Algebra - Number Theory - Previous Year CAT/MBA Questions
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The least common multiple of a number and 990 is 6930. The greatest common divisor of that number and 550 is 110. What is the sum of the digits of the least possible value of that number?
- (a)
6
- (b)
9
- (c)
14
- (d)
18
- (e)
None of the remaining options is correct
Answer: Option C
Text Explanation :
Workspace:
Consider a 4-digit number of the form abbb, i.e., the first digit is a (a > 0) and the last three digits are all b. Which of the following conditions is both NECESSARY and SUFFICIENT to ensure that the 4-digit number is divisible by a?
- (a)
b is divisible by a
- (b)
b is equal to 0
- (c)
21b is divisible by a
- (d)
9b is divisible by a
- (e)
3b is divisible by a
Answer: Option E
Text Explanation :
Workspace:
There are two numbers which are greater than 21 and their LCM and HCF are 3003 and 21 respectively. What is the sum of these numbers?
- (a)
504
- (b)
508
- (c)
514
- (d)
528
Answer: Option A
Text Explanation :
Workspace:
If the digit in the unit's place of a two-digit number is halved and the digit in ten's place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true?
- (a)
Sum of the digits is a two-digit number.
- (b)
Digit in the unit's place is half of the digit in the ten's place.
- (c)
Digit in the unit's place and the ten's place are equal.
- (d)
Digit in the unit's place is twice the digit in ten's place.
Answer: Option D
Text Explanation :
Workspace:
If abc > 0, such that a,b,c are integers then which of the following must be true?
- (a)
a/b < 0
- (b)
ab/c > 0
- (c)
bc < 0
- (d)
a > bc
Answer: Option B
Text Explanation :
Workspace:
Let x and y be two positive integers and p be a prime number. If x(x – p) – y(y + p) = 7p, what will be the minimum value of x – y?
- (a)
1
- (b)
3
- (c)
5
- (d)
7
- (e)
None of the above
Answer: Option E
Text Explanation :
Given, x(x – p) – y(y + p) = 7p
⇒ x2 - xp - y2 - yp = 7p
⇒ x2 - y2 - xp - yp = 7p
⇒ (x - y)(x + y) - p(x + y) = 7p
⇒ (x + y)(x - y - p) = 7p
Since, 7 and p are both prime numbers, we have
⇒ (x + y)(x - y - p) = 7 × p or 7p × 1
i.e., one (x + y) and (x - y - p) can be 7 and p or 7p and 1.
Case 1: one (x + y) and (x - y - p) can be 7 and p
⇒ (x + y) + (x - y - p) = 7 + p
⇒ 2x = 7 + 2p
⇒ x = 3.5 + 2p
This is not possible as x should be a integer.
Case 2: one (x + y) and (x - y - p) can be 7p and 1.
⇒ (x + y) + (x - y - p) = 7p + 1
⇒ 2x = 8p + 1
⇒ x = 4p + 0.5
This is not possible as x should be a integer.
∴ No value of integral value of x is possible.
Hence, option (e).
Workspace:
Three flower fields have lengths of 125, 175 and 65 square meters of width 3 meters each. If the farmer wants to grow flower beds of equal area then what is maximum length of each flower bed?
- (a)
4
- (b)
5
- (c)
6
- (d)
7
Answer: Option B
Text Explanation :
Workspace:
3 different types of rubber pipes of lengths 42 ft, 49 ft and 63 ft were ordered for some construction work. They needed to be cut in such a way that final pipes to be used are of same size for all types of rubber pipes. What is the least number of total pieces of all the rubber pipes taken together?
- (a)
22
- (b)
23
- (c)
24
- (d)
None of these
Answer: Option A
Text Explanation :
Workspace:
There are three light bulbs which blink at intervals of 24, 36 and 54 seconds. If they all blink at 7:15 pm, then at what time will they blink together again?
- (a)
7:16:48 am
- (b)
7:22:12 pm
- (c)
7:18:36 pm
- (d)
None of these
Answer: Option C
Text Explanation :
Workspace:
On dividing P and Q by a divisor x, the remainder obtained are 265 and 378 respectively. But when their sum is divided by the same divisor x, the remainder obtained is 138. Find x, if P, Q and x are all natural numbers?
- (a)
101
- (b)
113
- (c)
154
- (d)
None of these
Answer: Option D
Text Explanation :
Workspace:
If 1/x is a positive fraction and 1/y is a negative fraction then which of the following statements are true
- 1/x + 1/y is positive
- 1/x ‐ 1/y is positive
- x‐y/xy is negative
- 1/x2 – 1/y2 is positive
- (a)
Only (A) & (B) are true
- (b)
Only (B) & (C) are true
- (c)
Only (D) & (E) are true
- (d)
Only (A) & (E) are true
Answer: Option B
Text Explanation :
Workspace:
Given below are two statements:
Statement I : The set of numbers (7,8,9,a,b,10,8,7) has an arithmetic mean of 9 and mode(most frequently occurring number) as 8, Then a × b = 120.
