on the square = 2 + 3 + 4 + 5 + 4 = 18 (including points A, B, C and D)
Of these 18 points, 5 points are collinear on side AB, 6 points are collinear on side BC, 7 points are collinear on side CD and 4 points are collinear on side DA.
So, the number of triangles that can be formed is $$^{18}C_3 - ^5C_3 - ^6C_3 - ^7C_3 - 4C_3$$ = 816 - 10 - 20 - 35 - 4 = 747
So, a total of 747 triangles can be formed from these points.

Let the upstream distance be d.

Let the speed of the stream be x. Hence, speed of the boat in still water = 2x

s = $$\frac{d+1.5d}{\frac{d}{2x-x}+\frac{1.5d}{2x+x}}=\frac{2.5d}{\frac{d}{x}+\frac{d}{2x}}=\frac{2.5d}{\frac{3d}{2x}}=\frac{5dx}{3d}=\frac{5x}{3}$$

Hence, ratio = 5/3

XY = 13 cm = XV+VY = 5+VY Thus, VY = 8 cm
Thus, ZY = $$\sqrt{10^2-8^2}$$ = 6 cm
Thus, WV = $$\sqrt{5^2+6^2}$$ = $$\sqrt{61}$$ $$\approx 8 cm$$
Thus, WZ>ZV>WV.
Thus, b>a>c
Hence, option B is the correct answer

Let the speed of the person be S and the distance between point A and point B be d.
Usual time of travel = d/S
Time taken when the speed is 20 kmph = d/20 = d/S + 3/60
Time taken when the speed is 24 kmph = d/24 = d/S - 3/60
Solving the two equations, we get, d = 12 km and d/S = 11/20 hours = 33 minutes
So, usual time of travel of the person = 33 minutes

Length of the direct common tangent = $$\sqrt{d^2 - (r1-r2)^2}$$
Substituting r1, r2 values in the above formula we get distance between circles d = $$\sqrt{269}$$
==> Length of the transverse common tangent = $$\sqrt{d^2 - (r1+r2)^2}$$ = 13 cm

Let us say that the total number of apples are 18 and he bought them at Rupees 10 each. So total amount spent by him = 18*10 = 80.
Now he sells the first 6 apples for 15 each. Hence, the amount earned will be 15*6 = 90
The next 6 apples are sold at Rupees 13.5 each. Hence, the amount earned will be 13.5*6 = 81
The last six apples are sold at 12 rupees each. Hence, amount earned = 72
Thus, total amount earned = 90 + 81 + 72 = 243
Hence, profit is 63
Thus, profit percent = 63*100/180 = 630/18 = 35 %

Applying the angle bisector theorem, we can say that BD: DC = 6:8 = 3:4, and we know that BC = (BD+DC) = 7 cm.

Therefore, BD = n = 3 cm, and CD = m = 4cm. Let, the length of the angular bisector be x cm.

Now applying the cosine rule in triangle BAD, we can say that $$\cos A\ =\ \ \frac{\ b^2+c^2-a^2}{2bc}$$

In this case, for triangle BAD, a = 3 cm, c = 6 cm , and b = x cm

Hence, $$\cos A\ =\ \ \ \ \frac{\ x^2+36-9}{2\ \times\ x\times\ 6}=\ \frac{\ x^2+27}{12x}$$  ..... Eq(1)

Similarly, for triangle CAD, we can use the cosine rule, where a= 4 cm, c = 8 cm, and b = x cm

Therefore, $$\cos A\ =\ \ \ \ \frac{\ x^2+64-16}{2\ \times\ x\times\ 8}=\ \frac{\ x^2+48}{16x}$$  .... Eq(2)

Equating (1) and (2), we get:

$$\ \frac{\ x^2+48}{16x}=\ \frac{\ x^2+27}{12x}\ =>\ \ \frac{\ x^2+48}{4}=\ \frac{\ x^2+27}{3}$$

=> $$=>\ 4x^2+108=\ \ 3x^2+144$$

=> $$x^2=36\ =>x\ =\ 6$$ cm

The correct option is A

The ratio in which Abdul and Karim divides their profit if we subtract Abdul’s fixed share = 7 x 4 = 5 x 3 = 28 : 15
Let x be the total profit and y be Abdul’s share of profit for managing the business.
Then, the final profit amount paid to Abdul = 2/3 x = y + 28/43 (x - y)
On rearranging we get y = 2x/45
Thus, the required proportion = 2/45

Sum to n terms of the given geometric progression is $$\frac{4^n-1}{4-1}=\frac{4^n-1}{3}$$
As this is less than 8000, it means that $$4^n<24001$$
$$n=7$$

Let the money lent/deposited be denoted by P.

