Quantitative Aptitude (QA) Test - 12

Let ‘R’ be the rate of return on a y-o-y basis. It is given that his investment doubled in 2 years. Also, we have to calculate the rate of return semi-annually compounding.
Therefore, we can say that
$$\Rightarrow$$ $$2P = P(1 + \dfrac{R/2}{100})^{2*2}$$
$$\Rightarrow$$ $$2^{1/4}$$ = 1+$$\dfrac{R/2}{100}$$
$$\Rightarrow$$ $$1.1892$$ = 1+$$\dfrac{R/2}{100}$$
$$\Rightarrow$$ $$18.92$$ = $$R/2$$
$$\Rightarrow$$ $$R$$ = 37.84%

A(L)= probability of selecting a leap year. 

A(NL) = probability of selecting a non - leap year. 

A(M) = probability of getting 53 Mondays in a year.

the probability that it contains 53 Mondays = 2/7*1/4 + 1/7*3/4 = 5/28

Consider WXYZ. It is given that the rhombus side WX is given by the equation X+Y =3. Hence, the length of this side is 3√2 units as can be seen from the figure. Hence, we can straightaway reject option A where the length of that side would be 5√2 units. 

If we reflect X+Y=3 around the origin, we get X+Y=-3. This can be the equation of YZ. If we reflect X+Y=3 in the x-axis, we get X-Y=3 which is the equation of XY. If we reflect X+Y=3 in the y-axis, we get -X+Y=3 which is the equation of WZ. Hence, the right answer is option B X+Y=-3.

$$a_{k+1}=a_k^2+a_k$$
$$\dfrac{1}{a_{k+1}}=\dfrac{1}{a_k(a_k+1)}=\dfrac{1}{a_k}-\dfrac{1}{{a_k+1}}$$
$$\dfrac{1}{a_k+1}=\dfrac{1}{a_k}-\dfrac{1}{a_{k+1}}$$
$$\dfrac{1}{a_1+1}+\dfrac{1}{a_2+1}+....=\dfrac{1}{a_1}-\dfrac{1}{a_{2}}+\dfrac{1}{a_2}-\dfrac{1}{a_3}+....$$
= $$\dfrac{1}{a_1}=1/0.5 = 2$$

Let us assume that Ram's initial investment = Rs. 100x. This amount will grow by 50% in three years, therefore the total amount that Ram will get after 3 years = 1.5*100x = Rs. 150x.
It is given that his investment grew by 25%, 30% in the first year and second year respectively. Therefore, we can say that the total amount that Ram will get after 2 years = 100x*1.25*1.3 = Rs. 162.5x
Hence, we can say that return on investment in the third year = $$\frac{150x - 162.50x}{162.50}\times 100 = -7.69$$ percent.

We know that arithmetic mean of two numbers is greater than the geometric mean.
Let the two number be $$ab$$ and $$xy$$

Therefore, $$\frac{ab+xy}{2}>(ab \times xy)^{1/2}$$
$$ax+by>2(axby)^{1/2}$$
Multiply both equations
$$\frac{(ab+xy)(ax+by)}{abxy}$$>4
Hence only b option is less than 4

When x < -5, -x - 16 = -x-5
=> Has no solution
When -5<x<0

2x = -21

=> has no solution within the given range

When x>0, x- 16 = x + 5
=> Has no solution

Assuming the time taken is t if travels with 50 km/hr.
Hence, distance = 50t

Also, with 60 km/hr, distance = 60(t-1)
Both distances will be equal.
Hence, 50t =60t-60

=> t=6 hrs,

So the distance = 50*6 =300km
Hence, required speed = $$\ \frac{\ 300}{6-2}$$ = 75 km/h

Let the Cost Price of the item be Rs. 100. To get a profit of 20% he must have sold it at Rs. 120. But this is after giving a discount of 20%.
It means that the selling price is 20% lower than marked Price of the item. So the marked price is Rs. 150.
Now if he gives a discount of 30% he will sell the item at Rs. 105.
He earns a profit of Rs. 5 and the profit % is 5%.

