Algebra - Number Theory - Previous Year CAT/MBA Questions

Answer: 4195

Explanation :

Let the thousands, hundreds, tens and units digits be a, b, c and d.

Given, sum of its digits in the thousands, hundreds and tens places is 14
a + b + c = 14   …(1)
The sum of its digits in the hundreds, tens and units places is 15
b + c + d = 15   …(2)

(2) – (1) ⇒ a = d – 1   …(3)

The tens place digit is 4 more than the units place digit ⇒ c = d + 4   …(4)

For the number to be highest possible d should also be highest possible.

Highest possible of d is 5 (from (4)).
∴ highest possible value of c = 9 and that of a = 4

From (1), b = 14 – a – c = 1

∴ The highest such number is 4195

Hence, 4195.

Answer: Option A

Explanation :

Given x² – xy – x = 22 and y² – xy + y = 34.

Adding both the equations

⇒ x² + y² – 2xy - (x – y) = 56

⇒ (x - y)² – (x - y) = 56

⇒ (x - y)(x - y - 1) = 56

Let (x – y) = a [a > 0 since x > y]

⇒ a(a - 1) = 56

⇒ a² – a – 56 = 0

⇒ (a – 8)(a + 7) = 0

⇒ a = 8 or -7 (rejected)

∴ a = x - y = 8

Hence, option (a).

Answer: Option C

Explanation :

Let the three integers x, y and z be (a - d), a, (a + d) respectively.

[(a – d), a and (a + d) are all positive integers]

Since y – x > 2, hence d > 2.

Given, xyz = 5(x + y + z) 

⇒ (a – d) × a × (a + d) = 5 × 3a

⇒ (a – d)(a + d) = 15

Here (a – d) and (a + d) are positive integers

∴ We need to write 15 as product of 2 positive integers. This can be done in two ways, 1 × 15 or 3 × 5

Hence, (a, d) is either (8, 7) or (4, 1).

Since d > 0 hence, (4, 1) is rejected.

∴ (a, d) = (8, 7)

∴ z – x = (a + d) – (a - d) = 2d = 14

Hence, option (c).

Answer: Option B

Explanation :

Let the two sides of the rectangle be ‘l’ and ‘b’ ft.

⇒ l × b = 60,000

Cost of fencing = 200 × l + 100 × (l + 2b) = 100 (3l + 2b)

To minimize cost we need to minimize 3l + 2b

Now, we know AP ≥ GP

⇒ $\frac{3l+2b}{2}$ ≥ $\sqrt{3l\times 2b}$
⇒ 3l + 2b ≥ 2 × √$\sqrt{6\times 60,000}$
⇒ 3l + 2b ≥ 1200

∴ Minimum cost of fencing = 100 × 1200 = 1,20,000.

Hence, option (b).

Answer: 21

Explanation :

Let the digits of the 3-digit number be p, q, & r.
∴ 2 < p × q × r < 7
⇒ p × q × r can take the values 3, 4, 5, or 6.

Let's start with prime numbers 3 & 5.
Since they are prime, they can't be split, and hence if one of p, q or r is 3, the remaining two should be 1.

So, the possible combinations are

If p × q × r = 3
The number can be {113, 131, 311}

If p × q × r = 5
The number can be {115, 151, 511}

If p × q × r = 4
4 can be written as 1 × 2 × 2 or 1 × 1 × 4. 
Therefore, the possible combinations of p, q, r are {122, 212, 221, 114, 141, 411}

If p × q × r = 6
6 can be written as 1 × 3 × 2 or 1 × 1 × 6. 
Therefore, the possible combinations of p, q, r are {123, 132, 213, 231, 312, 321, 116, 161, 611}

Therefore, the total number of possibilities are 3 + 3 + 6 + 9 = 21.
Hence, 21.

Answer: Option C

Explanation :

Given ab = 432, bc = 96 and c < 9. 

To get the minimum value for a + b + c, the two bigger numbers should be as close as possible. 

So possible values are 

a = 36, b = 12, c = 8 ⇒ Sum = 58

a = 27, b = 16, c = 6 ⇒ Sum = 49

a = 18, b = 24, c = 4 ⇒ Sum = 46

Least possible value = 46

Hence, option (c).

