NUMBERS AND EQUATION FORMATION
- An automobile owner reduced his monthly petrol consumption when the prices went up.The price-consumption relationship is as follows:(2015)
Price (in Rs per litre) 40 50 60 75
Monthly consumption (in litres) 60 48 40 32
If the price goes up to Rs 80 per litre, his expected consumption (in litres) will be
(A) 30
(B) 28
(C) 26
(D) 24
Answer-A
Expenditure when the petrol price is Rs 40/L = 40 multiplied by 60 = Rs 2400
Expenditure when the petrol price is Rs 50/L = 50 multiplied by 48 = Rs 2400
Expenditure when the petrol price is Rs 60/L = 60 multiplied by 40 = Rs 2400
Expenditure when the petrol price is Rs 75/L = 75 multiplied by 32 = Rs 2400
If the price goes up to 80/L ,then the expenditure = 80 multiplied by x = Rs 2400,
So monthly consumption = 2400 divided by 80 = 30
- Each of A, B, C and D has Rs 100. A pays Rs 20 to B, who pays Rs 10 to C, who gets Rs 30 from D. In this context, which one of the following statements is not correct?(2015)
(A) C is the richest.
(B) D is the poorest.
(C) C has more than what A and D have together.
(D) B is richer than D.
Answer-C
Let's track the transactions step by step:
Initial amounts:
A: Rs 100
B: Rs 100
C: Rs 100
D: Rs 100
A pays Rs 20 to B:
A: Rs 80
B: Rs 120
C: Rs 100
D: Rs 100
B pays Rs 10 to C:
A: Rs 80
B: Rs 110
C: Rs 110
D: Rs 100
C gets Rs 30 from D:
A: Rs 80
B: Rs 110
C: Rs 140
D: Rs 70
Now, let's evaluate each statement:
(a) C is the richest.
Based on the final amounts, C has Rs 140, which is more than A, B, and D. Therefore, statement (a) is correct.
(b) D is the poorest.
Based on the final amounts, D has Rs 70, which is the lowest among A, B, C, and D. Therefore, statement (b) is correct.
(c) C has more than what A and D have together.
A has Rs 80, and D has Rs 70, so together they have Rs 150. However, C has Rs 140, which is less than Rs 150. Therefore, statement (c) is not correct.
(d) B is richer than D.
Based on the final amounts, B has Rs 110, and D has Rs 70. Therefore, statement (d) is correct.
- Out of 130 students appearing in an examination, 62 failed in English, 52 failed in Mathematics, whereas 24 failed in both English and Mathematics. The number of students who passed finally is (2015)
(A) 40
(B) 50
(C) 55
(D) 60
Answer-A
To find the number of students who passed the examination, we need to subtract the number of students who failed from the total number of students.
Let's calculate the number of students who failed in at least one subject:
Number of students who failed in English = 62
Number of students who failed in Mathematics = 52
Number of students who failed in both English and Mathematics = 24
Number of students who failed in at least one subject = (Number of students who failed in English) + (Number of students who failed in Mathematics) - (Number of students who failed in both English and Mathematics)
= 62 + 52 - 24
= 90
Number of students who passed = Total number of students - Number of students who failed
= 130 - 90
= 40
Therefore, the number of students who passed finally is 40.
- In a group of persons travelling in a bus, 6 persons can speak Tamil, 15 can speak Hindi and 6 can speak Gujarati. In that group none can speak any other language. If 2 persons in the group can speak two languages only and one person can speak all the three languages, then how many persons are there in the group?(2015)
(A) 21
(B) 22
(C) 23
(D) 24
Answer-C
Let us assume the two persons who can speak two languages, Hindi and Tamil. The third person then speaks all the three languages.
Number of persons who can speak Tamil is 6. Only Tamil 6 – 2 – 1 = 3
Number of persons who can speak Hindi is 15. Only Hindi 15 – 2 – 1 = 12
Number of persons who can speak Gujarati is 6. Only Gujarati 6 – 1 = 5
Thus the number of persons who can speak only one language is 3 + 12 + 5 = 20
Number of persons who can speak two languages = 2
Number of person who can speak all the languages = 1
Total number of persons in the group = 23.
- In a parking area, the total number of wheels of all the cars (four-wheelers) and scooters/ motorbikes (two-wheelers) is 100 more than twice the number of parked vehicles. The number of cars parked is?(2015)
(A) 35
(B) 45
(C) 50
(D) 55
Answer-C
Let's assume the number of cars parked is 'C' and the number of scooters/motorbikes parked is 'S'.
Each car has 4 wheels, so the total number of wheels contributed by cars is 4C.
Each scooter/motorbike has 2 wheels, so the total number of wheels contributed by scooters/motorbikes is 2S.
According to the given information, the total number of wheels is 100 more than twice the number of parked vehicles, which can be expressed as:
4C + 2S = 2(C + S) + 100
4C + 2S = 2C + 2S + 100
2C = 100
C = 50
Therefore, the number of cars parked is 50.
- The number of persons who read magazine X only is thrice the number of persons who read magazine Y. The number of persons who read magazine Y only is thrice the number of persons who read magazine X. Then, which of the following conclusions can be drawn?(2015)
- The number of persons who read both the magazines is twice the number of persons who read only magazine X.
- The total number of persons who read either one magazine or both the magazines is twice the number of persons who read both the magazines.
Select the correct answer using the code given below:
(A) 1 only
(B) 2 only
(C) Both 1 and 2
(D) Neither 1 nor 2
Answer-C
Let's assume the number of persons who read magazine X only as "x", and the number of persons who read magazine Y only as "y".
According to the given information, the number of persons who read magazine X only is thrice the number of persons who read magazine Y:
x = 3y -- (Equation 1)
Similarly, the number of persons who read magazine Y only is thrice the number of persons who read magazine X:
y = 3x -- (Equation 2)
Now, let's analyze the conclusions:
1.The number of persons who read both the magazines is twice the number of persons who read only magazine X.
To determine if this conclusion can be drawn, we need to find the relationship between the number of persons who read both magazines and the number of persons who read only magazine X.
Let's assume the number of persons who read both magazines as "z".
From Equation 1, we can substitute x with 3y:
z = 2x
So, the number of persons who read both magazines is twice the number of persons who read only magazine X. This conclusion can be drawn.
Now let's analyze statement 2:
"The total number of persons who read either one magazine or both the magazines is twice the number of persons who read both the magazines."
The total number of persons who read either one magazine or both is given by:
Total = X + Y + XY
The number of persons who read both magazines is simply XY.
According to statement 1:
Total = 2XY
Substituting the given equations:
X + Y + XY = 2XY
Let's simplify the equation:
X + Y = XY
statement 2 is true based on the given information.
- If ABC × DEED = ABCABC; where A, B, C, D and E are different digits, what are the values of D and E?(2015)
(A) D = 2, E= 0
(B) D = 0, E = 1
(C) D = 1, E = 0
(D) D = 1, E = 2
Answer-C
Let's analyze the given equation: ABC × DEED = ABCABC
From the equation, we can deduce that D × E must result in the last two digits of the product, which are the same as the first two digits of the product.
Looking at the options provided:
(a) D = 2, E = 0
(b) D = 0, E = 1
(c) D = 1, E = 0
(d) D = 1, E = 2
Let's substitute the values and check which option satisfies the equation:
(a) D = 2, E = 0:
ABC × 2002 = ABCABC
In this case, the first two digits of the product would be 2, but the last two digits are 0. This option does not satisfy the equation.
(b) D = 0, E = 1:
ABC × 1010 = ABCABC
In this case, the first two digits of the product would be 0, but the last two digits are 0. This option does not satisfy the equation.
(c) D = 1, E = 0:
ABC × 1001 = ABCABC
In this case, the first two digits of the product would be 0, which matches the last two digits of the product. This option satisfies the equation.
(d) D = 1, E = 2:
ABC × 1212 = ABCABC
In this case, the first two digits of the product would be 2, but the last two digits are 4. This option does not satisfy the equation.
Therefore, the correct values are D = 1 and E = 0. Hence, the answer is option (c) D = 1, E = 0.
- A cow costs more than 4 goats but less than 5 goats. If a goat costs between Rs 600 and Rs 800, which of the following is a most valid conclusion?(2015)
(a) A cow costs more than Rs 2,500.
(b) A cow costs less than Rs 3,600.
(c) A cow costs between Rs 2,600 and Rs 3,800.
(d) A cow costs between Rs 2,400 and Rs 4,000.
Answer-D
Let's assume the cost of a cow is "C" and the cost of a goat is "G." Based on the information given in the problem, we know that:
4G < C < 5G
We also know that a goat costs between Rs 600 and Rs 800, so we can write:
600 ≤ G ≤ 800
To find the possible range of values for C, we can substitute the upper and lower bounds of G into the inequality:
4(600) < C < 5(800)
2400 < C < 4000
Therefore, the most valid conclusion is (d) "A cow costs between Rs 2,400 and Rs 4,000."
- In a box of marbles, there are three less white marbles than the red ones and five more white marbles than the green ones. If there are a total of 10 white marbles, how many marbles are there in the box?(2015)
(A) 26
(B) 28
(C) 32
(D) 36
Answer-B
Let's assume the number of red marbles is "R," the number of green marbles is "G," and the number of white marbles is "W."
According to the given information, we have three equations:
1.W = R - 3
2.W = G + 5
3.W = 10
From equations 1) and 3), we can equate R - 3 and 10:
R - 3 = 10
R = 10 + 3
R = 13
From equations 2) and 3), we can equate G + 5 and 10:
G + 5 = 10
G = 10 - 5
G = 5
Now that we have the values of R and G, we can substitute them into equation 1) to find the value of W:
W = R - 3
W = 13 - 3
W = 10
Therefore, the number of marbles in the box is:
R + G + W = 13 + 5 + 10 = 28
Hence, the correct answer is (b) 28.
- In aid of charity, every student in a class contributes as many rupees as the number of students in that class. With the additional contribution of Rs. 2 by one student only, the total collection is Rs. 443. Then how many students are there in the class ?(2016)
(A) 12
(B) 21
(C) 43
(D) 45
Answer-B
Let's assume the number of students in the class is x.
According to the given information, each student contributes x rupees, so the total collection before the additional contribution is given by:
Total collection = x * x = x^2 rupees
Now, one student contributes an additional 2 rupees, so the total collection becomes:
x^2 + 2 rupees
According to the question, the total collection is 443 rupees, so we can write:
x^2 + 2 = 443
x^2 = 441
Taking the square root of both sides, we get:
x = 21
- In a class, there are 18 very tall boys. If these constitute three-fourths of the boys and the total number of boys is two-thirds of the total number of students in the class, what is the number of girls in the class ?(2016)
(A) 6
(B) 12
(C) 18
(D) 21
Answer-B
Let's assume the total number of boys in the class is B and the total number of students in the class is S.
