Quantitative Aptitude (QA) Test - 7

$$\frac{2 + \sqrt{2}}{ 5 + \sqrt{5}}$$ = $$\frac{10.23}{x}$$

The required consequent is $$ 10.23 * \frac{5 + \sqrt{5}}{ 2 + \sqrt{2}} =21.68$$

Number of chocolates left of each type is same => remainder of each type of chocolates when divided by number of children is same.
Let x be the remainder
=> Number of chocolates distributed of each type are 2627-x, 3752-x, 4877-x and 6377-x respectively.

Since all the 3 numbers leave the same remainder, the difference between them must be divisible by the divisor.

The maximum number of children to whom the chocolates can be distributed = HCF(3752-2627, 4877-3752 and 6377-4877).
HCF of the difference is equal to 375
Hence the maximum number of children to whom the chocolates can be distribute is 375.

Let the cost of a english 3D ticket be ₹$$100x$$ 

Thus, cost of an english ticket = $$\dfrac{60}{100} \times 100x$$ = ₹60x 

Cost of a hindi ticket = $$(1 - \dfrac{20}{100}) \times 60x $$ = ₹48x 

Now, total number of english tickets = $$\dfrac{3}{3+5+2} \times 300 = 90$$ tickets 

Thus, total value of the english tickets = $$90 \times 60x $$ = ₹5400x 

Now let us look at the options : 

Option A : 5400x = 6300 => $$x = \dfrac{7}{6} $$. 

In this case, cost of english 3D tickets = $$100 \times \dfrac{7}{6}$$ which is not an integer. 

Option B : 5400x = 9000 => $$x = \dfrac{5}{3} $$ 

In this case, cost of english 3D tickets = $$100 \times \dfrac{5}{3}$$ which is not an integer. 

Option C : 5400x = 1350 => $$x = \dfrac{1}{4} $$ 

In this case, cost of english 3D tickets = $$100 \times \dfrac{1}{4}$$ which is an integer. 

In this case, cost of hindi tickets = $$48 \times \dfrac{1}{4}$$ which is an integer. 

Option D : 5400x = 4500 => $$x = \dfrac{5}{6} $$ 

In this case, cost of english 3D tickets = $$100 \times \dfrac{5}{6}$$ which is not an integer.

Thus option C is the only possible option 

We have to find the probability that at least two of them fit in the correct slot.

Required probability = 1 - probability that none of them fit in the right slot - probability that exactly one fits in the right slot

The number of ways that none of them fit in the right slot = derangement of 5 

The number of ways that exactly one fits in the right slot = 5 * derangement of 4

Derangement of $$n = n! * [1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + (-1)^n \frac{1}{n!}]$$

Total number of ways of arranging = 5! = 120

The number of ways that none of them fit in the right slot = 44

The number of ways that exactly one fits in the right slot = 5 *9 = 45

Required probability = $$1 - \frac{44}{120} - \frac{45}{120}$$ = $$\frac{31}{120}$$

Let the total units of work in the 1st piece and 2nd piece be A and B, respectively.

In 4 days, total work done in 1st piece = 4( work done by A and C) = $$4*\left(\dfrac{A}{x}+\dfrac{3A}{x}\right)=\dfrac{16A}{x}$$

Total work done in 2nd piece in 4 days = 4( work done by B and C) = $$4*\left(\dfrac{B}{y}\ +\dfrac{4B}{y}\right)=\dfrac{20B}{y}$$

Percentage of work remaining is same => $$\dfrac{A - \dfrac{16A}{x}}{A} = \dfrac{B - \dfrac{20B}{y}}{B}$$

$$1 - \dfrac{16A}{Ax}=1 - \dfrac{20B}{By}$$

$$1-\dfrac{16}{x}=1-\dfrac{20}{y}$$

$$\dfrac{16}{x}=\dfrac{20}{y}$$

$$y=\dfrac{5}{4}x=\dfrac{125}{100}x$$.

Therefore, y is 125% of x.

Consider the total contributed by 7 in the unit's place.
The number of numbers with 7 in the unit's place is 6.
Hence, the sum contributed by in the unit's place is 7*6
Similarly, the sum contributed by 7 in the ten's place is 70*6
Hence, total contributed by 7 is 7000*6+700*6+70*6+7*6=7*1111*6
Similarly for other numbers.
So, the total sum is 1111*6*(7+5+3+2) = 1111*6*17=113322

Let A and B take a and b days respectively to complete the project alone.
$$\frac{1}{a} = \frac{1}{b} + \frac{1}{60}$$
Also, $$\frac{1}{a} + \frac{1}{b} = \frac{1}{20}$$
We have, $$\frac{1}{20} - \frac{1}{b} = \frac{1}{b} + \frac{1}{60}$$
$$\frac{1}{30} = \frac{2}{b}$$
b = 60 days
Hence, option B is the right choice.