Answer: Option A

Explanation :

$\frac{x}{\sqrt{1+x^4}}$ = $\frac{1}{\sqrt{\frac{1+x^4}{x^2} }}$ = $\frac{1}{\sqrt{\frac{1}{x^2}+x^2 }}$

We know x2 is positive and sum of a positive number and its reciprocal is always ≥ 2, i.e., 

$\frac{1}{x^2}+x^2$ ≥ 2.

⇒ $\frac{1}{\sqrt{\frac{1}{x^2}+x^2 }}$ will be maximum when 1$\frac{1}{x^2}+x^2$ is minimum i.e., 2.

∴ Maximum value of $\frac{1}{\sqrt{\frac{1}{x^2}+x^2 }}=\frac{1}{\sqrt{2}}$

Hence, option (a).

Answer: 23

Explanation :

Given,

x = z – 9 and
y = z – 1

Also, x + y < z + 5

⇒ (z – 9) + (z – 1) < z + 5

⇒ z < 15

∴ Highest integral value of z is 14.

Now, we have to calculate maximum value of 2x + y

= 2(z - 9) + (z - 1)

= 3z – 19

= 3 × 14 – 19 = 23

Hence, 23.

Answer: 2704

Explanation :

Let’s try to calculate the minimum possible value of $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$

⇒ $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$ = $\frac{1+x+y+xy}{xy}$ = $\frac{103}{xy}$ + 1

Minimum value of this expression will be when xy is maximum.

Now, when sum of two number is constant their product is maximum when both the number are equal to each other.

∴ x = y = 51

∴ Least possible value of 2601$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$ = 2601 × $\frac{52}{51}\times \frac{52}{51}$ = 2704

Hence, 2704.

Answer: 17

Explanation :

Given, 2x + 5y = 99

⇒ x = (99 - 5y)/2

For x to be an integer, y has to be odd.

Possible value of x and y

 x       y
97    -19
92    -17
…
17    13

∴ Total 17 possible values.

Hence, 17.

Answer: Option D

Explanation :

We know, 0.2 <  m/20 <0.5 ⇒ 4 < m < 10

Values m can take = {5, 6, 7, 8, 9}

We know, 0.2 < n/11 <0.5 ⇒ 2.2 < n < 5.5

Values n can take = {3, 4, 5}

Since n is even, the only possible value of n is 4.

Also, 0.2 < $\frac{\mathrm{m}}{20},\frac{\mathrm{n}}{\mathrm{m}},\frac{\mathrm{n}}{11}$ < 0.5 ⇒ m/5 < n < m/2

Since n = 4 ⇒ 20 > m > 8

The only possible value of m is 9.

∴ m – 2n = 9 – 2 × 4 = 1.

Hence, option (d).

Answer: 6

Explanation :

Given, 2 < x < 10

x can take any of the values from {3, 4, 5, 6, 7, 8, 9}

Also, 14 < y < 23

y can take any of the values from {15, 16, 17, 18, 19, 20, 21, 22}

The highest value N (i.e., x + y) can take = 9 + 22 = 31 (when x = 9; y = 22)

The lowest value N (i.e., x + y) can take = 3 + 15 = 18 (when x = 3; y = 15)

But, N = x + y > 25. Hence the different values of x + y are {31, 30, 29, 28, 27, 26}.

Hence, x + y, and thereby N can take 6 distinct values.

Hence, 6.

Answer: Option B

Explanation :

There are total 120 numbers given.

Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)

We know, n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C).

n(x) = number of multiples of x.

∴ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(2 ∩ 5) - n(5 ∩ 7) - n(7 ∩ 2) + n(2 ∩ 5 ∩ 7)

⇒ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(10) - n(35) - n(14) + n(70)

⇒ n(2 ∪ 5 ∪ 7) = 60 + 24 + 17 - 12 - 3 - 8 + 1 = 79.

Now, Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)

= 120 – 79 = 41

∴ Out of 120 numbers given, 41 numbers are not divisible by any of 2, 5 and 7.

Hence, option (b).

Answer: Option C

Explanation :

Let the two numbers be x and y.

So, xy = 616 and (x³ − y³)/(x − y)³ = 157/3. 