According to the given information, 18 boys constitute three-fourths of the total number of boys. This can be expressed as:
18 = (3/4) * B
To find the total number of boys, we can solve this equation:
B = (18 * 4) / 3
B = 24
The total number of boys in the class is 24.
The total number of boys is also given as two-thirds of the total number of students, which can be expressed as:
B = (2/3) * S
To find the total number of students, we can solve this equation:
S = (24 * 3) / 2
S = 36
The total number of students in the class is 36.
Now, to find the number of girls in the class, we subtract the number of boys from the total number of students:
Number of girls = S - B
Number of girls = 36 - 24
Number of girls = 12
- The sum of income of A and B is more than that of C and D taken together. The sum of income of A and C is the same as that of B and D taken together. Moreover, A earns half as much as the sum of the income of B and D. Whose income is the highest?(2017)
(A) A
(B) B
(C) C
(D) D
Answer-B
Let's assume the incomes of A, B, C, and D are represented by a, b, c, and d, respectively.
According to the given information, the sum of income of A and B is more than that of C and D taken together. This can be expressed as:
a + b > c + d
The sum of income of A and C is the same as that of B and D taken together. This can be expressed as:
a + c = b + d
Moreover, A earns half as much as the sum of the income of B and D. This can be expressed as:
a = (b + d) / 2
We can use these three equations to find the relative incomes of A, B, C, and D.
Substituting the third equation into the first equation, we get:
(b + d) / 2 + b > c + d
Simplifying this equation, we get:
3b/2 - c/2 > d/2
Multiplying both sides by 2, we get:
3b - c > d
Substituting the second equation into this inequality, we get:
3b - (a + c) > d
(From Equation one and two it is clear that 2a = b+d, Therefore a+c =2a------> a=c)
Substituting the third equation into this inequality, we get:
3b - 2a > d
Therefore, we can see that the income of D is less than three times the income of B minus twice the income of A.
Since A earns half as much as the sum of the income of B and D, we can see that the income of B is the highest.
- There are three pillars X, Y and Z of different heights. Three spiders A, B and C start to climb on these pillars simultaneously. In one chance, A climbs on X by 6 cm but slips down 1 cm ,B climbs on Y by 7 cm but slips down 3 cm. C climbs on Z by 6.5 cm but slips down 2 cm. If each of them requires 40 chances to reach the top of the pillars, what is the height of the shortest pillar?(2017)
(A) 161 cm
(B) 163 cm
(C) 182 cm
(D) 210 cm
Answer-B
Let's analyze the climbing process of each spider. We'll assume the height of pillar X is x cm, the height of pillar Y is y cm, and the height of pillar Z is z cm.
In one chance, spider A climbs 6 cm but slips down 1 cm, so it effectively climbs 6 - 1 = 5 cm.
Similarly, spider B climbs 7 cm but slips down 3 cm, so it effectively climbs 7 - 3 = 4 cm.
Spider C climbs 6.5 cm but slips down 2 cm, so it effectively climbs 6.5 - 2 = 4.5 cm.
Since each spider requires 40 chances to reach the top, we can calculate the effective climbing distance for each spider using the formula:
Effective Climbing Distance = (Total Height - Total Slipping Distance) / Total Chances
For spider A:
5 cm = (x - 1) / 40
For spider B:
4 cm = (y - 3) / 40
For spider C:
4.5 cm = (z - 2) / 40
Multiplying both sides of each equation by 40, we get:
200 = x - 1 ...(1)
160 = y - 3 ...(2)
180 = z - 2 ...(3)
Solving these equations, we find:
x = 201 cm
y = 163 cm
z = 182 cm
Therefore, the height of the shortest pillar is 163 cm.
So, the correct option is (b) 163 cm.
- Six boys A, B, C, D, E and F play a game of cards. Each has a pack of 10 cards. F borrows 2 cards from A and gives away 5 to C who in turn gives 3 to B while B gives 6 to D who passes on 1 to E. Then the number of cards possessed by D and E is equal to the number of cards possessed by ?(2017)
(A) A, B and C
(B) B, C and F
(C) A, B and F
(D) A, C and F
Answer-B
Let's track the changes in the number of cards for each person:
Initially:
A: 10 cards
B: 10 cards
C: 10 cards
D: 10 cards
E: 10 cards
F: 10 cards
F borrows 2 cards from A:
A: 8 cards
B: 10 cards
C: 10 cards
D: 10 cards
E: 10 cards
F: 12 cards
F gives away 5 cards to C:
A: 8 cards
B: 10 cards
C: 15 cards
D: 10 cards
E: 10 cards
F: 7 cards
C gives 3 cards to B:
A: 8 cards
B: 13 cards
C: 12 cards
D: 10 cards
E: 10 cards
F: 7 cards
B gives 6 cards to D:
A: 8 cards
B: 7 cards
C: 12 cards
D: 16 cards
E: 10 cards
F: 7 cards
D gives 1 card to E:
A: 8 cards
B: 7 cards
C: 12 cards
D: 15 cards
E: 11 cards
F: 7 cards
Now we can see that the number of cards possessed by D and E (15 and 11, respectively) is equal to the number of cards possessed by B, C, and F (7, 12, and 7, respectively).
Therefore, the answer is (b) B, C, and F.
- There are certain 2-digit numbers. The difference between the number and the one obtained on reversing it is always 27. How many such maximum 2-digit numbers are there?(2017)
(A) 3
(B) 4
(C) 5
(D) None of the above
Answer-D
Let's consider a 2-digit number in the form of AB, where A represents the tens digit and B represents the units digit. The number obtained by reversing it will be BA.
According to the given condition, the difference between the number AB and the reversed number BA is always 27. Mathematically, we can express this as:
(10A + B) - (10B + A) = 27
Simplifying this equation, we get:
9A - 9B = 27
Dividing both sides by 9, we have:
A - B = 3
From this equation, we can see that the difference between the tens digit (A) and the units digit (B) is always 3. We need to find the maximum 2-digit numbers that satisfy this condition.
The valid 2-digit numbers are 41, 52, 63, 74, 85, and 96. So, there are six maximum 2-digit numbers that satisfy the given condition.
Therefore, the answer is (d)
- What is the total number of digits printed, if a book containing 150 pages is to be numbered from 1 to 150?(2017)
(A) 262
(B) 342
(C) 360
(D) 450
Answer-B
To find the total number of digits printed in the book, we need to count the number of digits in the page numbers from 1 to 150.
We can observe that the number of digits required to represent a page number depends on the range of the page numbers. Let's analyze this:
For page numbers 1 to 9, we need 1 digit per page number (total of 9 digits).
For page numbers 10 to 99, we need 2 digits per page number (total of 90 x 2 = 180 digits).
For page numbers 100 to 150, we need 3 digits per page number (total of 51 x 3 = 153 digits).
Now, let's calculate the total number of digits by adding up the digits required for each range:
9 digits + 180 digits + 153 digits = 342 digits
Therefore, the total number of digits printed in the book is 342.
- A number consists of three digits of which the middle one is zero and their sum is 4. If the number formed by interchanging the first and last digits is greater than the number itself by 198, then the difference between the first and last digits is?(2018)
(A) 1
(B) 2
(C) 3
(D) 4
Answer-B
Let the three-digit number be ABC, where B is the middle digit and A and C are the first and last digits, respectively. We know that B = 0.
From the given information, we have:
A + B + C = 4 (Equation 1)
100C + 10B + A - (100A + 10B + C) = 198 (Equation 2)
Simplifying Equation 2, we get:
99C - 99A = 198
Dividing both sides by 99, we get:
C - A = 2
From Equation 1, we know that A + B + C = 4. Since B = 0, we have:
A + C = 4
Adding this equation to C - A = 2, we get:
2C = 6
Solving for C, we get:
C = 3
Substituting C = 3 in A + C = 4, we get:
A = 1
Therefore, the three-digit number is 103.
The number formed by interchanging the first and last digits is 301, which is greater than 103 by 198.
The difference between the first and last digits is:
C - A = 3 - 1 = 2
Therefore, the answer is (b) 2.
- A lift has the capacity of 18 adults or 30 children. How many children can board the lift with 12 adults?(2018)
(A) 6
(B) 10
(C) 12
(D) 15
Answer-B
To find the capacity of the lift, find LCM of 18 and 30 = 90x
1 adult = 90x/18 = 5x
1 child = 90x /30 = 3x
If, the lift is occupied by 12 adults, 12 × 5x = 60x
Remaining capacity will be = 90x - 60x = 30x
So, the number of children that be accommodated with 12 adults are: 30x/3x = 10
- If x - y= 8, then which of the following must be true?
- Both x and y must be positive for any value of x and y.
- If x is positive, y must be negative for any value of x and y.
- If x is negative, y must be positive for any value of x and y.
Select the correct answer using the code given below.(2018)
(A) 1 only
(B) 2 only
(C) Both 1 and 2
(D) Neither 1 nor 2 nor 3
Answer-D
To determine which statement must be true given the equation x - y = 8, we can analyze the possible scenarios.
Let's consider the given options:
Both x and y must be positive for any value of x and y.
This statement is not necessarily true. There are multiple solutions to the equation x - y = 8. For example, if x = 10 and y = 2, both x and y are positive. However, if x = 5 and y = -3, x is positive but y is negative. Therefore, statement 1 is not always true.
If x is positive, y must be negative for any value of x and y.
This statement is also not true. Using the same example as above, if x = 10 and y = 2, both x and y are positive. Therefore, statement 2 is not always true.
If x is negative, y must be positive for any value of x and y.
This statement is not always true. If x = -10 and y = -18, both x and y are negative. Therefore, statement 3 is not true.
Considering the analysis above, the correct answer is (d) Neither 1 nor 2 nor 3.
- Rakesh and Rajesh together bought 10 balls and 10 rackets. Rakesh spent Rs 1300 and Rajesh spent Rs 1500. If each racket costs three times a ball does, then what is the price of a racket?(2019)
(A) Rs 70
(B) Rs 90
(C) Rs 210
(D) Rs 240
Answer-C
Let's assume the price of a ball is 'b' rupees.
Since each racket costs three times as much as a ball, the price of a racket would be 3b rupees.
According to the given information:
Rakesh and Rajesh together bought 10 balls and 10 rackets. This implies that Rakesh bought 'r' rackets and Rajesh bought '10 - r' rackets, as the total number of rackets bought by both is 10.
Rakesh spent 1300 rupees, so we can set up the equation:
r * (3b) + 10 * b = 1300
Similarly, Rajesh spent 1500 rupees, so we can set up the equation:
(10 - r) * (3b) + 10 * b = 1500
Let's simplify these equations:
3br + 10b = 1300 --(1)
30b - 3br + 10b = 1500 --(2)
Now, let's solve these equations to find the price of a racket:
Adding equation (1) and equation (2):
3br + 30b - 3br + 10b = 1300 + 1500
40b = 2800
Dividing both sides by 40:
b = 2800 / 40
b = 70
Therefore, the price of a ball is 70 rupees.
