# CAT Questions on Perfect Squares

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## CAT Questions on Perfect Squares

Download important Perfect Squares Questions for CAT PDF based on previously asked questions in CAT exam. Practice Perfect Squares Questions PDF for CAT exam.

Question 1: Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

a) 3

b) 2

c) 4

d) 0

e) 1

Question 2: Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?

a) $1 \leq m \leq 3$

b) $4 \leq m \leq 6$

c) $7 \leq m \leq 9$

d) $10 \leq m \leq 12$

e) $13 \leq m \leq 15$

Question 3: The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?

a) 21

b) 25

c) 41

d) 67

e) 73

Question 4: A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of white tiles is the same as the number of red tiles. A possible value of the number of tiles along one edge of the floor is

a) 10

b) 12

c) 14

d) 16

Question 5: A rich merchant had collected many gold coins. He did not want anybody to know about him. One day, his wife asked, ” How many gold coins do we have?” After a brief pause, he replied, “Well! if I divide the coins into two unequal numbers, then 48 times the difference between the two numbers equals the difference between the squares of the two numbers.” The wife looked puzzled. Can you help the merchant’s wife by finding out how many gold coins the merchant has?

a) 96

b) 53

c) 43

d) None of these

Question 6: Mr.X enters a positive integer Y(>1) in an electronic calculator and then goes on pressing the square . root key repeatedly. Then

a) The display does not stabilize

b) The display becomes closer to 0

c) The display becomes closer to 1

d) May not be true and the answer depends on the choice of Y

Question 7: If n = 1 + x, where x is the product of four consecutive positive integers, then which of the following is/are true?
A. n is odd
B. n is prime
C. n is a perfect square

a) A and C only

b) A and B only

c) A only

d) None of these

Question 8: The smallest perfect square that is divisible by 7!

a) 44100

b) 176400

c) 705600

d) 19600

Question 9: If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

a) 1777

b) 1785

c) 1875

d) 1877

Question 10: While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is

Let the number be xxyy
xxyy = 1000x + 100x + 10y +y = 1100x+11y = 11(100x+y)
Since xxyy is a perfect square, and 11 is one of the factors, it should be a multiple of 121
So, xxyy = 121k, where k is also a perfect square.
For k = 4, xxyy is a 3 digit number and for k > 82, xxyy is a five digit number
Between 4 and 82, only for k = 64, the number is of the form xxyy
121*64 = 7744
So, there is only 1 number 7744 which is of the form xxyy and a perfect square.

Alternatively:

The number should be definitely more than 32 and less than 100 as the square is a two digit number.

A number of such form can be written as $(50 \pm a)$ and $100 – a$ where $0 \leq a \leq 100$
So, the square would be of form $(50 \pm a)^2 = 2500 + a^2 \pm 100a$ or $(100 – a)^2$ i.e. $10000 + a^2 + 200 a$

In both cases, only $a^2$ contributes to the tens and ones digit. Among squares from 0 to 25, only 12 square i.e. 144 has repeating tens and ones digit. So, the number can be 38, 62, or 88. Checking these squares only 88 square is in the form of xxyy i.e. 7744.

Let us assume that three positive consecutive integers are x, x+1, x+2. They are raised to first, second and third powers respectively.

$x^{1} + (x+1)^{2} + (x+2)^{3} = (x + (x+1) +(x+2))^{2}$

$x^{1} + (x+1)^{2} + (x+2)^{3}$ = $(3x + 3)^{2}$

$x^{3} + 7x^{2} + 15x + 9$ = $9x^{2} + 9 + 18x$

After simplifying you get,

$x^{3} – 2x^{2} – 3x = 0$

=> x=0,3,-1

Since x is a positive integer, it can only be 3.

So, the minimum of the three integers is 3. Option a) is the correct answer.

Sum of the four numbers < 396
396/10 = 39.6
So, the perfect square is a number less than 39.6
The possibilities are 36, 25, 16 and 9
For the sum to be 360, the numbers can be 87, 89, 91 and 93
The sum of four consecutive odd numbers cannot be 250
For the sum to be 160, the numbers can be 37,39,41 and 43
The sum of 4 consecutive odd numbers cannot be 90
So, from the options, the answer is 41.

Let C and R be no. of columns and rows respectively.

The number of red coloured tiles would be given by (R-2)(C-2). This is because two outer rows made of white tiles and the two outer columns made up of outer columns are removed.

Similarly the number of white tiles would be given by R*2 + (C-2)*2. Two tiles are removed from columns because the corner tiles would have already been included while considering the rows.

So according to given condition we have (C-2)*2 + 2*R = (C-2)(R-2).
Now start putting value of c from options into this equation. Only for one option B we get an integer value of R .

Let x = y+z such that z > y.

We know that $48*(z – y)^2 = z^2 – y^2$

Solving the above equation, we get z + y = 48

So, option d) is the correct answer.

Take 64 for example.

$\sqrt{64}$ = 8

$\sqrt{8}$ = 2.828

$\sqrt{2.828}$ is close to 1.7

So, we can see that the result is tending towards 1.

Let the four consecutive positive integers be $a,a+1,a+2$ and $a+3$.
Therefore, $n=1+a(a+1)(a+2)(a+3)$
Or, $n = 1+(a^2+3a)*(a^2+3a+2)$
Or, $n = (a^2+3a)^2 + 2*(a^2+3a)+1 = (a^2+3a+1)^2$
Hence, n is a perfect square and therefore not a prime.

The product of four consecutive positive integers is always even. Hence, n is always odd.
Therefore, from the given statements, only A and C are true.

7! can be broken into prime factors.

7! = 1*2*3*4*5*6*7 = $2^4*3^2*5*7$

Hence, the smallest perfect square which is divisible by 7! will $2^4*3^2*5^2*7^2$ = 5040*5*7 =  176400

$(x -1)x(x+1) = 15600$
=> $x^3 – x= 15600$
The nearest cube to 15600 is 15625 = $25^3$
We can verify that x = 25 satisfies the equation above.
Hence the three numbers are 24, 25, 26. Sum of their squares = 1877

We know that one of the 3 numbers is 37.
Let the product of the other 2 numbers be x.
It has been given that 73x-37x = 720
36x = 720
x = 20

Product of 2 real numbers is 20.
We have to find the minimum possible value of the sum of the squares of the 2 numbers.
Let x=a*b
It has been given that a*b=20

The least possible sum for a given product is obtained when the numbers are as close to each other as possible.
Therefore, when a=b, the value of a and b will be $\sqrt{20}$.

Sum of the squares of the 2 numbers = 20 + 20 = 40.

Therefore, 40 is the correct answer.

We hope this Perfect Squares Questions PDF for CAT with Solutions will be helpful to you.