Algebra - Number System - Previous Year CAT/MBA Questions

Answer: Option A

Explanation :

The difference between the numbers xyz and zyx will be a multiple of 99. therefore, 99(x - z) = 99*7
x - z = 7
No to maximise the LCM of x, y, z we can take x = 9, y = 7 and z = 2
Therefore LCM = 126

Answer: Option A

Explanation :

S = $\frac{3}{7}$ + $\frac{4}{7^2}$ + $\frac{5}{7^3}$ + $\frac{3}{7^4}$ + $\frac{4}{7^5}$ + $\frac{5}{7^6}$ + ...

Now here S is a summation of 3 distinct infinite geometric series
And we know that sum of an infinite geometric progression = (a/1-r) where a is the first term and r is the common ratio .

Therefore we can say S will be:

$\frac{3}{{\displaystyle \frac{7}{1-{\displaystyle \frac{1}{7^3}}}}}$ + $\frac{4}{{\displaystyle \frac{7^2}{1-{\displaystyle \frac{1}{7^3}}}}}$ + $\frac{5}{{\displaystyle \frac{7^3}{1-{\displaystyle \frac{1}{7^3}}}}}$

solving we get S = $\frac{180}{7^3-1}$ = $\frac{180}{342}$

Answer: Option D

Explanation :

10⁶⁷ is nothing but 10...00000 (67 times) i.e. 1 followed by 67 0s. 
When a two-digit number is subtracted from it, the last two 0s get replaced by (100 − that two-digit number number) while the remaining 0s get replaced by 9s due to carry of the subtraction. 
Hence, when 87 is subtracted from a number having 67 0s, we get 67 − 2 = 65 9s and the last two digits as 100 − 87 = 13. 
∴ Sum of digits = 65(9) + 1 + 3 = 589
Hence, option (d).

Answer: Option C

Explanation :

(2061)_B = (1 + 6B + 2B3)₁₀

(601)_B = (1 + 6B2)₁₀

∴ 2B³ + 6B² + 6B + 2 = 432

⇒ B³ + 3B² + 3B + 1 = 216

⇒ (B + 1)³ = 216

⇒ B = 5

∴ 1010₅ = (1 × 53 + 1 × 5)₁₀ = 13010

Hence, option (c).

Answer: Option A

Explanation :

The total number of 8-digit landline telephone numbers that can be formed having at least one of their digits repeated = The total number of 8-digit landline numbers – The number of 8-digit landline numbers in which no digit is repeated.

The total number of 8-digit landline numbers = 10⁸

The number of 8-digit landline numbers in which no digit is repeated = 10!/2

∴ Number of required landline numbers = 98185600

Hence, option (a).

Answer: Option C

Explanation :

The unit’s digit of (8267)¹⁵³ is same as unit digit of 7¹⁵³.

Since cyclicity of 7 is 4 and the remainder obtained when 153 is divided by 4 is 1,

∴ Unit’s digit of (8267)¹⁵³ = unit’s digit of 7¹⁵³ = unit’s digit of 7¹ = 7

Similarly,

Unit’s digit of (341)⁷² is same as unit’s digit of 1⁷² = 1

Hence the unit’s digit of the product = 7 × 1 = 7

Hence, option (c).

Answer: Option D

Explanation :

Z = 31!

Z is divisible by all numbers less than 32.

X = 31! + 1

X + 1 = 31! + 2, will be divisible by 2,

X + 2 = 31! + 3 will be divisible by 3,

X + 3 = 31! + 4 will be divisible by 4 and so on.

Hence none of the numbers will be prime.

Hence, option (d).

Answer: Option B

Explanation :

7429 = 17 × 19 × 23

17, 19 and 23 are consecutive primes.

Also, 12673 = 19 × 23 × 29

Hence, first number = 17

Hence, option (b).

Answer: Option C

Explanation :

(1 – x⁶)⁴ (1 – x)^–4

= (1 – 4x⁶ + 6x¹² – 4x¹⁸ + x²⁴)(1 – x)^–4

To find the coefficient of x¹² in the given expression, we need to find the coefficients of x¹², x⁶ and x⁰ terms in (1 – x)^–4

We use the Binomial Theorem for negative coefficients.

Coefficient of x¹²
= $\frac{(-4)(-4-1)(-4-1-2)..(-4-11)}{12!}$

= $\frac{4\times 5\times 6\times ...\times 15}{12!}$ = $\frac{13\times 14\times 15}{1\times 2\times 3}$

= 35 × 13

Coefficient of x⁶
= $\frac{(-4)(-4-1)...(-4-5)}{6!}$

= $\frac{7\times 8\times 9}{3!}$ = 84

The coefficient of x⁰ is 1.

∴ The coefficient of x¹² is 35 × 13 + (– 4) × 84 + 6

= 125

Hence, option (c).