Since each racket costs three times as much as a ball, the price of a racket is:
3b = 3 * 70 = 210 rupees
Hence, the price of a racket is 210 rupees.
- If x is greater than or equal to 25 and y is less than or equal to 40, then which one of the following is always correct?(2019)
(A) x is greater than y
(B) (y -x) is greater than 15
(C) (y - ×) is less than or equal to 15
(D) (x + y) is greater than or equal to 65
Answer-C
Given conditions x ≥ 25 and y ≤ 40, we analyze each option:
(a) x is greater than y
This statement is not always correct. If x = 25 and y = 40, x is not greater than y. Therefore, statement (a) is not always true.
(b) (y - x) is greater than 15
This statement is not always correct. If x = 25 and y = 40, (y - x) = (40 - 25) = 15, which is not greater than 15. Therefore, statement (b) is not always true.
(c) (y - x) is less than or equal to 15
This statement is always correct. Since y ≤ 40 and x ≥ 25, the maximum difference between y and x would occur when y is at its maximum value of 40 and x is at its minimum value of 25. In this case, (y - x) = (40 - 25) = 15, which is less than or equal to 15. Therefore, statement (c) is always true.
(d) (x + y) is greater than or equal to 65
This statement is not always correct. If x = 25 and y = 40, (x + y) = (25 + 40) = 65, which is equal to 65. Therefore, statement (d) is not always true.
- A printer numbers the pages of a book starting with 1 and uses 3089 digits in all. How many pages does the book have?(2019)
(A) 1040
(B) 1048
(C) 1049
(D) 1050
Answer-C
To solve this problem, we need to determine the number of pages in the book given that the printer uses 3089 digits in total.
Let's consider the number of digits required to represent the page numbers.
For single-digit page numbers (from 1 to 9), we need 1 digit per page, which totals 9 digits.
For double-digit page numbers (from 10 to 99), we need 2 digits per page, which totals 90 * 2 = 180 digits.
For three-digit page numbers (from 100 to 999), we need 3 digits per page, which totals 900 * 3 = 2700 digits.
To find the number of pages, we can subtract the total number of digits used for single and double-digit page numbers from the total number of digits used:
3089 - (9 + 180 + 2700) = 3089 - 2889 = 200
This means that the remaining 200 digits are used for four-digit page numbers.
Since each four-digit page number requires 4 digits, we divide the remaining digits by 4:
200 / 4 = 50
Therefore, the book has 50 four-digit pages.
To find the total number of pages, we add the number of single, double, three, and four-digit pages:
9 + 90 + 900 + 50 = 1049
Hence, the correct answer is (c) 1049.
- One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1.The sum of the numbers on the remaining pages is 195. The torn page contains which of the following numbers?(2020)
(A) 5, 6
(B) 7, 8
(C) 9, 10
(D) 11, 12
Answer-B
Sum of 1st n natural numbers = (n/2)(n + 1)
According to the question, the sum is 195 But there is a page which is torn up So we consider the sum of number of pages in book be x So, total sum will be 195 + x
So, 195 + x = (n/2)(n + 1)
Now at 20, sum of pages = 210 [(20/2) × 21 = 210)]
So the number of pages must be 20
Now we can assume that 195 + x = 210 ⇒ x = 15
Here in option (b) the sum of pages is 15 (page 7 and 8)
- Let A3BC and DE2F be four-digit numbers where each letter represents a different digit greater than 3. If the sum of the numbers is 15902, then what is the difference between the values of A and D?(2020)
(A) 1
(B) 2
(C) 3
(D) 4
Answer-C
A3BC + DE2F = 15902
Since A and D are the thousands digits of the two numbers, we can write:
A000 + 3BC + D000 + E2F = 15902
A + D + 3 + E + B + C = 15.9
We know that all the digits are greater than 3 and different from each other. So, the minimum possible value of A and D is 4 and the maximum possible value is 8.
Try different values of A and D to see which ones satisfy the above equation. Starting with A = 4, we get:
4 + D + 3 + E + B + C = 15.9
D + E + B + C = 8.9
Since all the digits are different and greater than 3, we can see that the possible values of D and E are 5 and 6 in some order. If D = 5 and E = 6, we get:
5 + B + C = -1.9
This is not possible since B and C are digits greater than 3. So, we can try D = 6 and E = 5:
6 + B + C = -0.9
This gives us B + C = 2.9. The only digits greater than 3 that add up to 2.9 are 7 and 6. So, we have:
B = 7 and C = 6
Substituting these values, we get:
A + D = 2.9
The only possible values of A and D that satisfy this equation are A = 7 and D = 4. Therefore, the difference between A and D is:
A - D = 7 - 4 = 3
- Let x, y be the volumes; m, n be the masses of two metallic cubes P and O respectively. Each side of Q is two times that of P and the mass of Q is two times that of P.
Let u = m /x and v = n / y.
Which one of the following is correct?(2020)
(a) u = 4v
(b) u = 2u
(c) v= u
(d) v= 4u
Answer-A
Let x, y be the volumes of cubes P and Q respectively.
Let m, n be the masses of P and Q respectively.
Each side of Q is two times that of P and the mass of Q is two times that of P
The volume of a cube = side^3
Let the side of cube P be 'a'
Let the side of cube Q be '2a'
Volume of cube P(x) = a3
Volume of cube Q(y) = (2a)3 = 8a3
Mass of P(m) = c
Mass of Q(n) = 2c (mass of Q is 2 times the mass of P)
Now, u = m/x = c/a3 and v = n/y = (2c)/8a3 = (c)/4a3
Here, v = u/4
Option 1 is the correct answer.
- A vessel full of water weighs 40 kg. If it is one-third filled, its weight becomes 20 kg. What is the weight of the empty vessel?(2020)
(A) 10 kg
(B) 15 kg
(C) 20 kg
(D) 25 kg
Answer-A
Let's assume the weight of the empty vessel is "x" kg.
When the vessel is completely filled with water, its weight would be the weight of the empty vessel plus the weight of the water, which is 40 kg. Therefore, we can write the equation:
x + weight of water = 40 kg
Now, we are given that when the vessel is one-third filled with water, its weight is 20 kg. We can set up another equation using this information:
x + (1/3) * weight of water = 20 kg
To find the weight of the empty vessel, we need to solve these two equations simultaneously.
Let's substitute the value of weight of water from the second equation into the first equation:
x + (1/3) * (40 kg - x) = 20 kg
3x + 40 kg - x = 60 kg
2x + 40 kg = 60 kg
2x = 20 kg
x = 10 kg
- A frog tries to come out of a dried well 4.5 m deep with slippery walls. Every time the frog jumps 30 cm, slides down 15 cm. What is the number of jumps required for the frog to come out of the well?(2020)
(A) 28
(B) 29
(C) 30
(D) 31
Answer-B
Each jump, the frog covers a distance of 30 cm, but then slides down 15 cm. This means that in each jump, the net distance covered by the frog is 30 cm - 15 cm = 15 cm.
Depth of the well is 4.5 meters * 100 centimeters/meter = 450 centimeters.
After 28 jumps, the frog climbs
28 x (30 – 15) = 28 x 15 = 420 cm.
In the 29th jump, the frog will reach the top of the well, so it won’t slip back on slippery walls.
So, The correct option is B 29
- A person P asks one of his three friends X as to how much money he had. X replied, "If Y gives me Rs 40, then Y will have half of as much as Z, but if Z gives me Rs 40, then three of us will have equal amount." What is the total amount of money that X, Y and Z have ?(2021)
(A) Rs 420
(B) Rs 360
(C) Rs 300
(D) Rs 270
Answer-B
Let's assume that X initially has 'x' amount of money, Y has 'y' amount of money, and Z has 'z' amount of money.
According to the given information:
"If Y gives me Rupees 40, then Y will have half as much as Z."
This can be represented as: y - 40 = (1/2)(z) i.e, y=(1/2)(z) + 40—-------(1)
"If Z gives me Rupees 40, then three of us will have an equal amount."
This can be represented as: x+40 = y = (z - 40)---------------(2)
Comparing y = 1/2z + 40 from equation (1) and y = z - 40 from equation (2)
⇒ 1/2z + 40 = z - 40
⇒ z - 1/2z = 40 + 40
⇒ z/2 = 80
⇒ z = 160
Putting value of z in equation (2)
⇒ x + 40 = y = 160 - 40
⇒ x + 40 = y = 120
Hence, y = 120
and x = y - 40 = 120 - 40 = 80
Therefore, total amount of money with that X, Y and Z have = x + y + z
= 80 + 120 + 160
= Rs. 360
- In an objective type test of 90 questions, 5 marks are allotted for every correct answer and 2 marks are deducted for every wrong answer. After attempting all the 90 questions, a student got a total of 387 marks. What is the number of incorrect responses ?(2021)
(A) 9
(B) 13
(C) 27
(D) 43
Answer-A
Let's assume the number of correct answers the student gave is 'x'. The number of incorrect answers can be represented as '90 - x'.
For each correct answer, the student receives 5 marks, so the total marks for correct answers would be 5x.
For each incorrect answer, the student loses 2 marks, so the total marks deducted for incorrect answers would be 2(90 - x).
According to the given information, the student got a total of 387 marks. We can form the equation:
5x - 2(90 - x) = 387
Expanding and simplifying the equation, we have:
5x - 180 + 2x = 387
7x - 180 = 387
7x = 567
x = 81
Therefore, the student gave 81 correct answers.
To find the number of incorrect responses:
Number of incorrect responses = Total number of questions - Number of correct responses
Number of incorrect responses = 90 - 81
Number of incorrect responses = 9
Therefore, the correct answer is (a) 9.
- Consider the following addition problem: 3P + 4P + PP + PP = RQ2; where P, Q and R are different digits.What is the arithmetic mean of all such possible sums?(2021)
(A) 102
(B) 120
(C) 202
(D) 220
Answer-C
P is added 4 times on LHS resulting 2 in units place on RHS
The values of P such that P × 4 will give 2 in units place are 3 and 8.
If P = 3, then on LHS we have 33 + 43 + 33 + 33 = 142
When R = 1, Q = 4 and the number is 142
If P = 8, then LHS = 38 + 48 + 88 + 88 = 262
when R = 2, Q = 6 and the number is 262
arithmetic mean = (142 + 262) divided by 2 = 202.
- There are three points P, Q and R on a straight line such that PQ: QR =3: 5. If n is the number of possible values of PQ: PR, then what is n equal to ?(2021)
(A) 1
(B) 2
(C) 3
(D) 4
Answer-B
There are three points P, Q and R on a straight line such that PQ : QR = 3:5.
n is the number of possible values of PQ : PR,
To Find : Value of n
Solution:
PQ : QR = 3:5.