Statement II : Let a and b be two positive integers such that a + b + a x b = 84, then a + b =20.
In the light of the above statements, choose the most appropriate answer from the options given below
- (a)
Both Statement I and Statement II are correct
- (b)
Both Statement I and Statement II are incorrect
- (c)
Statement I is correct but Statement II is incorrect
- (d)
Statement I is incorrect but Statement II is correct
Answer: Option D
Text Explanation :
Workspace:
If (x + y) = 3, xy = 2, then what is the value of x3 + y3 :
- (a)
6
- (b)
7
- (c)
8
- (d)
9
Answer: Option D
Text Explanation :
Workspace:
A supplier receives orders from 5 different buyers. Each buyer places their order only on a Monday. The first buyer places the order after every 2 weeks, the second buyer, after every 6 weeks, the third buyer, after every 8 weeks, the fourth buyer, every 4 weeks, and the fifth buyer, after every 3 weeks. It is known that on January 1st, which was a Monday, each of these five buyers placed an order with the supplier.
On how many occasions, in the same year, will these buyers place their orders together excluding the order placed on January 1st?
- (a)
1
- (b)
5
- (c)
2
- (d)
4
- (e)
3
Answer: Option C
Text Explanation :
The supplier receives his orders from the five buyers once every 2 weeks, once every 6 weeks, once every 8 weeks, once every 4 weeks, and once every 3 weeks.
The number of occasions where all the five buyers place the order on the same day is :
The LCM of the 5-time frames during which the 5 buyers place their orders :
Hence the LCM is :
(2, 6, 8, 4, 3).
= 24 weeks.
Once every 24 weeks, all five of them place the order simultaneously.
A year has 53 weeks in total :
Hence all five of them place the orders after 24 weeks, 48 weeks.
Workspace:
The sum of the cubes of two numbers is 128, while the sum of the reciprocals oftheir cubes is 2. What is the product of the squares of the numbers?
- (a)
64
- (b)
256
- (c)
16
- (d)
48
- (e)
32
Answer: Option C
Text Explanation :
Considering the two numbers to a, b :
We were given that:
a3 + b3 = 128
+ = 2
= 2 =
k = 64.
Hence a3 . b3 = 64
a*b = 4 and a2 .b2 = 16
Workspace:
Nadeem’s age is a two-digit number X, squaring which yields a three-digit number,whose last digit is Y. Consider the statements below:
Statement I: Y is a prime number
Statement II: Y is one-third of X
To determine Nadeem’s age uniquely:
- (a)
either of I and II, by itself, is sufficient.
- (b)
only II is sufficient, but I is not.
- (c)
only I is sufficient, but II is not.
- (d)
it is necessary and sufficient to take I and II together.
- (e)
even taking I and II together is not sufficient.
Answer: Option D
Text Explanation :
The age of Nadeem is a two-digit number. When squared yields a three-digit number whose last digit is Y. Y is a prime number.
Using statement 1:
When a number is squared :
The last digit of the number can be :
1, 2, 3, 4, 5, 6, 7, 8, 9. When squared the last digit can be :
For a number ending with 1: 1
For a number ending with 2: 4
For a number ending with 3: 9
For a number ending with 4: 6
For a number ending with 5: 5
For a number ending with 6: 6
For a number ending with 7: 9
For a number ending with 8: 4
For a number ending with 9: 1
The only possible prime number is 5.
Hence the last digit of X is 5 and Y is 5.
Using statement 2 : Y = X/3.
This alone cannot be sufficient to determine the possibilities for Y and X.
Combining both the statements :
Since Y = 5, then the value of X = 15.
The age is equal to 15.
Workspace:
Wilma, Xavier, Yaska and Zakir are four young friends, who have a passion for integers. One day, each of them selects one integer and writes it on a wall. The writing on the wall shows that Xavier and Zakir picked positive integers, Yaska picked a negative one, while Wilma’s integer is either negative, zero or positive. If their integers are denoted by the first letters of their respective names, the following is true:
W4 + X3 + Y2 + Z ≤ 4
X3 + Z ≥ 2
W4 + Y2 ≤ 2
Y2 + Z ≥ 3
Given the above, which of these can W2 + X2 + Y2 + Z2 possibly evaluate to?
- (a)
9
- (b)
0
- (c)
4
- (d)
6
- (e)
1
Answer: Option D
Text Explanation :
Given that X, Z are positive Y is negative and W can be either positive or zero or negative.
