Algebra - Number System - Previous Year CAT/MBA Questions
The best way to prepare for Algebra - Number System is by going through the previous year Algebra - Number System OMET questions. Here we bring you all previous year Algebra - Number System OMET questions along with detailed solutions.
It would be best if you clear your concepts before you practice previous year Algebra - Number System OMET questions.
“xyz” and “zyx” are three digit numbers where x, y, z are distinct digits from 0 to 9. Differenceof xyz and zyx has a factor of 7.
What is the maximum possible value of the LCM of x, y and z?
- (a)
126
- (b)
72
- (c)
90
- (d)
56
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
Workspace:
Find the sum of following series:
+ + + + + + ...
- (a)
- (b)
- (c)
- (d)
Answer: Option A
Explanation :
S = + + + + + + ...
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:
+ +
solving we get S = =
Workspace:
If 1067 – 87 is written as an integer in base 10 notation, what is the sum of digits in that integer?
- (a)
683
- (b)
489
- (c)
583
- (d)
589
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).
Workspace:
Two number in the base system B are 2061B and 601B. The sum of these two numbers in decimal system is 432. Find the value of 1010B in decimal system.
- (a)
110
- (b)
120
- (c)
130
- (d)
140
- (e)
150
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).
Workspace:
The total number of eight – digit landline telephone numbers that can be formed having at least one of their digits repeated is:
- (a)
98185600
- (b)
97428800
- (c)
100000000
- (d)
None of the above
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).
Workspace:
The unit digit in the product of (8267)153 × (341)72 is
- (a)
1
- (b)
2
- (c)
7
- (d)
9
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).
Workspace:
Z is the product of first 31 natural numbers. If X = Z + 1, then the numbers of primes among X + 1, X + 2, ..., X + 29, X + 30 is
- (a)
30
- (b)
2
- (c)
Cannot be determined
- (d)
None of the above
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).
Workspace:
There are four prime numbers written in ascending order of magnitude. The product of the first three is 7429 and last three is 12673. Find the first number.
- (a)
19
- (b)
17
- (c)
13
- (d)
None of the above
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).
Workspace:
Find the coefficient of x12 in the expansion of (1 – x6)4(1 – x)– 4
- (a)
113
- (b)
119
- (c)
125
- (d)
132
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
=
= =
= 35 × 13
Coefficient of x6
=
= = 84
The coefficient of x0 is 1.
∴ The coefficient of x12 is 35 × 13 + (– 4) × 84 + 6
= 125
Hence, option (c).
Workspace:
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