PQ = 3x and QR = 5x
Case 1 : Q is in between P and R
P ---3x --- Q --- 5x ---- R
PR = PQ + QR = 3x + 5x = 8x
PQ : PR = 3x : 8x
=> PQ:QR = 3 : 8
Case 2 :
R is between P and Q
not possible as QR > PQ
Case 3 :
P is between Q and R
Q --- 3x --- P --2x -- R
QR = PQ + PR
=> 5x = 3x + PR
=> PR = 2x
PQ : QR = 3x : 2x = 3: 2
PQ:QR = 3 : 8 or 3:2
Hence two possible values
n = 2
- An amount of money was distributed among A, B and C in the ratio p: q: r.
Consider the following statements:
- A gets the maximum share if p is greater than (q + r).
- C gets the minimum share if r is less than (p + q).
Which of the above statements is/are correct?(2021)
(A) 1 only
(B) 2 only
(C) Both 1 and 2
(D) Neither 1 nor 2
Answer-A
To analyze the given statements, let's consider the ratio of distribution as p:q:r, where p, q, and r represent the respective shares of A, B, and C.
1.A gets the maximum share if p is greater than (q + r).
This statement is correct. If p is greater than the sum of q and r, it means that A's share is the largest among the three. In other words, A receives a larger portion of the total amount compared to B and C.
2.C gets the minimum share if r is less than (p + q).
This statement is incorrect. If r is less than the sum of p and q, it does not necessarily imply that C receives the minimum share. The distribution depends on the specific values of p, q, and r, and their respective ratios. C's share could still be larger than or equal to B's share.
- The difference between a 2-digit number and the number obtained by interchanging the positions of the digits is 54.
Consider the following statements :
- The sum of the two digits of the number can be determined only if the product of the two digits is known.
- The difference between the two digits of the number can be determined.
Which of the above statements is/are correct?(2021)
(A) 1 only
(B) 2 only
(C) Both 1 and 2
(D) Neither 1 nor 2
Answer-B
To solve this problem, let's assume the original 2-digit number is represented as "10a + b," where "a" and "b" represent the tens and units digits, respectively.
According to the problem, the number obtained by interchanging the positions of the digits is "10b + a."
The given information states that the difference between these two numbers is 54:
(10a + b) - (10b + a) = 54
Expanding the equation, we have:
10a + b - 10b - a = 54
9a - 9b = 54
Dividing both sides of the equation by 9:
a - b = 6 ----(1)
Now, let's evaluate the given statements:
The sum of the two digits of the number can be determined only if the product of the two digits is known.
This statement is incorrect. From equation (1), we know that a - b = 6. It implies that the difference between the two digits is known (which is 6), but the product of the digits is not required to find their sum.
The difference between the two digits of the number can be determined.
This statement is correct. Equation (1) shows that the difference between the two digits is 6.
Therefore, the correct answer is (b) 2 only.
- A biology class at high school predicted that a local population of animals will double in size every 12 years. The population at the beginning of the year 2021 was estimated to be 50 animals. If P represents the population after n years, then which one of the following equations represents the model of the class for the population?(2021)
(A) P = 12 + 50n
(B) P = 50 + 12n
(C) P = 50 (2)(power)12n
(D) P = 50 *(2)(power)n/12
Answer-D
The equation that represents the model for the population growth according to the given information is:
P = 50 * (2^(n/12))
Let's break down the equation to understand it better:
P represents the population after n years.
The initial population in 2021 is 50 animals, so we start with 50.
The population doubles every 12 years, which is represented by 2^(n/12). This is because the exponent n/12 accounts for the number of 12-year periods that have passed, and raising 2 to that power reflects the doubling of the population.
Therefore, the equation that represents the model for the population growth is:
P = 50 * (2^(n/12))
- A boy plays with a ball and he drops it from a height of 1.5 m. Every time the ball hits the ground, it bounces back to attain a height 4/5th of the previous height. The ball does not bounce further if the previous height is less than 50 cm. What is the number of times the ball hits the ground before the ball stops bouncing?(2021)
(A) 4
(B) 5
(C) 6
(D) 7
Answer-B
To solve this problem, we can start by determining the maximum number of bounces the ball can make. Since the ball stops bouncing when its height is less than 50 cm, we need to find the point at which the height becomes less than 50 cm.
Let's write down the heights of the ball after each bounce:
1st bounce: 4/5 * 1.5 m = 1.2 m
2nd bounce: 4/5 * 1.2 m = 0.96 m
3rd bounce: 4/5 * 0.96 m = 0.768 m
4th bounce: 4/5 * 0.768 m = 0.6144 m
5th bounce: 4/5 * 0.6144 m = 0.49152 m
After the 5th bounce, the height is 0.49152 m, which is less than 50 cm (0.5 m). Therefore, the ball stops bouncing after the 5th bounce.
The correct answer is (b) 5
- On one side of a 1:01 km long road, 101 plants are planted at equal distance from each other. What is the total distance between 5 consecutive plants?(2022)
(A) 40 m
(B) 40.4 m
(C) 50 m
(D) 50.5 m
Answer-B
To find the total distance between 5 consecutive plants, we need to determine the distance between each plant and then add them up.
The total length of the road is 1.01 km, which is equivalent to 1010 meters (1 km = 1000 m). Since there are 101 plants planted at equal distances from each other, the distance between each plant can be calculated by dividing the total length of the road by the number of intervals between plants:
Distance between each plant = Total length / (Number of plants - 1)
= 1010 m / (101 - 1)
= 1010 m / 100
= 10.1 m
Now that we know the distance between each plant is 10.1 meters, we can calculate the total distance between 5 consecutive plants by multiplying the distance between each plant by 4 (as there are 4 intervals between 5 consecutive plants):
Total distance between 5 consecutive plants = Distance between each plant * 4
= 10.1 m * 4
= 40.4 m
- A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two-digit number. How many coins does A have in the beginning?(2022)
(A) 76
(B) 68
(C) 60
(D) 52
Answer-D
Let's assume that A starts with "x" number of coins.
A gives half of the coins and 2 more to B:
A has (x/2 - 2) coins left, and B now has (x/2 + 2) coins.
B gives half of the coins and 2 more to C:
B has ((x/2 + 2)/2 - 2) coins left, and C now has ((x/2 + 2)/2 + 2) coins.
C gives half of the coins and 2 more to D:
C has (((x/2 + 2)/2 + 2)/2 - 2) coins left, and D now has (((x/2 + 2)/2 + 2)/2 + 2) coins.
The problem states that the number of coins D has is the smallest two-digit number. So, we can set up the following inequality:
(((x/2 + 2)/2 + 2)/2 + 2) ≥ 10
Now, let's solve this inequality to find the minimum value of x:
(((x/2 + 2)/2 + 2)/2 + 2) ≥ 10
((x/2 + 2)/2 + 2)/2 ≥ 8
(x/2 + 2)/2 + 2 ≥ 16
(x/2 + 2)/2 ≥ 14
x/2 + 2 ≥ 28
x/2 ≥ 26
x ≥ 52
So, the minimum number of coins A must have in the beginning is 52.
- Five friends P, Q, X, Y and Z purchased some notebooks.
The relevant information is given below :
- Z purchased 8 notebooks more than X did.
- P and Q together purchased 21 notebooks.
- Q purchased 5 notebooks less than P did.
- X and Y together purchased 28 notebooks.
- P purchased 5 notebooks more than X did.
If each notebook is priced Rs 40, then what is the total cost of all the notebooks?(2022)
(A) Rs 2,600
(B) Rs 2,400
(C) Rs 2,360
(D) Rs 2,320
Answer-A
Let's assign variables to represent the number of notebooks purchased by each friend:
Let the number of notebooks purchased by P be Pn.
Let the number of notebooks purchased by Q be Qn.
Let the number of notebooks purchased by X be Xn.
Let the number of notebooks purchased by Y be Yn.
Let the number of notebooks purchased by Z be Zn.
We need to find the total cost of all the notebooks, which is equal to the sum of the cost of notebooks purchased by each friend.
Given information:
Z purchased 8 notebooks more than X did: Zn = Xn + 8.
P and Q together purchased 21 notebooks: Pn + Qn = 21.
Q purchased 5 notebooks less than P did: Qn = Pn - 5.
X and Y together purchased 28 notebooks: Xn + Yn = 28.
P purchased 5 notebooks more than X did: Pn = Xn + 5.
We can use this information to solve for the values of Pn, Qn, Xn, Yn, and Zn.
From equation 2, we have Qn = 21 - Pn.
Substituting this into equation 3, we get 21 - Pn = Pn - 5.
Simplifying, we find 2Pn = 26, which gives us Pn = 13.
Using equation 5, we have Pn = Xn + 5.
Substituting Pn = 13, we get 13 = Xn + 5.
Solving for Xn, we find Xn = 8.
From equation 1, we have Zn = Xn + 8.
Substituting Xn = 8, we get Zn = 8 + 8 = 16.
Using equation 4, we have Xn + Yn = 28.
Substituting Xn = 8, we find 8 + Yn = 28.
Solving for Yn, we get Yn = 20.
Now we know the number of notebooks purchased by each friend:
Pn = 13, Qn = 21 - Pn = 21 - 13 = 8,
Xn = 8, Yn = 20, and Zn = 16.
The total cost of all the notebooks is given by:
Total cost = (Pn + Qn + Xn + Yn + Zn) * Notebook price
= (13 + 8 + 8 + 20 + 16) * 40
= 65 * 40
= 2600.
Therefore, the total cost of all the notebooks is Rs 2,600. The correct option is (a) Rs 2,600.
- A person X wants to distribute some pens among six children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number?(2022)
(A) 147
(B) 150
(C) 294
(D) 300
Answer-C
Let the number of pens received by child B be x. Then A will receive 2x pens, C will receive (1/3)*2x pens, D will receive (1/4)*2x pens, E will receive (1/5)*2x pens and F will receive (1/6)*2x pens.
We want to find the minimum number of pens X should buy so that each child receives an even number of pens. This means that x should be an even number.
Let's find the least common multiple (LCM) of 2, 3, 4, 5 and 6:
LCM(2,3,4,5,6) = 60
This means that x must be a multiple of 60 in order for each child to receive an even number of pens.
Now, we can substitute x = 60 in the expressions for the number of pens received by each child:
B ha 60 pens
A will receive 2x = 120 pens,
C will receive (1/3)*2x = 40 pens,
D will receive (1/4)*2x = 30 pens,
E will receive (1/5)*2x = 24 pens, and
F will receive (1/6)*2x = 20 pens.
Total 294 pens
- What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?(2022)
(A) 1063
(B) 1073
(C) 1083
(D) 1183
Answer-C
To find the smallest number greater than 1000 that leaves a remainder of 3 when divided by any one of the numbers 6, 9, 12, 15, and 18, we need to find the least common multiple (LCM) of these numbers and then add 3 to it.