The given conditions are:
W4 + X3 + Y2 + Z ≤ 4
X3 + Z ≥ 2
W4 + Y2 ≤ 2
Y2 + Z ≥ 3
For W4 + Y2 ≤ 2. Since Y is negative but Y2 is always positive and must be less than 2 because W4 is a nonnegative value. Hence Y = -1 is the only possibility. For W this can take any value among -1, 0, 1.
Y2 + Z ≥ 3. Since Y = -1, Z must be at least equal to 2 so the value of Y2 + Z ≥ 3 is greater than 2.
X is a positive value and must at least be equal to 1.
The condition: W2 + X2 + Y2 + Z2 here has all the independent values:
X2, Y2, Z2, W2 are non-negative.
W4 + X3 + Y2 + Z ≤ 4:
Since the value of Z is at least equal to 2 the value of Y2 is equal to 1.
Since X is a positive number in order to have the condition of W4 + X3 + Y2 + Z ≤ 4 satisfied. The value of Z must be the minimum possible so that X3 + Y2 + Z to have a value equal to 4 when X takes the minimum possible positive value equal to 1.
Hence X must be 1. W must be equal to 0 so that:
W4 + X3 + Y2 + Z ≤ 4. The sum = (0 + 1 + 1 + 2) = 4. The only possible case.
The value of W2 + X2 + Y2 + Z2 = (0 + 1 + 1 + 4) = 6.
Workspace:
Fatima found that the profit earned by the Bala dosa stall today is a three-digit number. She also noticed that the middle digit is half of the leftmost digit, while the rightmost digit is three times the middle digit. She then randomly interchanged the digits and obtained a different number. This number was more than the original number by 198.
What was the middle digit of the profit amount?
- (a)
1
- (b)
2
- (c)
6
- (d)
This cannot be solved with only the given information
- (e)
8
Answer: Option B
Text Explanation :
From the given conditions :
Considering the three-digit number to be a b c.
With the given conditions :
a = 2b, c = 3b.
Hence the number is of the form : 2b b 3b.
Since all three of the values must be less than 10 and non-negative:
This takes values : b = 1, b = 2, b = 3.
Hence the possible numbers are : (213, 426, 639) :
The interchanged number must be greater than the original by 198.
Hence the different rearrangements for the three numbers are :
213 : (312, 321, 132, 123, 231).
426 : (462, 624, 642, 246, 264)
639 : ( 693, 963, 936, 396, 369)
The only possible value which is higher than the original by 198 is :
(426, 624).
The middle digit is 2.
Workspace:
The Madhura Fruits Company is packing four types of fruits into boxes. There are 126 oranges, 162 apples, 198 guavas and 306 pears. The fruits must be packed in such a way that a given box must have only one type of fruit and must contain thesame number of fruit units as any other box.
What is the minimum number of boxes that must be used?
- (a)
21
- (b)
18
- (c)
44
- (d)
42
- (e)
36
Answer: Option C
Text Explanation :
The number of oranges, apples, guavas, and pears = 126, 162, 198, and 306.
Each box must contain an equal number of fruits with only one type of fruit. The additional condition provided is that there should be a minimum number of boxes in total.
The distribution is possible in multiple ways in such a way that distribution in each box is placed in such that each box contains a certain number of fruits n which is a factor for all the four given number of fruits :
Arrangement of 1 fruit of one kind in a basket.
2 is a factor of 126, 162, 198, and 306. So we can place 2 fruits of a particular kind in a basket.
Since we were asked for the minimum number of boxes this is possible when a maximum number of fruits of a kind are placed in a box.
Hence each box must contain the Highest common factor for the four numbers :
The prime factorization for the four numbers:
126 : 2 ∙ 7 ∙ 9, 162 : 2 ∙ 9 ∙ 9, 198 : 2 ∙ 9 ∙ 11, 306 = 2 ∙ 9 ∙ 17
The HCF is 18.
The number of boxes required for each :
, , ,
7 + 9 + 11 + 17 = 44.
Workspace:
Determine the number of positive integer 'n' where 1 ≤ n ≤ 100 and 'n' is not divisible by 2, 3 or 5.
- (a)
23
- (b)
25
- (c)
26
- (d)
29
Answer: Option C
Text Explanation :
The number of numbers divisible by 2 = 50
The number of numbers divisible by 3 = 33
The number of numbers divisible by 5 = 20
The number of numbers divisible by 2 and 3 = 16
The number of numbers divisible by 2 and 5 = 10
The number of numbers divisible by 3 and 5 = 6
The number of numbers divisible by 2, 3 and 5 = 3
The number of numbers divisible by 2, 3 or 5 = 50 + 33 + 20 - 16 - 10 - 6 + 3 =74
The number of numbers not divisible by 2, 3 or 5 = 100 - 74 = 26
Answer is option C.