3(x³ − y³) = 157(x − y)³

∴ 3(x³ − y³) =157 [x³ − y³ − 3xy(x − y)]

∴ 154(x³ − y³) = 3 × 157 × xy(x − y)

∴ 154(x³ − y³) = 3 × 157 × 616 × (x − y)  [Using xy = 616]

∴ (x − y)(x² + y² + xy) = 1884(x − y)

∴ (x² + y² + xy) = 1884 (If x ≠ y)

Adding xy both sides, we get;

(x²+ y² + 2xy) = 1884 + xy = 1884 + 616 = 2500.

∴ x + y = 50.

Hence option 3.

Answer: Option A

Explanation :

Given: a² + b² = 25 and x² + y² = 169

We know 5² = 25 and 13² = 169

Multiply both equations to get (a² + b²) (x² + y²) = 25 × 169

(a² + b²) × (x² + y²) = 4225

We know, 6225 = 65²

We also know that ax + by = 65

So, numerically (Not algebraically), 

(a² + b²) × (x² + y²) = (ax + by)²

Expanding the equation,

⇒ (ax)² + (ay)² + (bx)² + (by)² = (ax)² + (by)² + 2axby

⇒ (ay)² + (bx)² = 2axby

⇒ (ay)² + (bx)² - 2axby = 0

This is of the form, (p - q)²

(ay - bx)² = 0

⇒ ay - bx = 0 = k

Hence, option (1).

Answer: 44

Explanation :

N = 2⁴ × 3⁵ × 10⁴ = 2⁴ × 3⁵ × 2⁴ × 5⁴ = 2⁸ × 3⁵ × 5⁴

Now, for any number to be a perfect square, it must have even powers.

So, if we consider 2⁸, only powers of 0, 2, 4, 6, 8 can lead to perfect squares i.e. 5 ways

Now, 

If we consider 3⁵, only powers of 0, 2, 4 can lead to perfect squares i.e. 3 ways

If we consider 5⁴, only powers of 0, 2, 4 can lead to perfect squares i.e. 3 ways

So, total number of possibilities = 5 × 3 × 3 ways = 45 ways.

Since we need to find the number of factors greater than 1,

Required number of ways = 45 -1 = 44 ways

Hence, 44.

Answer: Option D

Explanation :

Given : $\frac{(n^2+7n+12)}{(n^2-n-12)}$

$\frac{(n^2+7n+12)}{(n^2-n-12)}$ = $\frac{(n^2+4n+3n+12)}{(n^2-4n+3n-12)}$ = $\frac{(n+4)(n+3)}{(n-4)(n+3)}$

Now, n cannot be equal to -3, since denominator cannot be 0

⇒ $\frac{(n^2+7n+12)}{(n^2-n-12)}$ = $\frac{(n+4)}{(n-4)}$ = $\frac{(n-4+8)}{(n-4)}$ = $1 + \frac{8}{(n-4)}$

For $\frac{(n^2+7n+12)}{(n^2-n-12)}$ to be an integer, 8/(n-4) should also be an integer.

Largest value of n – 4 = 8

∴ largest possible value of n = 12.

Hence, option (4).

Answer: 4

Explanation :

We know that m² + 105 = n²

105 = n² - m²

105 = (n + m) × (n - m)

105 can be written as 3 × 5 × 7, which can be written as product of two numbers in 4 ways i.e. (105 × 1), (35 × 3), (21 × 5), (15 × 7)

Solving each of these 4 cases will give corresponding values of m and n.

Hence, 4.

Answer: 5

Explanation :

For x = (−2), 2^x = 0.25 and x = 7 is the largest value that x can take for 2^x ≤ 200

For x = (−2) and (−1), (2^x + 2) is not an integer.

For x = 0, (2^x + 2) = 3 i.e., divisible by 3.

For x = 1, (2^x + 2) = 4 i.e., divisible by 4.

For x = 2, (2^x + 2) = 6 i.e., divisible by 3.

For x = 3, (2^x + 2) = 10 i.e., neither divisible by 3 nor by 4.

For x = 4, (2^x + 2) = 18 i.e., divisible by 3.

For x = 5, (2^x + 2) = 34 i.e., neither divisible by 3 nor by 4.

For x = 6, (2^x + 2) = 66 i.e., divisible by 3.

For x = 7, (2^x + 2) = 130 i.e., neither divisible by 3 nor by 4.