The LCM of 6, 9, 12, 15, and 18 is 180. Therefore, any number that is a multiple of 180 will leave a remainder of 3 when divided by any one of these numbers.
To find the smallest number greater than 1000 that is a multiple of 180, we can start by dividing 1000 by 180:
1000 ÷ 180 = 5 remainder 100
Since the remainder is not zero, we need to add the difference between the divisor (180) and the remainder (100) to 1000 to find the next multiple of 180:
1000 + (180 - 100) = 1080
So, the smallest number greater than 1000 that is a multiple of 180 is 1080.
To get the smallest number greater than 1000 that leaves a remainder of 3 when divided by any one of the given numbers, we add 3 to 1080:
1080 + 3 = 1083
- An Identity Card has the number ABCDEFG, not necessarily in that order, where each letter represents a distinct digit (1, 2, 4, 5, 7, 8, 9 only). The number is divisible by 9. After deleting the first digit from the right, the resulting number is divisible by 6. After deleting two digits from the right of original number, the resulting number is divisible by 5. After deleting three digits from the right of original number, the resulting number is divisible by 4. After deleting four digits from the right of original number, the resulting number is divisible by 3. After deleting five digits from the right of original number, the resulting number is divisible by 2. Which of the following is a possible value for the sum of the middle three digits of the number?(2022)
(A) 8
(B) 9
(C) 11
(D) 12
Answer-A
The identity Card number ABCDEFG is a 7 digit number.
Step 1: The number is divisible by 9, which means the sum of digits is divisible by 9.
1 + 2 + 4 + 5 + 7 + 8 + 9 = 36.
Step 2: After deleting the first digit from the right, the resulting number is divisible by 6.
The number is divisible by 6, which means the number that is divisible by 3 and 2 both. If we delete any single-digit number from a number divisible by 9, the number is only divisible by 3 if the deleting digit is also divisible by 3, so that sum is divisible by 3.
Out of (1, 2, 4, 5, 7, 8, 9) only 9 is divisible by 3, So, the first digit from the right is 9 and the second digit from the right must be even as it is also divisible by 2.
Step 3: After deleting two digits from the right of the original number, the resulting number is divisible by 5.
The number is divisible by 5, which means the last digit is 0 or 5. So, the third digit from the right must be 5.
Step 4: After deleting three digits from the right of the original number, the resulting number is divisible by 4.
The number is divisible by 4, which means the last two digits are divisible by 4.
Step 5: After deleting four digits from the right of the original number, the resulting number is divisible by 3.
The number is divisible by 3, which means the sum of digits is divisible by 3. So the sum of the first three digits of the number from the right is divisible by 3.
Step 6: After deleting five digits from the right of the original number, the resulting number is divisible by 2 which means the second digit is even.
Possible outcomes are 7412589 and 1472589. Possible values for the sum of the middle 3 digits of the number are 8 (7412589) and 14 (1472589).
- If R and S are different integers both divisible by 5, then which of the following is not necessarily true?(2016)
(A) R - S is divisible by 5
(B) R + S is divisible by 10
(C) R × S is divisible by 25
(D) R2 + S2 is divisible by 5
Answer-B
Let's consider the options one by one:
(A) R - S is divisible by 5:
Since both R and S are divisible by 5, their difference (R - S) will also be divisible by 5. So, option (a) is necessarily true.
(B) R + S is divisible by 10:
Since both R and S are divisible by 5, their sum (R + S) will also be divisible by 5. However, it may not necessarily be divisible by 10. For example, if R = 10 and S = -10, both divisible by 5, then R + S = 10 + (-10) = 0, which is not divisible by 10. So, option (b) is not necessarily true.
(C) R × S is divisible by 25:
Since both R and S are divisible by 5, their product (R × S) will also be divisible by 5. However, it may not necessarily be divisible by 25. For example, if R = 5 and S = -5, both divisible by 5, then R × S = 5 × (-5) = -25, which is divisible by 25. So, option (c) is necessarily true.
(D) R^2 + S^2 is divisible by 5:
Since both R and S are divisible by 5, their squares (R^2 and S^2) will also be divisible by 5. So, option (d) is necessarily true.
Therefore, the option that is not necessarily true is (b) R + S is divisible by 10.
- How many numbers are there between 100 and 300 which either begin with or end with 2?(2016)
(A) 110
(B) 111
(C) 112
(D) None of the above
Answer-A
To find the numbers between 100 and 300 that either begin with or end with 2, we can consider two cases:
Numbers that begin with 2:
The numbers in this case range from 200 to 299, and there are 100 numbers in total (from 200 to 299).
Numbers that end with 2:
The numbers in this case range from 102 to 292, and there are 10 numbers in total (102, 112, 122, ..., 292).
Therefore, the total number of numbers between 100 and 300 that either begin with or end with 2 is 100 + 10 = 110.
- There are thirteen 2-digit consecutive odd numbers. If 39 is the mean of the first five such numbers, then what is the mean of all the thirteen numbers?(2017)
(A) 47
(B) 49
(C) 51
(D) 45
Answer-A
The first five consecutive odd numbers have a mean of 39. Let's denote the first number as x. Then the five consecutive odd numbers are x, x+2, x+4, x+6, and x+8.
The mean of these five numbers is given by:
(1/5) * (x + (x+2) + (x+4) + (x+6) + (x+8)) = 39
5x + 20 = 195
5x = 175
x = 35
So, the first number in the sequence is 35, and the thirteen consecutive odd numbers are:
35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59.
To find the mean of all thirteen numbers, we sum them up and divide by 13:
(35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55 + 57 + 59) / 13 = 47
Therefore, the mean of all thirteen numbers is 47.
The correct answer is (a) 47.
- Certain 3-digit numbers have the following characteristics:
- All the three digits are different.
- The number is divisible by 7.
- The number on reversing the digits is also divisible by 7.
How many such 3-digit numbers are there?(2017)
(A) 2
(B) 4
(C) 6
(D) 8
Answer-B
We are looking for a 3-digit number of the form "100x + 10y + z," where x, y, and z are distinct digits. The number should satisfy the following conditions:
All three digits are different.
The number is divisible by 7.
The number, when its digits are reversed, is also divisible by 7.
To find the possible values for x, y, and z, we can use the divisibility rule for 7:
The number "100x + 10y + z" is divisible by 7 if and only if the number obtained by removing the last digit (z) and subtracting twice this digit from the remaining number (100x + 10y) is divisible by 7.
Mathematically, we can express this as:
(100x + 10y) - 2z = 7k, where k is an integer.
Simplifying this equation, we get:
99x + 10y - 2z = 7k.
Since 99 is not divisible by 7, the quantity (10y - 2z) must be divisible by 7. The only way this is possible is if (y - z) is equal to 7, as both y and z are single-digit numbers.
Now let's consider the possible values for (y - z):
If (y - z) = 7, the only possible values for (y, z) are (8, 1) and (9, 2).
Now 1_8 number at the unit place is 8 we should get carryover of 2 from the division of the number formed by the digits at hundreds and tens place to make the number become completely divisible by 7 (we want to make it 28 which is divisible by 7)
Thus the ten's digit should be 6 and the number shall be 168 the reverse number 861 also divisible by 7. Thus two of our required number are 168 and 861
Now in 2_9, the number at the unit's place is 9 so we should get a carryover of 4 from the division of the number formed by the digits at hundreds and tens place to make the number become completely divisible by 7 (We want to make it 49 which is divisible by 7).
Thus ten's digit should be 5 and number shall be 259 and the reverse number is 952 is also divisible by 7. Thus the other two number are 259 and 952
- How many numbers are there between 99 and 1000 such that the digit 8 occupies the units place?(2017)
(A) 64
(B) 80
(C) 90
(D) 104
Answer-C
To find the numbers between 99 and 1000 where the digit 8 occupies the units place, we need to consider the possible values for the hundreds and tens digits.
The hundreds digit can take values from 1 to 9, and the tens digit can take values from 0 to 9 (including 8).
Case 1: When the hundreds digit is less than 9 (1 to 8)
For each value of the hundreds digit, there is only one possible value for the units digit, which is 8. The tens digit can take any value from 0 to 9, including 8. Therefore, there are 10 possibilities for the tens digit.
Total numbers in this case = Number of possible hundreds digits × Number of possible tens digits × Number of possible units digits = 8 × 10 × 1 = 80
Case 2: When the hundreds digit is 9
In this case, the tens digit can take any value from 0 to 9, including 8, and the units digit is fixed as 8. Therefore, there are 10 possibilities for the tens digit.
Total numbers in this case = Number of possible tens digits × Number of possible units digits = 10 × 1 = 10
Adding the numbers from both cases, we get:
Total numbers = Total numbers in Case 1 + Total numbers in Case 2 = 80 + 10 = 90
Therefore, the correct answer is (c) 90.
- A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible remainder?(2017)
(A) 9
(B) 27
(C) 36
(D) 45
Answer-D
To maximize the remainder, we should make one of the two digits as large as possible, and it should be 9. This means the two numbers will be of the form 9x and x9, where x represents the other digit.
To ensure that the smaller number does not divide the larger number more than once, x should not be very small or very large. Let's consider the values x = 4 and x = 5.
If x = 5: The numbers are 95 and 59. When we divide 95 by 59, the remainder is 36.
If x = 4: The numbers are 94 and 49. When we divide 94 by 49, the remainder is 45.
Therefore, the highest possible remainder is 45, which is achieved when the numbers are 94 and 49.
- If X is between -3 and -1, and Y is between -1 and 1, then X^2 - y^2 is in between which of the following?(2018)
(A) -9 and 1
(B) -9 and -1
(C) 0 and 8
(D) 0 and 9
Answer-D
Given that X lies between -3 and -1, we can represent it as -3 ≤ X ≤ -1.
Squaring both sides of the inequality, we have 9 ≥ X^2 ≥ 1.
Similarly, for Y lying between -1 and 1, we have -1 ≤ Y ≤ 1.
Squaring both sides, we get 1 ≥ Y^2 ≥ 0.
Now, let's consider the expression X^2 - Y^2.
The maximum value of X^2 is 9, and the minimum value of Y^2 is 0. So, the maximum value of X^2 - Y^2 is 9 - 0 = 9.
Likewise, the minimum value of X^2 is 1, and the maximum value of Y^2 is 1. So, the minimum value of X^2 - Y^2 is 1 - 1 = 0.
Therefore, we can conclude that 0 ≤ X^2 - Y^2 ≤ 9.
- X and Y are natural numbers other than 1, and Y is greater than X. Which of the following represents the largest number?(2018)
(A) XY
(B) X/Y
(C) Y /X
(D) (X+Y) / XY
Answer-A
To determine which option represents the largest number, we can analyze each option separately.
(A) XY: This represents the product of X and Y. Since Y is greater than X, multiplying the two numbers will result in a larger number.
(B) X/Y: This represents the division of X by Y. Since Y is greater than X, the division will result in a fraction less than 1. Therefore, this option represents a number smaller than the product XY.