Workspace:
What is the remainder if 1920 – 2019 is divided by 7?
- (a)
5
- (b)
1
- (c)
6
- (d)
0
- (e)
3
Answer: Option A
Text Explanation :
= = = = = = = 4
= = -1
= 4 - (-1) = 5
Hence, option (a).
Workspace:
When expressed in a decimal form, which of the following numbers will be nonterminating as well as non-repeating?
- (a)
- (b)
sin21° + sin22° + … + sin289°
- (c)
- (d)
- (e)
Answer: Option E
Text Explanation :
We need to check options.
Option (a):
= = = 0.5 (terminating)
Option (b): sin21° + sin22° + … + sin288° + sin289°
= sin21° + sin22° + … + sin245° + … + cos22° + cos21°
= 44 + ½ = 44.5 (terminating)
Option (c):
= = 3 + 22/7 (terminating)
Option (d):
= = 4 (non-terminating)
Option (e):
= 6 - 4 + √3 = 2 + √3 (non-terminating and non-repeating)
Hence, option (e).
Workspace:
Consider the four variables A, B, C and D and a function Z of these variables, Z = 15A2 − 3B4 + C + 0.5D. It is given that A, B, C and D must be non-negative integers and that all of the following relationships must hold:
i) 2A + B ≤ 2
ii) 4A + 2B + C ≤ 12
iii) 3A + 4B + D ≤ 15
If Z needs to be maximised, then what value must D take?
- (a)
15
- (b)
12
- (c)
0
- (d)
10
- (e)
5
Answer: Option B
Text Explanation :
To maximise Z, we need to maximise A, C and D and minimise B.
Given, 2A + B ≤ 2
A can be either 1 or 0.
Case 1: If A = 1, minimum value of B = 0
From (ii): Maximum value of C = 8
From (iii): Maximum value of D = 12
⇒ Z = 15A2 − 3B4 + C + 0.5D = 15 – 0 + 8 + 6 = 29
Case 2: If A = 0, minimum value of B = 0
From (ii): Maximum value of C = 12
From (iii): Maximum value of D = 15
⇒ Z = 15A2 − 3B4 + C + 0.5D = 0 – 0 + 12 + 7.5 = 19.5
We can see that maximum value of Z = 29, when D = 12.
Hence, option (b).
Workspace:
When opening his fruit shop for the day a shopkeeper found that his stock of apples could be perfectly arranged in a complete triangular array: that is, every row with one apple more than the row immediately above, going all the way up ending with a single apple at the top.
During any sales transaction, apples are always picked from the uppermost row, and going below only when that row is exhausted. When one customer walked in the middle of the day she found an incomplete array in display having 126 apples totally. How many rows of apples (complete and incomplete) were seen by this customer? (Assume that the initial stock did not exceed 150 apples.)
- (a)
14
- (b)
13
- (c)
15
- (d)
11
- (e)
12
Answer: Option E
Text Explanation :
In the given triangular array, 1st row will have 1 apple, 2nd row will have 2 apples and so on.
∴ nth row will have n apples.
⇒ Total number of apples initially = 1 + 2 + 3 + … + n =
Initially total number of apples should be greater than 126 but less than or equal to 150
∴ 126 < ≤ 150
Only integral value of n satisfying this is 16.
⇒ = 136
∴ Initially there are 16 rows and 136 apples arranged in a triangular array in these 16 rows.
As the customer observes 126 apples, it would mean that 10 apples were removed.
Now, apples are removed starting from Row 1, which has 1 apple. So apples must have been removed in such a way with 1 apple from Row 1, 2 apples from Row 2, 3 apples from Row 3 and 4 apples from Row 4.
∴ Total 10 apples were removed the first 4 rows.
As out of 16 rows, apples are removed till Row 4, it would mean that only 16 - 4 or 12 rows of apples are visible to the customer.
Hence, option (e).
Workspace:
If x2 + x + 1 = 0, then x2018 + x2019 equals which of the following:
- (a)
-x
- (b)
x
- (c)
None of the others
- (d)
x - 1
- (e)
x + 1
Answer: Option A
Text Explanation :
x2 + x + 1 = 0 …(1)
⇒ x2 = - x – 1 …(2)
Multiplying x in (2)
⇒ x3 = - x2 – x
⇒ x3 = 1 …(3) [From (1)]
Now,
x2018 + x2019 = x2018(1 + x)
= x2018 × -x2
= -x2020
= - (x3)673 × x
= - 1 × x
= - x
∴ x2018 + x2019 = - x
Hence, option (a).
Workspace:
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