CAT Quantitative Aptitude- 3 - Free MCQ Practice Test with solutions, Quant

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 %

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.

Given:

Amar drives at a speed of 70 km/h = 2 hours

Amar drives at a speed of 80 km/h = 3 hours

Amar drives at a speed of 40 km/h = 1 hour

Formula used:

Distance = Speed × Time

Average speed = Total distance/Total time

Calculation:

Distance traveled at a speed of 70 km/h = (2 × 70) km = 140 km

Distance traveled at a speed of 80 km/h = (3 × 80) km = 240 km

Distance traveled at a speed of 40 km/h = (1 × 40) km = 40 km

Total time = (2 + 3 + 1) hours = 6 hours

Total distance (140 + 240 + 40) km = 420 km

The average speed = 420/6 km/h = 70 km/h

∴ Amar's average speed (in km/h) is 70 km/h