(C) Y/X: This represents the division of Y by X. As Y is greater than X, the division will result in a number greater than 1. This option represents a larger number compared to the division in option (b) but is still smaller than the product XY.
(D) (X+Y) / XY: This represents the sum of X and Y divided by their product. To determine if this option represents the largest number, let's analyze it further:
Considering that X and Y are natural numbers other than 1, we can assume that X = 2 and Y = 3. Substituting these values into the option, we get:
(X+Y) / XY = (2+3) / (2*3) = 5/6
Now, let's compare this value to the previous options:
XY = 2*3 = 6 (largest option so far)
X/Y = 2/3 (smaller than 5/6)
Y/X = 3/2 (greater than 5/6)
Since option (a) XY represents the largest number among the given options, the correct answer is (a) XY.
- While writing all the numbers from 700 to 1000, how many numbers occur in which the digit at hundreds place is greater than the digit at ten's place, and the digit at ten's place is greater than the digit at unit's place?(2018)
(A) 61
(B) 64
(C) 85
(D) 91
Answer-C
(i) Numbers from 700 to 800 where the hundreds digit is greater than the ten's digit:
In this range, the hundreds digit can be 7, and the tens and unit's digits can vary from 0 to 9. So, we have numbers from 700 to 769 that satisfy this condition.
(ii) Numbers from 801 to 900 where the hundreds digit is greater than the ten's digit:
In this range, the hundreds digit can be 8 or 9, and the tens and unit's digits can vary from 0 to 9. So, we have numbers from 801 to 879 that satisfy this condition.
Now, let's consider the numbers from the previous group where the ten's digit is greater than the unit's digit. We can continue this pattern for numbers from 801 to 869.
In addition to those numbers, we also have the following numbers where the ten's digit is greater than the unit's digit: 870, 871, 872, 873, 874, 875, and 876.
Therefore, in the range from 801 to 900, we have a total of 28 numbers that meet both conditions.
(iii) Numbers from 901 to 1000 where the hundreds digit is greater than the ten's digit:
In this range, the hundreds digit can be 9, and the tens and unit's digits can vary from 0 to 9. So, we have numbers from 901 to 989 that satisfy this condition.
Now, let's consider the numbers from the previous group where the ten's digit is greater than the unit's digit. We can continue this pattern for numbers from 901 to 969.
In addition to those numbers, we also have the following numbers where the ten's digit is greater than the unit's digit: 970, 971, 972, 973, 974, 975, 976, 980, 981, 982, 983, 984, 985, 986, and 987.
Therefore, in the range from 901 to 1000, we have a total of 36 numbers that meet both conditions.
Finally, summing up the numbers from each range:
Total numbers from 700 to 800 = 21
Total numbers from 801 to 900 = 28
Total numbers from 901 to 1000 = 36
Adding them together, we get a total of 85 numbers from 700 to 1000 that satisfy the given conditions.
- In a school every student is assigned a unique identification number. A student is a football player if and only if the identification number is divisible by 4, whereas a student is a cricketer if and only if the identification number is divisible by 6. If every number from 1 to 100 is assigned to a student, then how many of them play cricket as well as football ?(2019)
(A) 4
(B) 8
(C) 10
(D) 12
Answer-B
To find the number of students who play both cricket and football, we need to determine the common multiples of 4 and 6 within the range of 1 to 100.
First, let's find the multiples of 4 from 1 to 100: 4, 8, 12, 16, 20, ..., 96, 100.
Second, let's find the multiples of 6 from 1 to 100: 6, 12, 18, 24, 30, ..., 96.
To find the common multiples of 4 and 6, we can list the multiples that appear in both lists:
12, 24, 36, 48, 60, 72, 84, 96.
Therefore, there are 8 students who play both cricket and football within the given range (1 to 100).
So, the answer is 8.
- The ratio of a two-digit natural number to a number formed by reversing its digits is 4: 7. The number of such pairs is(2019)
(A) 5
(B) 4
(C) 3
(D) 2
Answer-B
Let's assume the two-digit natural number as AB, where A represents the tens digit and B represents the units digit. The number formed by reversing its digits would be BA.
According to the given information, the ratio of AB to BA is 4:7, which can be expressed as AB/BA = 4/7.
Now, let's express AB and BA in terms of their digit values:
AB = 10A + B
BA = 10B + A
Substituting these values into the ratio equation, we have:
(10A + B)/(10B + A) = 4/7
To simplify this equation, we can cross-multiply:
7(10A + B) = 4(10B + A)
70A + 7B = 40B + 4A
66A = 33B
2A = B
From this equation, we see that B must be an even number since it is twice A. The possible values for B are 2, 4, 6, and 8.
For each value of B, we can determine the corresponding value of A by substituting it into the equation 2A = B.
For B = 2, A = 1
For B = 4, A = 2
For B = 6, A = 3
For B = 8, A = 4
Thus, we have found four pairs of two-digit natural numbers that satisfy the given condition: (12, 21), (24, 42), (36, 63), (48, 84).
Therefore, the number of such pairs is 4.
Hence, the answer is (b) 4.
- If the numerator and denominator of a proper fraction are increased by the same positive quantity which is greater than zero, the resulting fraction is(2019)
(A) always less than the original fraction
(B) always greater than the original fraction
(C) always equal to the original fraction
(D) such that nothing can be claimed definitely
Answer-B
Let's consider a proper fraction as a/b, where a is the numerator and b is the denominator, with a < b.
If we increase both the numerator and denominator by the same positive quantity, let's say x, the resulting fraction would be (a+x)/(b+x).
To determine the relationship between the original fraction a/b and the resulting fraction (a+x)/(b+x), we need to consider the magnitude of x.
Case 1: x = 0
If x = 0, then the numerator and denominator remain the same, and the resulting fraction is a/b, which is equal to the original fraction.
Case 2: x > 0
If x is a positive quantity greater than zero, it means we are increasing both the numerator and denominator by the same amount. In this case, we can compare the fractions by considering their differences:
(a+x)/(b+x) - a/b = [(a+x)b - a(b+x)] / b(b+x)
= (ab + xb - ab - ax) / b(b+x)
= xb - ax / b(b+x)
= x(b-a) / b(b+x)
Since x > 0 and b > a (because a is the numerator and b is the denominator of a proper fraction), the numerator x(b-a) is positive, and the denominator b(b+x) is positive.
Therefore, the resulting fraction (a+x)/(b+x) will be greater than the original fraction a/b.
In conclusion, if we increase both the numerator and denominator of a proper fraction by the same positive quantity greater than zero, the resulting fraction will always be greater than the original fraction.
Hence, the answer is (b) always greater than the original fraction.
- Consider two statements S1 and S2 followed by a question :
S1: p and q both are prime numbers.
S2: p + q is an odd integer.
Question: Is pq an odd integer?
Which one of the following is correct?(2019)
(A) S1 alone is sufficient to answer the question
(B) S2 alone is sufficient to answer the question
(C) Both S1 and S2 taken together are not sufficient to answer the question
(D) Both S1 and S2 are necessary to answer the question
Answer-D
S1: p and q both are prime numbers.
S2: p + q is an odd integer.
Question: Is pq an odd integer?
Let's consider each statement separately:
S1: p and q both are prime numbers.
If p and q are both prime numbers, it means they are greater than 1 and only divisible by 1 and themselves. However, this statement does not provide any information about whether pq is odd or even.
S2: p + q is an odd integer.
If p + q is an odd integer, it means that the sum of p and q is not divisible by 2. This implies that at least one of p and q is odd, and the other is even. However, this statement does not provide any specific information about pq being odd or even.
Now, let's consider both statements together:
If both p and q are prime numbers and their sum p + q is odd, it means that one of the primes is even and the other is odd.
Considering this information, the product pq will always be even. This is because when we multiply an even number by any other number (odd or even), the result is always even.
Therefore, both statements together are necessary to determine that pq is an even integer.
- Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3.
The only possible value of B is(2019)
(A) 2
(B) 5
(C) 7
(D) 8
Answer-D
To solve this, we first note that for any number to be divisible by 3, the sum of its digits must be divisible by 3.
So, in the given equation, 7A3 is divisible by 3, which means that the sum of its digits, 7 + A + 3, is also divisible by 3. This implies that 10 + A is divisible by 3.
Now, we can try the possible values of A, which are 2, 5, and 8, and see which ones make 10 + A divisible by 3. We find that only A = 2 satisfies this condition.
Substituting A = 2 in the given equation, we get:
136 + 5B7 = 7A3
136 + 5B7 = 723
5B7 = 587
Now, we need to find the value of B. We can see that the only possible values for B are 6, 7, 8, and 9. However, we can eliminate 6 and 9 because if B = 6, then the sum of digits of 7A3 would be 13, which is not divisible by 3. Similarly, if B = 9, the sum of digits of 7A3 would be 17, which is not divisible by 3.
Thus, we are left with B = 7 and B = 8. However, if B = 7, then the sum of digits of 7A3 would be 13, which is not divisible by 3. Therefore, we can conclude that the only possible value of B is 8.
Hence, the answer to the question is (d) 8.
- Sunita cuts a sheet of paper into three pieces. Length of the first piece is equal to the average of the three single digit odd prime numbers. Length of the second piece is equal to that of the first plus one-third the length of the third. The third piece is as long as the other two pieces are together. The length of the original sheet of paper is(2019)
(A)13 units
(B) 15 units
(C) 16 units
(D) 30 units
Answer-D
Let the length of the third piece be x. Then, from the third condition, we can say that the length of the first and second piece is 2x.
From the first condition, the average of the three single-digit odd prime numbers is:
(3+5+7)/3 = 5
So, the length of the first piece is 5 units.
From the second condition, the length of the second piece is:
5 + (1/3)x
So, the total length of the three pieces is:
5 + (5 + (1/3)x) + x
Simplifying this, we get:
10 + (4/3)x
But we know that the total length of the paper is equal to the sum of the lengths of the three pieces. Therefore, we have:
10 + (4/3)x = 3x
Solving this equation, we get:
x = 15
So, the total length of the paper is:
10 + (4/3)x = 10 + (4/3)(15) = 30
Therefore, the answer is (d) 30 units.
- How many triplets (x, y, z) satisfy the equation x + y + z = 6, where x, y and z are natural numbers?(2019)
(A) 4
(B) 5
(C) 9
(D) 10
Answer-D
10 triplets of (x, y, z) =(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1), (4,1,1), (1,1,4), (1,4,1), (2,2,2)
- An 8-digit number 4252746B leaves remainder 0 when divided by 3. How many values of B are possible?(2019)
(A) 2
(B) 3
(C) 4
(D) 6
Answer-C
To determine the possible values of B in the 8-digit number 4252746B that leaves a remainder of 0 when divided by 3, we can apply the divisibility rule for 3.
According to the divisibility rule for 3, a number is divisible by 3 if the sum of its digits is divisible by 3.