If the sides of the triangle are in GP, let these sides have a common ratio. The sides can be represented as: $$a,\ ar,\ ar^2$$ where $$r\ge\ 1$$

The sum of two sides of a triangle must be greater than the third: (a+b) > c 

$$\therefore\ \left(a+ar\right)>\ ar^2$$

$$\ \left(1+r\right)>\ r^2\ ;\ \ r^2-r-1<0\ $$

$$\therefore\ r\ =\frac{1\pm\sqrt{\ 5}}{2}$$

Hence, $$r\in\ \left(\frac{\ 1-\sqrt{\ 5}}{2},\frac{\ 1+\sqrt{\ 5}}{2}\right)$$

Since $$r\ge\ 1$$, the final interval will be $$r\in\ \left(1,\ \frac{1+\sqrt{\ 5}}{2}\right)$$

Of the given choices, only Option C falls outside this range.

$$\simeq\ 67.5$$Let the strengths of P, Q, R be p,q,r respectively.

62p+71q=66(p+q) and 71q+79r=74(q+r)

=> 5q=4p and 5r=3q

=> p:q:r = 25:20:12

Average marks of P and R = $$\frac{\left(62p+79r\right)}{p+r}$$

=$$\left(62\cdot\frac{25}{37}+79\cdot\frac{12}{37}\right)$$ 

$$\simeq\ 67.5$$

SP of 1kg of mixture =Rs.100

Profit = 25%

Therefore, CP of 1 kg of mixture = 100/1.25 = Rs.80

Let a, b and c kg of $$W_1,W_2\ and\ W_3$$ respectively are mixed together

Total weight of mixture = (a+b+c) kg

Therefore, we can say that

20a + 50b + 100c = 80(a+b+c)

or, 2a +5b +10c = 8a + 8b + 8c

or, 6a + 3b = 2c

If we put a=3, b=2 and c=12 then it satisfies the above equation.

Therefore we can say that a:b:c = 3:2:12

Let Ravi's speed and Ramu's speed be 4x and x respectively => Speed of escalator = 8x - 5x = 3x.
Ravi's effective speed upwards = 4x - 3x = x
Ramu's effective speed downards = x + 3x = 4x

Hence, the distance covered by them when they meet will be in the ratio of their speeds i.e. 1:4

Let Ravi cover a steps, hence Ramu will have covered 4a steps. Total steps = a + 4a = 250. Hence, a = 50

Hence, the two meet at the 50th step from bottom or the 250 - 50 =200th step from the top.

Let’s assume the total amount of work to be LCM (8, 12,18) = 72 units.

The units of work done by ‘A’ in a day = $$\frac{72}{8}$$ = 9 units   

Similarly, the units of work done by ‘B’ in a day = $$\frac{72}{12}$$ = 6 units  

Similarly, the units of work done by ‘C’ in a day = $$\frac{72}{18}$$ = 4 units

So, working together, they all can complete = 9 + 6 + 4 = 19 units of work in a day.

Hence, in three days, they will complete = 19 * 3 = 57 units of work

The amount of work that will be completed by ‘C’ alone = 72 - 57 = 15 units

So, ‘C’ will take = $$\frac{15}{4}$$ = $$3\frac{3}{4}$$ days to complete the remaining work.
Therefore, option A is the right answer.

A -> 35

B + C -> 15

D + F -> 10

D + E + F -> 5

Let the work be equivalent to LCM(35,15,10,5) = 210 units.

So, in one day,

A -> 6

B + C -> 14

D + F -> 21

D + E + F -> 42

Therefore E -> 21.

Hence, on day 1, A, B and C together does 20 units.

On day 2, D and F together does 21 units.

On day 3, E does 21 units.

Hence, in 3 days, units completed = 20 + 21 + 21 = 62.

Hence, in 9 days, units completed = 62 x 3 = 186.

On the 10th day, units completed = 20.

So total remaining units = 210 - 206 = 4

These 4 units will be completed on the 11th day.

f(5) = -4*f(2) 25a + 5b + c = -4(4a + 2b + c) => 41a + 13b +5c =0

f(7) =0 => 49a + 7b + c =0

Two equations, three unknowns without any special condition. Hence the answer cannot be determined.