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.

Distance travelled by the ship travelling north from the point (till 6 PM on Tuesday) = 29 hours * 150 nautical miles per hour = 4350 nautical miles
Distance travelled by the ship travelling west from the point (till 6 PM on Tuesday) = 27 hours * 100 nautical miles per hour = 2700 nautical miles
So, distance between the two ships at 6 PM on Tuesday = $$\sqrt{4350^2 + 2700^2}$$ = 5100 nautical miles approximately

Given that the points C and Q are vertically above the points E and S respectively.
So, $$\angle CES = \angle QSE = 90^{\circ}$$
Since, the pentagon is regular, $$\angle CDE$$ = $$\frac{2 * 5 - 4}{5} * 90 = 108^{\circ}$$.
$$\angle DUP = \angle EUS$$ = $$180^{\circ} - (\angle USE + \angle UES)$$ = $$180^{\circ} - (45^{\circ} + 90^{\circ} - \angle CED)$$ = $$81^{\circ}$$
=> $$\angle CTP$$ = $$180^{\circ} - \angle PTD$$ = $$180^{\circ} - (360^{\circ} - (\angle TPU + \angle TDU + \angle PUD))$$ = $$99^{\circ}$$

Let the number of steps on a stationary escalator be L steps.
Ram took 25 steps to reach the top. So, the escalator took the other L-25 steps.
Shyam took 20 steps to reach the top. So, the escalator took the other L-20 steps.
Time taken by Ram to climb 25 steps = t
So, Ram’s speed = 25/t
Shyam’s speed = (2/3)*(25/t) = 50/3t
So, time taken by Shyam = 20*3t/50 = 6t/5

Since the speed of the escalator is same in both the cases, (L-25)/t = (L-20)*5/6t

=> 6*(L-25) = 5*(L-20) => 6L - 150 = 5L - 100
=> L = 50

So, there are 50 steps on the escalator

The equation of a circle with a centre at the origin is 

$$x^2+y^2=r^2$$, where r is the radius of the circle.

If we shift the center to (a,b), the equation changes to

$$\left(x-a\right)^2+\left(y-b\right)^2=r^2$$

$$x^2+y^2-2ax-2by+\left(a^2+b^2-r^2\right)=0$$

Hence comparing it with the given equation, we find a = 7, b = 4.5.

Hence,

$$7^2+4.5^2-r^2=-41$$

$$r^2=\ 49\ +\ 20.25+41\ =\ 110.25$$

r = 10.5

Eqn of S = 

$$\left(x-4\right)^2+\left(y+3\right)^2=10.5^2$$

For points to satisfy the condition that they lie inside S, they must satisfy the condition

$$\left(x-4\right)^2+\left(y+3\right)^2<10.5^2$$

The following points satisfy the equation,

(10,-11),(5,7),(14,-6),(14,0)

Hence, the correct answer is 4.

Let the price of a teapot be $$P = 100$$. We can trace the change in price over time as follows: 

After the first three days, the price change follows a set pattern: we can identify these as cyclic changes wherein, after every three days, the price drops below the lowest point observed in the previous cycle. For instance, at the end of cycle 1 (Wednesday), the price of the teapot was 89.10 - multiplying this by a factor of 0.891 [=$$\left(1.1\right)^1\left(0.9\right)^2$$] renders us with the value at the end of cycle 2 (Saturday) and so on. We notice that at the end of cycle 2 (Saturday), the teapot sells at a loss of 20% or more for the first time.

Hence, the correct answer is Option C.   

As XYZ is an isosceles triangle with XY = XZ the altitude to side YZ bisects YZ. Let XW be the altitude to side YZ.

YW = WZ = 15 cm.

Triangle XYW is a right triangle and using Pythagoras theorem we can find that XW = $$ 5\sqrt{7}$$

Area of the triangle XYZ = $$\frac{1}{2}\times XW\times YZ = \frac{1}{2}\times 5\sqrt{7}\times 30 = 75\sqrt{7}$$

Let the length of the altitude to the side XZ be ‘h’.
Area of the triangle XYZ = $$ \frac{1}{2}\times h\times XZ = 75\sqrt{7}$$
=> h = $$ 7.5\sqrt{7}$$.