Thus, for x = 0, 1, 2, 4 and 6, 2^x + 2 is perfectly divisible by either 3 or 4.

Hence, 5.

Answer: 40

Explanation :

Let X and Y are the two real numbers.

X × Y × 73 − X × Y × 37 = 720

X × Y × (73 – 37) = 720

X × Y × 36 = 720

X × Y = 20

AM(X², Y²) ≥ GM(X², Y²)

∴ $\frac{X^2+Y^2}{2}$ ≥ $\sqrt{X^2\times Y^2}$

∴ (X² + Y²) ≥ 40

Answer : 40

Alternately, 

We know, X × Y = 40

Also, (X - Y)² ≥ 0 [Square of any number is always greater than or equal to zero]

∴ X² + Y² - 2XY ≥ 0

∴ X² + Y² ≥ 2XY

∴ X² + Y² ≥ 40

Answer: Option A

Explanation :

If ‘x’ is the digit in the ten’s place and ‘y’ is the digit in the unit’s place of the two digit number, we have the following:
$$\begin{gathered} 10x+y>3(10y+x) \\ \therefore x>\frac{29}{7}y \end{gathered}$$

When y = 1, x = 5, 6, 7, 8 or 9 (Total 5 values)

When y = 2, x = 9 (Total 1 value)

Therefore the required answer is 5 + 1 = 6.

Hence option 1.

Answer: Option A

Explanation :

For n = 1, A₁ = 0

For n = 2, A₂ = 1225 = 35 × 35

Also, 6²ⁿ- 35n - 1 = (36n-1) - 35n.

The term in the bracket 36ⁿ - 1 is always divisible by (36 - 1)or by 35. Similarly 35n is also always divisible by 35. Therefore each term in the set A is divisible by 35. However, not all positive multiples of 35 are present in set A.

For n = 1, B₁ = 35(1 - 1) = 0

For n = 2, B₂ = 35(2 - 1) = 35 and so on.

Thus all whole number multiples of 35 are present in set B.

Thus every member of set A is present in every member of set B but at least one member of set B is not present in set A.

Hence, option 1.

Answer: Option D

Explanation :

x + 1 = x²

⇒ x² - x – 1 = 0

⇒ x = (1+√5)/2 (∵ x > 0)

Now, squaring the original equation, we get

(x  +1)² = x⁴

⇒ x² + 1 + 2x = x⁴

⇒ x⁴ = x + 1 + 1 + 2x (Since, x² = x + 1)

⇒ x⁴ = 3x + 2

⇒ 2x⁴ = 6x + 4

⇒ 2x⁴ = 6((1+√5)/2) + 4

⇒ 2x⁴ = 7 + 3√5

Alternately,

We know, x = (1+√5)/2

⇒ x ≈ 1.62

∴ 2x⁴ ≈ 13.8

option I ≈ 14.8,
option II ≈ 11.6,
option IV ≈13.6

Hence, option 4.

Answer: Option D

Explanation :

Since 15600 is product of 3 consecutive we can start by taking the cube of an integer closest to 15600, which is actually 15625

Now $\sqrt[3]{15625}=25$

Dividing 15600 by 25 we get 624

624 can be expressed as a product of 24 and 26. So the 3 consecutive positive integers are 24, 25 and 26

24² + 25² + 25² = 576 + 625 + 676

= 1877

Hence, option 4.

Answer: 3

Explanation :

Given $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$

$\frac{b+a}{ab}=\frac{1}{9}$

⇒ 9b + 9a = ab
⇒ ab + 9b – 9a + 81 = 81
⇒ b(a – 9) – 9(a – 9) = 81
⇒ (a – 9) = (b – 9) = 81

Now 81 as a product of 2 positive integers can be written as 1 × 81, 3 × 21 or 9 × 9

If a - 9 = 1 ⇒ a = 10. Also b - 9 = 81

⇒ b = 90

Further, if a – 9 = 3, ⇒ a = 12 Also b – 9

= 27 ⇒ b = 36

If a – 9 = 9 ⇒ a = 18. Also b – 9 = 9 ⇒ b = 18

∴ 3 pairs of number i.e., (10, 90), (12, 36) and (18, 18) will satisfy the given equation.

Answer: 3