The tangents drawn to a circle from a point are of equal lengths.

Hence EC = EF = 4 cm.

AD = AF = 6 cm.

Let DB = BE = kcm.

The lengths of the sides of the triangle are AB = 6+k.

BC = 4+k cm.

AC = 10 cm.

Applying cosine rule for angle ACB

$$Cos\alpha\ \ =\ \frac{\left(b^2+c^2-a^2\right)}{2\cdot b\cdot c}$$.

Now looking at the sides b =k+4, c = 10cm, and a = k+6 cm.

Cos(60) = $$\frac{\left(\left(\left(k+4\right)^2\right)+100-\left(k+6\right)^2\right)}{2\cdot\left(k+4\right)\left(10\right)}$$

Solving this we get

10k + 40 = $$k^2+8k+16+100-\left(k^2+12k+36\right)$$

k = 20/7 cm.

Now the sides are 4+k = 48/7 cm, 10 cm, 6+k = 62/7 cm.

The circum radius of the triangle is given by 

R = abc/4(Area of the triangle)

R = $$\frac{abc}{4\cdot\left(\frac{1}{2}a\cdot b\cdot SinC\right)}$$

= c/2Sin(C)

c = 62/7 cm, Sin(C) = $$\frac{\sqrt{\ 3}}{2}$$

$$\frac{62}{7\sqrt{\ 3}}\ =\ \frac{62\sqrt{\ 3}}{21}$$ cm

Let the total annual profit be Rs.$$‘x’$$

The ratio of annual profit shared refects $$\left(investment\times\ time\right)$$

hence, the annual profit obtained by Akshat $$(60,000\times\ 12 months)$$ : Nikhil $$(40,000\times\ 9 months)$$ = $$2:1$$

We know that Nikhil’s share includes a salary element equal to 19% of the annual profit  = $$\frac{19x}{100}$$

Also, he get $$\left(\frac{1}{3}\right)^{rd}$$ of the remaining $$\frac{81x}{100}$$

$$\therefore\ \frac{1}{3}\left(\frac{81x}{100}\right)+\left(\frac{19x}{100}\right)=27,600$$

On solving, we get  $$x = 60,000$$

$$\therefore\ $$ Akshat’s share = $$60,000 - 27,600 = 32,400$$

Hence, the correct answer is Option D.

Going by the formula the total number of triangles if the perimeter is give by :

N is the perimeter of the triangle.

Since N is even

{$$N^2$$/48} if the expressed value is not an integer then then the value is rounded off to the closest integer

= 72*72/48 = 108.

Number of equilateral triangles is 1with sides 24 cm , 24 cm, 24 cm

Number of scalene triangles is  {$$\frac{\left(N-6\right)^2}{48}$$}. Here the value is rounded off to the closest integer.

         $$\frac{\left(66\right)^2}{48}$$ = 90.75 = 91.

108-91 = 17.

17 triangles include isosceles and equilateral of which one is equilateral and hence 16 isosceles triangles.

a = 1, b = 16, c = 91

3a+2b+c =3(1)+2(16)+91 = 126.

The total number of such five digit numbers is $$4^5=1024$$ and the number of five digit numbers formed using (2,3 and 4 only) is $$3^5=243$$.
The number of five digit numbers formed using exactly one 1 is $$5*3^4=405$$.
So, the answer is $$1024-243-405=376$$

Differentiating area function and putting it equal to zero gives minima value at x=1.
at x=1; area= $$ 6  cm^2 $$
Volume of pyramid= (area of base * height)/3
Putting values;
Volume = 6*2.5/3
=$$ 5  cm^3 $$

Alternate explanation:

$$ 6x^2-12x+12 $$ =6($$x^2-2x+2$$) =6($$(x-1)^2 + 1$$)

The minimum possible value of ($$(x-1)^2 + 1$$) =1.

The minimum value of area of base =6.

Volume of pyramid= (area of base * height)/3 = 6*2.5/3 = 5 $$cm^3 $$

Number of trailing zeroes in 1000 factorial would be
[1000/5] + [1000/25] + [1000/125] + [1000/625]
=> 200 + 40 + 8 + 1 = 249