In the given number, the sum of the first seven digits is 4 + 2 + 5 + 2 + 7 + 4 + 6 = 30, which is divisible by 3.
To find the possible values of B, we need to determine the values that, when added to the sum of the first seven digits, result in a multiple of 3.
We know that the sum of the first seven digits is divisible by 3 (as it leaves a remainder of 0). Therefore, we need to find the values of B that, when added to 30, also give a multiple of 3.
The possible values of B that satisfy this condition are 0, 3, 6, and 9. Adding any of these values to 30 will result in a multiple of 3.
Therefore, the answer is (c) 4, as there are 4 possible values of B (0, 3, 6, and 9) that satisfy the given condition.
- The number of times the digit 5 will appear while writing the integers from 1 to 1000 is (2019)
(A) 269
(B) 271
(C) 300
(D) 302
Answer-C
To determine the number of times the digit 5 will appear while writing the integers from 1 to 1000, we can break it down into three cases:
Units digit: The digit 5 appears in the units place 100 times (from 5 to 995) because there are 100 multiples of 10 in the given range.
Tens digit: The digit 5 appears in the tens place 100 times (from 50 to 599) because there are 10 multiples of 10 in the tens place, and each of those can be combined with the digits 0 to 9 in the units place.
Hundreds digit: The digit 5 appears in the hundreds place 100 times (from 500 to 599) because there is only one multiple of 100 in the hundreds place, and it can be combined with the digits 0 to 9 in the tens and units places.
Therefore, the total number of times the digit 5 appears is 100 + 100 + 100 = 300.
Therefore, the correct option is (c) 300.
- How many zeroes are there at the end of the following product?(2020)
1x 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60
(A) 10
(B) 14
(C) 12
(D) 15
Answer-A
To determine the number of zeros at the end of the given product, we need to count the number of factors of 10 in the product. Since 10 = 2 × 5, we need to count the number of factors of 2 and 5.
Given product: 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60
If we analyze the product, we can simplify it as follows:
1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60
= (1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60)
= (1 × 5 × 2 × 3 × 2 × 5 × 2 × 3 × 7 × 2 × 3 × 5 × 2 × 3 × 11)
= (2^4 × 3^6 × 5^6 × 7 × 11).
We need to determine the number of factors of 10 in this product, which means we need to count the minimum power of 2 and 5.
From the prime factorization, we have 2^4 and 5^6. This means we have a minimum of 4 factors of 2 and 6 factors of 5.
So total 10.
- Let XYZ be a three-digit number, where (X + Y + 2) is not a multiple of 3. Then
(XYZ + YZX + ZXY) is not divisible by(2020)
(A) 3
(B) 9
(C) 37
(D) (X+Y+Z)
Answer-B
The three-digit number XYZ in its general form: XYZ = 100X + 10Y + Z.
Similarly, YZX can be expressed as: YZX = 100Y + 10Z + X.
And ZXY can be expressed as: ZXY = 100Z + 10X + Y.
XYZ + YZX + ZXY = (100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y).
XYZ + YZX + ZXY = 100X + 10Y + Z + 100Y + 10Z + X + 100Z + 10X + Y.
XYZ + YZX + ZXY = (100X + X) + (10Y + Y) + (100Z + Z).
XYZ + YZX + ZXY = 101X + 11Y + 101Z.
Factoring out the common factor of 111, we have:
XYZ + YZX + ZXY = 111(X + Y + Z).
It is clear that 111 is a common factor of XYZ + YZX + ZXY. Therefore, (XYZ + YZX + ZXY) is divisible by 111.
However, we are given that (X + Y + Z) is not a multiple of 3. This means that the sum of the digits is not divisible by 3.
Since 111 is divisible by 3 (3 x 37 = 111), but (X + Y + Z) is not, it follows that (XYZ + YZX + ZXY) is not divisible by 3.
Additionally, since 111 is not divisible by 9, (XYZ + YZX + ZXY) is also not divisible by 9.
Therefore, the correct answer is option (b) 9.
- Let p,q, r and s be natural numbers such that p - 2016 = q+2017 = r - 2018 = s + 2019
Which one of the following is the largest natural number?(2020)
(A) p
(B) q
(C) r
(D) s
Answer-C
Given the conditions:
p - 2016 = q + 2017 = r - 2018 = s + 2019
Let's solve the equations one by one to find the values of p, q, r, and s.
From p - 2016 = q + 2017, we can rewrite it as:
p - q = 4033 ---- (Equation 1)
From q + 2017 = r - 2018, we can rewrite it as:
r - q = 4035 ---- (Equation 2)
From r - 2018 = s + 2019, we can rewrite it as:
r - s = 4037 ---- (Equation 3)
Now, let's analyze the equations:
Equation 1 tells us that p - q = 4033.
Equation 2 tells us that r - q = 4035.
Equation 3 tells us that r - s = 4037.
From Equation 2, we can conclude that r > q.
From Equation 3, we can conclude that r > s.
Therefore, among the given options, (c) r is the largest natural number.
Hence, the correct answer is (c) r.
- How many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits?(2020)
(A) Zero
(B) One
(C) Nine
(D) Ten
Answer-A
Using divisibility rules, we can determine that none of these combinations will result in a prime number:
If the sum of the digits is divisible by 3, the number is also divisible by 3. Since the sum of the digits (1 + 2 + 3 + 4 + 5) is 15, which is divisible by 3, none of the combinations will result in a prime number.
Therefore, the answer is (a) Zero. There are no five-digit prime numbers that can be formed using the digits 1, 2, 3, 4, and 5 without repetition.
- A simple mathematical operation in each number of the sequence 14,18,20, 24,30, 32, ... results in a sequence with respect to prime numbers. Which one of the following is the next number in the sequence?(2020)
(A) 34
(B) 36
(C) 38
(D) 40
Answer-C
To find the next number in the sequence, we need to understand the pattern of the given sequence and how it relates to prime numbers.
Given sequence: 14, 18, 20, 24, 30, 32, ...
Let's analyze the sequence:
The first number, 14, is not a prime number.
The second number, 18, is not a prime number.
The third number, 20, is not a prime number.
The fourth number, 24, is not a prime number.
The fifth number, 30, is not a prime number.
The sixth number, 32, is not a prime number.
Looking at the given sequence, it seems that each number is obtained by performing a mathematical operation on a prime number.
To find the pattern, let's examine the differences between consecutive numbers in the sequence:
18 - 14 = 4
20 - 18 = 2
24 - 20 = 4
30 - 24 = 6
32 - 30 = 2
We can observe that the differences between consecutive numbers are alternating between 2 and 4.
Now, let's consider prime numbers that are close to the given numbers:
The prime numbers close to 14 are 13 and 17.
The prime numbers close to 18 are 17 and 19.
The prime numbers close to 20 are 19 and 23.
The prime numbers close to 24 are 23 and 29.
The prime numbers close to 30 are 29 and 31.
The prime numbers close to 32 are 31 and 37.
Based on the pattern and the prime numbers close to the given numbers, we can deduce that the next number in the sequence should have a difference of 4 and be close to a prime number.
Among the given options, the only number that satisfies these conditions is (c) 38.
Hence, the correct answer is (c) 38.
- Two Statements are given followed by two Conclusions:
Statements
All numbers are divisible by 2.
All numbers are divisible by 3.
Conclusion-I:
All numbers are divisible by 6.
Conclusion-II:
All numbers are divisible by 4.
Which of the above Conclusions logically follows/follow from the two given Statements?(2020)
(A) Only Conclusion-I
(B) Only Conclusion-II
(C) Neither Conclusion-I nor Conclusion-II
(D) Both Conclusion-I and Conclusion-II
Answer-A
The given statements are:
1.All numbers are divisible by 2.
2.All numbers are divisible by 3.
Let's evaluate the conclusions:
Conclusion-I: All numbers are divisible by 6.
Conclusion-II: All numbers are divisible by 4.
To determine if the conclusions follow logically from the given statements, we need to check if they are true in all possible cases.
For Conclusion-I:
If a number is divisible by both 2 and 3, it will be divisible by their least common multiple, which is 6. Therefore, Conclusion-I follows logically from the given statements.
For Conclusion-II:
The given statements do not guarantee that all numbers are divisible by 4. For example, numbers that are divisible by 3 but not divisible by 2 (such as 3, 9, 15) are not divisible by 4. Therefore, Conclusion-II does not follow logically from the given statements.
Based on the above analysis:
Conclusion-I (All numbers are divisible by 6) follows logically from the given statements.
Conclusion-II (All numbers are divisible by 4) does not follow logically from the given statements.
Hence, the correct answer is (a) Only Conclusion-I.
- Consider the following sequence of numbers
5 1 4 7 3 9 8 5 7 2 6 3 15 8 6 3 8 5 2 2 4 3 4 9 6
How many odd numbers are followed by the odd number in the above sequence?(2020)
(A) 5
(B) 6
(C) 7
(D) 8
Answer-B
Let's analyze the given sequence to determine the number of odd numbers followed by an odd number:
5 1 4 7 3 9 8 5 7 2 6 3 15 8 6 3 8 5 2 2 4 3 4 9 6
Starting from the beginning, we look at each number and its adjacent number to determine if both are odd:
5 1 - The number 5 is odd, and it is followed by the number 1, which is also odd.
4 7 - Neither of these numbers is odd.
3 9 - The number 3 is odd, and it is followed by the number 9, which is also odd.
8 5 - Neither of these numbers is odd.
7 2 - Neither of these numbers is odd.
6 3 - The number 6 is not odd, but the number 3 is odd.
15 8 - The number 15 is odd, and it is followed by the number 8, which is not odd.
6 3 - The number 6 is not odd, but the number 3 is odd.
8 5 - Neither of these numbers is odd.
2 2 - Neither of these numbers is odd.
4 3 - Neither of these numbers is odd.
4 9 - Neither of these numbers is odd.
6 - There are no more numbers after 6.
From the analysis above, we can see that there are a total of 6 odd numbers that are followed by an odd number in the given sequence.
Therefore, the correct answer is (b) 6.
- How many integers are there between 1 and 100 which have 4 as a digit but are not divisible by 4?(2020)
(A) 5
(B) 11
(C) 12
(D) 13
Answer-C
The correct answer is (c) 12.
There are 19 integers between 1 and 100 which have 4 as a digit, namely: 4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84, and 94.
Out of these, 7 integers are divisible by 4, namely: 4, 24, 40, 44, 48, 64, and 84.
Therefore, there are 19 - 7 = 12 integers between 1 and 100 which have 4 as a digit but are not divisible by 4.
- What is the largest number among the following?(2020)
(A) (1/2)^-1
(B) (1/4)^-3
(C) (1/3)^-4
(D) (1/6)^-2
Answer-C
To determine the largest number among the options provided, we can simplify each expression by applying the exponent and comparing the resulting values.
