Quantitative Aptitude (QA) Test - 3

Let the number of people in the group be n
For the maximum possible age of A, 
24(n+1) = 25n + A
A = 24 - n
For minimum possible age of B,
26(n+1) = 25n + B
B = n + 26
Hence, average = (24 - n + n + 26)/2 = 25 years
Option (D)

Let the number of 50 p coin be n.
Hence value = n/2.
Number of 25 p coins = n/2.
Number of Rs 1 coins = n/8 - 14
Hence,

n = 352
Hence, required percentage = 176/250 x 100 = 70.40 %

b = 20% of a = a/5
c = 30% of b = 3b/10
So, 'a' must be a multiple of 10.
Let us assume that a = 10n, where n = 1, 2, 3 etc
when a = 10,

When a = 20, b = 4 and c = 6, this is not possible, since b and c should be co-prime
Similarly, when a = 30, b = 6 and c = 9, again they are not co-prime,
So, we can infer when a = 10n, b = 2n and c = 3n.
Hence, only n the case when n = 1, they satisfy the condition.
Hence, 1 fraction is possible only.

I + II + III + IV = 100%
I + 2 II + 3 III + 4 IV = 60 + 50 + 30 + 40 = 180%
Hence, the additional 80% should be distributed among II + 2 III + 3 IV.
It is mentioned that II = 0, III = 0
Hence, IV = 80/3%
Now, IV has to be natural number since it is the number of students.

Hence, n must be a multiple of 15.
But, we have not yet considered the people who only follow one newspaper.
People who only follow A = 
People who only follow B = 
People who only follow C =
People who only follow D = 
Hence, n must be divisible by 30 as well.
Hence, 30030 is the only possible alternative.

Let us start by calculating

We observe that when n is odd  then 
when n is even  then fⁿ (12) = 12

[g(12)] = 13

then f(x) has to be of form
this can be written as

Multiplying 1 and 2 we get

Assuming f(x)-1 = g(x) gives us 
As per the conditions 
Since f(x) is a polynomial function g(x) is a polynomial function.
g(x) can be satisfied only by ± xⁿ
Hence f(x) = 1 + xⁿ
f(-2) = 1 + (−2)ⁿ = 9 or + (−2)ⁿ = 8
Thus n=3 and function has to be 1 - x³
f(4) = 1 - 4³ = 1 - 64 = -63
Modulus value = 63

5{k}+2 = 2[k-1]

We know, 0 < {k} < 1
Least value of [k] which satisfies is [k] = 2 at which {k} = 0. Thus k ={k}+[k] = 2.0
When [k]=3,  Thus k ={k}+[k] = 
When [k]=4, {k} = Thus k ={k}+[k] = 
When [k]=5, {k} = Which is not possible
Sum of all possible value of k = 2+3.4+4.8 = 10.2 which is equal to p
2p - 1 = 2 x 10.2 - 1 = 20.4 - 1 = 19.4
[2p-1] = 19

f(x) = max(x² - 5x + 6,x²)
The difference between the 2 expressions is -5x + 6.
Whenever this part is greater than or equal to 0, f(x) is the first function, else, the second function.
-5x + 6 >= 0
x <= 6/5
Hence, for x <= 6/5, f(x) = x² - 5x + 6
For x > 6/5, f(x) = x²
Since they both are quadratic, they will have the same value at x = 6/5.
Also, at x < 6/5, f(x) is the left side of a U - shaped quadratic curve(part of the quadratic curve before the leftmost root)
At x >= 6/5, f(x) is the right side of a U - shaped quadratic curve (part of the quadratic curve after the rightmost root)
Hence, the minimum value of the function is at x = 6/5.
Value = x² = 36/25

Initial volume of 10 Spheres = 
Let N be the number of cones which can be made by  melting the sphere.
Volume of a cone = 
As per question = 

Initial curved surface area = 10×4πR² = 40πR² ~ 125.6R²
Curved surface of the cone = πrl where r is radius of the base and l is the slant height of the cone.

Total curved surface area for the Cone = 
Change in Curved Surface Area = 1270.05R² - 125.6R² =1144.45R²

In the above figure, we need to find the Area of BO'CO
Since B and C are on the circumference of the circle, BO=CO=BO'=CO' =r . Also OO' =r
We can say that BOO' and COO' are 2 equilateral triangles with sides r
Thus area of BO'CO = area BOO' + area COO'
Thus area of BO'CO 
Area of 1 circle = πr²
Ratio = 

Let the diameter of the circle be 2r
△SPT is a right-angle triangle.
tan(30^º) = 6/2r
1/√3 = 3/r
or r = 3√3. Since PT and PQ are the tangents from same point P, PT = PQ
Area PTOQ = Area PTO + Area PQO