Let the cost price of 6 apples be 66(LCM of 6 and 11).
Thus, the cost price of 1 apple and 1 mango will be 11 and 6, respectively.

Cost price of 11 apples = $$11\times\ 11=121$$
Thus, the Selling price of 9 mangoes will be 121
Thus, the selling price of 1 mango will be $$\frac{121}{9}$$, and for 12 mangoes, it will be $$\frac{484}{3}$$
Similarly, the selling price of 1 apple will be $$\frac{484}{21}$$

Now, the cost price of 3 mangoes and 14 apples will be $$3\times\ 6\ +\ 14\times\ 11=172$$
Selling price =$$3\times\ \frac{121}{9}\ +\ 14\times\ \frac{484}{21}=363$$

Thus, profit percentage = $$\frac{\left(363-172\right)}{172}\times\ 100\ \approx\ \ 111\%$$

Thus, the correct option is D.

Let the first term be a, the second term be b and the third term c.
First term = a
Second term = b = sum of its neighbors = a + c

=> Third term = c = b-a = second term + fourth term

=> Fourth term = (b -a) - b = -a
Fifth Term = -b
Sixth term = a-b
Seventh term = a
This cycle repeats
Sum of first six terms = 0
Sum of first 2000 terms = Sum of first 1998 terms + Sum of last two terms = a+b = 9

Option A:

Total amount obtained from compound interest = 10000$$(1.1)^3$$ = 13310

Total amount obtained from simple interest = 10000(1 + 0.45) = 14500

Total interest obtained = 13310 + 14500 - 20000 = Rs. 7810

Option B:

Total amount obtained = 20000$$(1.05)^6$$ = 26802

Interest obtained = 26802 - 20000 = Rs. 6802

Option C:

Total amount obtained from simple interest = 20000(1.2) = 24000

Total amount obtained from compound interest = 24000$$(1.1)^2$$ = 29040

Total interest obtained = 29040 - 20000 = Rs. 9040

Option D:

Amount obtained from simple interest = 15000(1.2) = 18000

Amount obtained from compound interest = 5000$$(1.1)^2$$ = 6050

Total amount obtained from the second simple interest scheme = [18000 + 6050](1.2) = 28860

Total interest obtained = 28860 - 20000 = Rs. 8860

Thus, the maximum return is in option C.

$$2^{\log_7 |a+9|} = \log_7 2401$$
=> $$2^{\log_7 |a+9|} = 4$$
=> $$\log_7 |a+9| = 2$$
=> |a+9| = 49
=> a+9 = 49 or a+9 = -49
=> a = 40 or a = -58
So, required sum = 40 - 58 = -18

The proportion of water in solution A is $$\ \frac{\ 3}{7}$$.

The proportion of water in solution B is $$\ \frac{\ a}{a+b}$$.

The proportion of water in the final solution is $$\ \frac{\ 1}{2}$$

Now, using alligations, $$\ \frac{\ \frac{\ a}{a+b}-\ \frac{1\ }{2}\ }{\ \ \frac{\ 1}{2}-\frac{\ 3}{7}}\ =\ \ \frac{\ 7}{3}$$

=> $$\ \ \frac{\ a}{a+b}=\ \frac{\ 2}{3}$$

=> $$\ \ \frac{\ a}{b}=\ \frac{\ 2}{1}$$

=> a+b = 2+1=3

For any two natural numbers 'n' and 'r', the below inequality is true
$$(1^r + 2^r + ... + n^r)^n > n^n * (n!)^r$$
This is because $$\frac{1^r + 2^r + ... + n^r}{n} > (1^r*2^r*...*n^r)^{\frac{1}{n}}$$ [AM $$\geq$$ GM]
From the above inequality, we can deduce that the first option is false and the third option is true.

Let the number of hundred rupee notes by h
number of original five hundred rupee notes by f
and number of original thousand rupee notes be t

100h + 500f + 1000t = 27200 ... (1)
After the black money, 100h + 1000f + 2000t = 53700 .... (2)

Multiplying (1) by 2, we get 200h+ 1000t + 2000t = 54400 ... (3)

Subtracting (2) from (3), we get 100 h = 700

64a+8b+c+12a+12b+12c=0 or 76a+20b+13c=0  ----- (1)
F(2)=0 or 4a+2b+c=0 ------- (2)

(1) - 13 * (2)    =>. 24a - 6b= 0 => b=4a

(1) - 10 * (2)   =>  36a + 3c = 0 => c = -12a

So, F(x) = $$a(x^{2} + 4x -12)$$.
Hence, the other root is -6