Money earned by the bank in 3 years = $$P(1 + \frac{r}{100})^3$$ - P

Let $$\frac{r}{10}$$ = R

Money earned = $$P[1 + R^3 + 3R^2 + 3R] - P=PR^3 + 3PR^2 + 3PR$$

Money Spent = $$\frac{3\times P\times r}{100}$$ = $$3PR$$

Money earned should be 58.33% more than money spent

$$PR^3 + 3PR^2 + 3PR = 1.5833\times 3PR$$

$$PR[R^2 + 3R] = 0.5833 \times 3PR$$

$$R^2 + 3R = 1.75$$

$$4R^2 + 12R - 7 = 0$$

Solving this we get,

$$R = 0.5 \ or -3.5$$

Since, the rate is positive

$$R = 0.5$$

Therefore, $$r = 50%$$

The answer is option C.

Since $$(x +2)$$ is a factor of $$2x^2 + ax - c$$, therefore, at $$x = -2 $$
$$\Rightarrow$$ $$2(-2)^2 - 2a - c = 0$$
$$\Rightarrow$$ $$a = 4 - \dfrac{c}{2}$$
So, $$a \times c = 4c - \dfrac{c^2}{2}$$
It is given that $$a \times c$$ is a natural number and $$4c$$ is an integer, so $$\dfrac{c^2}{2}$$ is an integer or $$c$$ is an integer.
Since it is given that $$4c$$ is divisible by 6, $$c$$ must be a multiple of 3.
Also, $$a \times c$$ is a natural number, $$'a'$$ must be a natural number.
For $$a$$ being a natural number, $$c$$ must be 6.
$$\therefore$$ $$a = 4 - 3 = 1$$
Hence, $$(a + c)$$ = 1 + 6 = 7.

Let us take 27 as base to reduce the calculations.
Average weight of seven experiments = 0.345
Average weight of first three experiments = 0.085
Average weight of last two experiments = 0.285
Let the values in fourth and fifth experiments be 'a' and 'b'
It is given, a = b + 0.024
7(0.345) = 3(0.085) + 2(0.285) + a + b
a + b = 1.59
2a - 0.024 = 1.59
2a = 1.614
a = 0.807
Therefore, weight calculated in fourth experiment = 27 + 0.807 = 27.807
The answer is option B.

$$a^2(b+c)+b^2(a+c)+c^2(a+b)-6abc = abc(a/c+a/b+b/c+b/a+c/b+c/a) - 6abc$$

For any positive number x, $$x+(\frac{1}{x}) \geq 2$$  (This expression can be derived from AM $$\geq$$ GM)

$$\frac{\frac{a}{b}+\frac{b}{a}}{2} \geq {(\frac{a}{b}\times\frac{b}{a})}^{\frac{1}{2}}$$     AM $$\geq$$ GM

$$\frac{a}{b}+\frac{b}{a} \geq 2 $$                     $$\frac{b}{c}+\frac{c}{b} \geq 2 $$                                  $$\frac{a}{c}+\frac{c}{a} \geq 2 $$ 

Hence, $$(a/c+a/b+b/c+b/a+c/b+c/a) \geq 6$$ and the given expression is always greater than or equal to 0

Let the digits of x be a b
=> x= 10a+b. no. of bottles sold in record=08.
Hence the diff in money collected and recorded = 80(10a+b)-8(10b+a)=792a = 3168.
Hence a=4.
Hence the min price of the bottles = 40.
Hence min amount collected = 40*80=3200

$$x = \frac{1}{5\sqrt{2} - 7} = \frac{5\sqrt{2} + 7}{(5\sqrt{2} - 7)(5\sqrt{2} + 7)}$$
$$x = 5\sqrt{2} + 7$$
$$x-7 = 5\sqrt{2}$$
Cubing both sides
$$(x - 7)^{3} = (5\sqrt{2})^{3}$$
$$x^{3} - 343 - 21x^{2} + 147x = 250\sqrt{2} $$
$$x^{3} - 343 - 21x^{2} + 147x - 3x^2 - 8x = 250\sqrt{2} - 3x^{2} - 8x $$
$$x^{3} - 24x^{2} + 139x + 11 = 354 + 250\sqrt{2} - 3x^{2} - 8x = 1 (3x^{2} + 8x = 250\sqrt{2} + 353)$$