Since it is given that xx and yy are integers then |x - 5|×|y - 7| also needs to be an integer.
Thus |x − 5| × |y − 7| = 5 or 6
Case I: |x − 5| × |y − 7| = 5
a) |x - 5| = 5 and |y - 7| = 1 It gives x = 0 or 10 and y = 6 or 8. (10, 6) and (10, 8) are only solution as we are given that x, yx, y are positive
b) |x - 5| = 1 and |y - 7| = 5. It gives x = 6 or 4 and y = 12 or 2. Total of 4 such pairs are possible
Total 6 such pairs possible for case I
Case II: |x − 5| × |y − 7| = 6
a) |x - 5| = 1 and |y - 7| = 6. It gives x = 4 or 6 and y = 1 or 13 Thus 4 pairs are possible
b) |x - 5| = 2 and |y - 7| = 3. It gives x = 3 or 7 and y = 4 or 10. 4 pairs are possible
c) |x - 5| = 3 and |y - 7| = 2. It gives x =  2 or 8 and y = 5 or 9. 4 pairs are possible
d) |x - 5| = 6 and |y - 7| = 1. It gives x = -1 or 11 and y = 6 or 8. Here 2 pairs are possible
Total of 4 + 4 + 4 + 2 = 14 such pairs are possible
Overall 6 + 14 = 20 cases are possible.

Therefore, factorising the numerator, we get

Factorising the quadratic terms in the numerator and denominator, we get

In the denominator, x² + 1 > 0, hence,

The boundary points are x = -5, -1, 1, 2, 5
For x > 5, this inequality holds true.
Now, we know that if the powers of the terms are odd, the inequality alternates between them.
Hence, the inequality holds true for
x ∈ (−5, −1) U (1, 2) U (5, ∞)
Hence, integer values of variable x <= 10 satisfying the condition are -4, -3, -2, 6, 7, 8, 9, 10.
Hence, count = 8.

Let the cost of pen be xx, a pencil be y and a markers be z
x + 3y + 5z = 100.......(I)
6x + 18y + 2z = 320....(II)
6 × (I) − (II) gives us,
28z = 280 or z = 10
putting z = 10 in (I) we get x + 3y = 50
We have to get the minimum value of x + y + z. z = 10 thus we need to minimize x + y
when y = 16 then x = 2 and x + y + z = 2 + 16 + 10 = 28
When y = 15 then x = 5 and x + y + z = 5 + 15 + 10 = 30
We see that as we keep on decreasing y the sum will increase.
Minimum possible value = 28

Take t-5 = k
Equation modifies into
log₃ 78 + k − 1 = log₃(1 − 3k)
Taking all the log terms to one side

Taking power 3 both sides

Or 26 × 3^k = 1 − 3^k
27 × 3^k =1
k = -3
Thus t - 5 = -3 or t = 2

According to the question
2000000(1.2)³ = x(1.2)² + x(1.2) + x + 180000 
3456000 − 180000 = x(1.44 + 1.2 + 1)
3276000 = x(1.44 + 1.2 + 1)
3276000 = x × 3.6
3276000/​3.64 = x
x = 900000.

CP of 1 litre = 50
MP of 1 litre = 1.2 x 50 = 60
Also, he mixes water to milk in the ratio of 1:4, so in 1 litre of milk, he mixes 250 mL of water.
Hence, the volume of liquid he is selling = 1.25
MP of 1.25 litres = 60 x 1.25 = 75 Rs.
So, to have a 10% profit, SP = Rs 50 x 1.1 = 55 Rs.
Hence, discount percent = 20/75 x 100 = 80/3%

Let us calculate the actual sum which Ram would have got if he did not have had made the mistake.
We can use the formula to calculate the sum = 30/2 ​ (2(2) + (30 − 1)7)
Thus the actual sum = 15(4 + 29 × 7) = 3105
Excess sum calculated = 3210 - 3105 = 105
Let last n terms be the terms where the common difference was taken as 14 instead of 7
Thus 7 + 14 + ...n terms = 105
n/2(2(7) + (n − 1)7) = 105
n(n + 1)7 = 210
Or 2 + n − 30 = 0 or (n + 6)(n − 5) = 0.
Since n can not be negative, n = 5, Thus last 5 terms had incorrect common differece. x = 30 - n = 30 - 5 = 25
x − 4 = 21 = 3 × 7
Total 4 factors are there.

Both of the series are an Arithmetic Progression
Common difference of the Series are 6 and 7 respectively.
Upon looking, we can see that the first common term is 16.
The series which will have common terms will have first term as 16 and common difference = lcm(6, 7) = 42
Let a_n ​be the n^th and final term of the series
a_{n ​}= 16 + (n − 1)42 ≤ 610
42(n − 1) = 594
n − 1 = 14.14 or n = 15.14. Thus their are total 15 number of common terms.

144 = 16 x 9 = 2⁴3²


Highest power of 16 = 5, highest power of 9 = 5.
Hence, highest power of 144 = 5.
96 = 32 x 3 = 2⁵3


Highest power of 32 = 9, highest power of 3 = 22.
Hence, the highest power of 96 = 9.
Y - X = 4. It is a perfect square and even.

CP of 1 good = 105/14 = Rs. 7.5
MP of 1 good = 140/7 = Rs. 20
Hence, markup % = 12.5/7.5 x 100 = 500/3 %
Hence, d1 = 100/3% = 1/3
d2 = 50/3% = 1/6


Hence, profit on 90 goods = 65 x 5 = Rs 325.

Speed of the car is twice the speed of the auto. So, time taken by the car is half the time taken by the auto. Amar goes by car for 10 min and by auto for 30 min. Total time taken by him if he goes all the way be car = 10 + 1/2(30) = 10 + 15 = 25 min