Test: CAT Quantitative Aptitude- 3


22 Questions MCQ Test CAT Mock Test Series | Test: CAT Quantitative Aptitude- 3


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Attempt Test: CAT Quantitative Aptitude- 3 | 22 questions in 40 minutes | Mock test for CAT preparation | Free important questions MCQ to study CAT Mock Test Series for CAT Exam | Download free PDF with solutions
QUESTION: 1

The average age of a group of travellers is 25 years. For a certain trip to Australia, the average age has to be a maximum of 24 years, so they included A in their group to satisfy the criteria. For a certain trip to South Africa, the average age of the group has to be a minimum of 26 years. To satisfy this, the original group (excluding A) included B in their group. What is the average of the highest possible age of A and the lowest possible age of B?

Solution:

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)

QUESTION: 2

A bag contains Rs. 1, 50 p and 25 p coins. The number of 25 p coins is numerically equal to the total value of all the 50 p coins (in Rs). Aso, the total amount in the bag is equivalent to one thousand 25 p coins. The number of Rs. 1 coins is 14 less than the total value of all 25 p coins(in Rs). What is the total value of all the 50 p coins as a percentage of the total amount of money present in that bag?

Solution:

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 %

*Answer can only contain numeric values
QUESTION: 3

A mixed fraction is as follows:
 where a>=0, b>0, c>0 and a,b,c are whole numbers.
Also, b/c cannot be further reduced, that is, b and c are co-prime, and b<c.
It is known that b is 20% of a and c is 30% of a.
How many such mixed fractions are possible?


Solution:


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.

QUESTION: 4

In Cracku, 60% of the enrolled students read newspaper A, 50% of the enrolled students read newspaper B, 30% of the enrolled students read newspaper C, 40% of the enrolled students read newspaper D. While a particular student may read more than one newspaper, each student must read at least one newspaper. It is known that no one reads 2 or 3 newspapers, which of the following is a possible number of enrolled students in Cracku?

Solution:

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.

QUESTION: 5

A function f(x) is defined as 
fn (x) is defined as f(f(...n times f(x))))

What is value of [g(12)]. [x] represents the greatest integer less than or equal to x 

Solution:

Let us start by calculating

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

[g(12)] = 13

QUESTION: 6

If a polynomial function exists such that what is the absolute value of f(4)?

Solution:

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

*Answer can only contain numeric values
QUESTION: 7

Let p be sum of all the values of kk which satisfy the equation:
5{k}+2 = 2[k-1]
What is the value of [2p-1]?
Here {x} denotes the fractional part of x and [x] denotes the greatest integer less than or equal to x


Solution:

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

QUESTION: 8

Find the minimum value of the following function:
f(x) = max (x - 5x + 6,x2)

Solution:

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

QUESTION: 9

10 spheres of radius "R" are melted into making right cones of radius and height  What is approximate change in the curved surface area?

Solution:

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

QUESTION: 10

2 circles of the same radius are on a 2-D plane that exists such that the centre of one circle lies on the circumference of another circle. A quadrilateral is constructed by taking the two points of intersection of the circles and their centres. What is the ratio of the area of the quadrilateral to the area of 1 circle?

Solution:


 

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 = 

QUESTION: 11


In the above figure, PT and RS are 2 parallel lines which are tangents to the circle with centre O. PQ is also a tangent to the circle 
∠ PSO = 30º and PT = 6 units. What is the value of the area of PTOQ?

Solution:

Let the diameter of the circle be 2r
△SPT is a right-angle triangle.
tan(30º) = 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

QUESTION: 12

How many positive integral pairs (x, y) exists such that 4 < ∣x − 5∣ × ∣y − 7∣ < 7

Solution:

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.

*Answer can only contain numeric values
QUESTION: 13

How many non-zero integer values of x satisfy the equation given below such that x <= 10.


Solution:


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.

QUESTION: 14

Cost of 1 pen, 3 pencils and 5 markers is Rs 100. Cost of 18 pencils, 2markers and 6 pens is Rs 320. What is the minimum possible cost of 1 marker, 1 pen and 1 pencils if it is given that the cost of each of the items is a natural number?

Solution:

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

*Answer can only contain numeric values
QUESTION: 15

If t satisfies the following equation:
log3 ​78 + t − 6 = log3 ​(1 − 3t−5)
What is the value of |t|


Solution:

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

*Answer can only contain numeric values
QUESTION: 16

An entrepreneur borrows an amount of Rs 20 lakhs from a local bank for a duration of 3 years at a 20% interest rate, compounded annually. At the end of each of these 3 years, he returns a certain amount of money to the bank. The amount returned at the end of each of these 3 years is the same. At the end of 3 years, the bank additionally demanded a cheque of Rs. 1.8 lakhs to settle the loan. What was the amount he used to pay at the end of each year?


Solution:

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.

QUESTION: 17

A milk merchant buys milk at Rs 50 per litre, but while selling, he marks up the price by 20% and also mixes water to the milk such that the milk-water mixture he sells has 80% milk and 20% water. Assuming that the water has zero cost price, what discount percentage should he give to the mixture such that by selling it, he earns a 10% profit?

Solution:

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%

*Answer can only contain numeric values
QUESTION: 18

Ram was asked to calculate the sum of an arithmetic progression whose first term was 2 and common difference was 7. The total number of terms was 30. He wrote down all the 30 terms in ascending order and then started adding from smallest number. After calculating the sum of first x terms correctly, Ram started taking common difference as 14 instead of 7 and ended up with sum of 3210. How many factors does x − 4 has?


Solution:

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.

QUESTION: 19

What is the number of common terms between the following series
A1: 4,10,16,22,28,..........,610
A2 ​: 2,9,16,23,............... 632

Solution:

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 an ​be the nth and final term of the series
an ​= 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.

*Answer can only contain numeric values
QUESTION: 20

X = Highest power of 144 in 25!
Y = Highest power of 96 in 50!
How many of the following sentences are correct about the value Y - X?
1. It is a perfect square
2. It is a perfect cube
3. It is prime
4. It is even


Solution:

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.

QUESTION: 21

A merchant purchases goods at the rate of 14 for Rs 105, marks them at the rate of 7 for Rs 140. If he offers 2 successive discounts on the marked price, d1% and d2%, such that d1 is one-fifth the markup percentage and d2 is half that of the value of d1, what is the profit/loss made by selling 90 goods?

Solution:

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.

*Answer can only contain numeric values
QUESTION: 22

Amar drives his car twice the speed of an auto. One day, he drove the car for 10 min and the car broke mid-way. He left the car and took an auto to reach the office. It took him 30 min from there to reach the office. What would be the total time taken by car to reach his office?


Solution:

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

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