Comparing the results, we can see that the largest number is 81, which corresponds to option c) (1/3)^(-4).
- What is the greatest length x such that 3 1/2 m and 8 3/4 m are integral multiples of x?(2020)
(A) 1 1/2 m
(B) 1 1/3 m
(C) 1 1/4 m
(D) 1 3/4 m
Answer-D
To find the greatest length x that is a common factor of 3 1/2 m and 8 3/4 m, we need to find the greatest common divisor (GCD) of the two lengths.
First, let's convert 3 1/2 m and 8 3/4 m to improper fractions:
3 1/2 m = (7/2) m
8 3/4 m = (35/4) m
Now, let's find the GCD of (7/2) m and (35/4) m.
To simplify calculations, we can ignore the unit (m) and work with the numbers alone.
The GCD of (7/2) and (35/4) can be found by finding the GCD of their numerators and denominators separately.
Numerator GCD: GCD(7, 35) = 7
Denominator GCD: GCD(2, 4) = 2
The GCD of (7/2) and (35/4) is (7/2) / 2 = 7/4.
Now, we need to find the greatest length x that corresponds to 7/4. We can express 7/4 as a mixed number:
7/4 = 1 3/4
Therefore, the greatest length x that is a common factor of 3 1/2 m and 8 3/4 m is 1 3/4 m, which corresponds to option (d).
- The recurring decimal representation 1.272727..... is equivalent to (2020)
(A) 13/11
(B) 14/11
(C) 127/99
(D) 137/99
Answer-B
To determine the equivalent fraction of the recurring decimal representation 1.272727..., we can use the concept of geometric series.
Let x = 1.272727...
Then, 100x = 127.272727...
Now, subtracting the two equations:
100x - x = 127.272727... - 1.272727...
99x = 126
Dividing both sides by 99:
x = 126/99 = 14/11
Therefore, the recurring decimal representation 1.272727... is equivalent to the fraction 14/11.
Hence, the correct answer is (b) 14/11.
- What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?(2020)
(A) 1012
(B) 1022
(C) 1122
(D) 1222
Answer-B
Let N be the required number. Since N leaves a remainder of 2 when divided by 3, 4, 5, and 6, it must be of the form:
N = lcm (3, 4, 5, 6) * k + 2
where lcm(3, 4, 5, 6) is the least common multiple of 3, 4, 5, and 6, and k is an integer. We can simplify this expression by finding the value of lcm(3, 4, 5, 6):
lcm(3, 4, 5, 6) = lcm(3, lcm(4, 5, 6))
= lcm(3, 60)
= 60
So we have:
N = 60k + 2
To find the smallest four-digit number of this form, we need to find the smallest value of k that makes N greater than or equal to 1000. We can solve for k as follows:
60k + 2 >= 1000
60k >= 998
k >= 16.63
Since k must be an integer, we round up to k = 17. Plugging this value into the expression for N, we get:
N = 60*17 + 2 = 1022
Therefore, the smallest four-digit number that leaves a remainder of 2 when divided by 3, 4, 5, and 6 is 1022, which corresponds to option (b).
- What is the remainder when 51×27 × 35 × 62 × 75 is divided by 100?(2020)
(A) 50
(B) 25
(C) 5
(D) 1
Answer-A
To find the remainder when the expression 51 × 27 × 35 × 62 × 75 is divided by 100, we need to consider the last two digits of each number and calculate their product.
Let's find the last two digits of each number:
51: The last two digits are 51.
27: The last two digits are 27.
35: The last two digits are 35.
62: The last two digits are 62.
75: The last two digits are 75.
Now, multiply the last two digits of all the numbers:
51 × 27 × 35 × 62 × 75 = 386,490,750
When you divide any number by 100, the remainder is always the last two digits.
The last two digits of this product, 386,490,750, are 50.
Therefore, when the expression 51 × 27 × 35 × 62 × 75 is divided by 100, the remainder is 50.
- How many pairs of natural numbers are there such that the difference of whose squares is 63?(2020)
(A) 3
(B) 5
(C) 4
(D) 2
Answer-A
To find the pairs of natural numbers whose squares have a difference of 63, we can use algebraic equations. Let's assume the two natural numbers are x and y.
According to the given information, the difference of their squares is 63. Mathematically, we can represent this as:
x^2 - y^2 = 63
We can factorize the left side of the equation:
(x + y)(x - y) = 63
Now, we need to find pairs of natural numbers whose product is 63. The pairs of factors of 63 are: (1, 63), (3, 21), and (7, 9).
Using these pairs, we can set up the equations:
x + y = 63, x - y = 1
x + y = 21, x - y = 3
x + y = 9, x - y = 7
Solving these equations, we find:
Adding the equations: 2x = 64 -> x = 32, y = 31
Adding the equations: 2x = 24 -> x = 12, y = 9
Adding the equations: 2x = 16 -> x = 8, y = 1
Therefore, there are three pairs of natural numbers (x, y) that satisfy the given conditions: (32, 31), (12, 9), and (8, 1).
Hence, the correct answer is option (a) 3.
- Which one of the following will have minimum change in its value if 5 is added to both numerator and the denominator of the fractions 2/3, 3/4, 4/5 and 5/6?(2020)
(A) 2/3
(B) 4/5
(C) 3/4
(D) 5/6
Answer-A
Let's calculate the change in value for each fraction when 5 is added to both the numerator and the denominator:
2/3:
Original value: 2/3 = 0.667
New value: (2 + 5)/(3 + 5) = 7/8 = 0.875
Change: 0.875 - 0.667 = 0.208
3/4:
Original value: 3/4 = 0.75
New value: (3 + 5)/(4 + 5) = 8/9 = 0.889
Change: 0.889 - 0.75 = 0.139
4/5:
Original value: 4/5 = 0.8
New value: (4 + 5)/(5 + 5) = 9/10 = 0.9
Change: 0.9 - 0.8 = 0.1
5/6:
Original value: 5/6 = 0.833
New value: (5 + 5)/(6 + 5) = 10/11 = 0.909
Change: 0.909 - 0.833 = 0.076
As we can see, the fraction 5/6 has the smallest change in value, which is 0.076. Therefore, option (d) 5/6 is the correct answer,
- A digit n> 3 is divisible by 3 but not divisible by 6. Which one of the following is divisible by 4?(2020)
(A) 2n
(B) 3n
(C) 2n+4
(D) 3n + 1
Answer-D
Given that n is a number greater than 3, divisible by 3 but not divisible by 6, let's consider the number 9 as an example that satisfies these conditions.
If we substitute n = 9 into each option:
Option 1: 2n = 2(9) = 18, which is not divisible by 4.
Option 2: 3n = 3(9) = 27, which is not divisible by 4.
Option 3: 2n + 4 = 2(9) + 4 = 22, which is not divisible by 4.
Option 4: 3n + 1 = 3(9) + 1 = 28, which is divisible by 4.
Hence, we can conclude that option 4, 3n + 1, is divisible by 4 when n is a digit greater than 3 that is divisible by 3 but not divisible by 6.
- Integers are listed from 700 to 1000. In how many integers is the sum of the digits 10?(2021)
(A) 6
(B) 7
(C) 8
(D) 9
Answer-D
There are 9 possible combinations of three digits whose sum is 10, and the numbers formed with these combinations that are greater than 700 and less than 1000 are:
0, 1, 9 → 901, 910
0, 2, 8 → 802, 820
0, 3, 7 → 703, 730
1, 1, 8 → 811
1, 2, 7 → 712, 721
So, there are 9 integers in the range 700 to 1000 whose sum of digits is 10.
- Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum.
Which of the following is/are correct?
- S is always divisible by 74.
- S is always divisible by 9.
Select the correct answer using the code given below:(2021)
(A) 1 only
(B) 2 only
(C) Both 1 and 2
(D) Neither 1 nor 2
Answer-C
Non-zero digits and multiples of 3 i.e. 3, 6, 9
No repetition of digits.
The numbers thus formed are 369, 396, 639, 693, 936, 963
Sum of these numbers = 3996
It is divisible by both 74 as well as 9.
- Consider the following statements :
1.The sum of 5 consecutive integers can be 100.
- The product of three consecutive natural numbers can be equal to their sum.
Which of the above statements is/are correct?(2021)
(A) 1 only
(B) 2 only
(C) Both 1 and 2
(D) Neither 1 nor 2
Answer-C
Statement 1: The sum of 5 consecutive integers can be 100.
To find the consecutive integers, let's assume the middle integer is x.
Then, the five consecutive integers can be expressed as (x-2), (x-1), x, (x+1), and (x+2).
The sum of these integers is: (x-2) + (x-1) + x + (x+1) + (x+2) = 5x.
For the sum to be 100, 5x = 100, which implies x = 20.
Therefore, the five consecutive integers are 18, 19, 20, 21, and 22.
This confirms that statement 1 is correct.
Statement 2: The product of three consecutive natural numbers can be equal to their sum.
Let's assume the three consecutive natural numbers as (n-1), n, and (n+1).
The product of these numbers is: (n-1) * n * (n+1) = n^3 - n.
The sum of these numbers is: (n-1) + n + (n+1) = 3n.
To satisfy the condition that the product is equal to the sum, we have:
n^3 - n = 3n.
Simplifying the equation, we get:
n^3 - 4n = 0.
Factoring out n, we have:
n(n^2 - 4) = 0.
This gives us two possible solutions: n = 0 or n = ±2.
However, we are considering natural numbers, which exclude zero. Therefore, the only valid solution is n = 2.
So, the three consecutive natural numbers are 1, 2, and 3, which satisfy the condition.
- When a certain number is multiplied by 7, the product entirely comprises ones only (1111...). What is the smallest such number?(2021)
(A) 15713
(B) 15723
(C) 15783
(D) 15873
Answer-D
To find the smallest number that, when multiplied by 7, results in a product consisting entirely of ones, we can start by trying each option provided and checking if it satisfies the condition.
Let's evaluate each option:
(A) 15713 x 7 = 109991
The product is not entirely composed of ones, so option (a) is not the correct answer.
(B) 15723 x 7 = 110061
Again, the product is not entirely composed of ones, so option (b) is not the correct answer.
(C) 15783 x 7 = 110481
Once more, the product is not entirely composed of ones, so option (c) is not the correct answer.
(D) 15873 x 7 = 111111
Finally, with option (d), the product is entirely composed of ones.
Therefore, the smallest number that, when multiplied by 7, results in a product consisting entirely of ones is 15873, and the correct answer is (d) 15873.
- If 3^2019 is divided by 10, then what is the remainder?(2021)
(A) 1
(B) 3
(C) 7
(D) 9
Answer-C
To find the remainder when 3^2019 is divided by 10, we can observe the pattern of the units digits of powers of 3:
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
...
We notice that the units digits of powers of 3 repeat in a cycle: 3, 9, 7, 1. The cycle length is 4. Therefore, we can conclude that the units digit of 3^n, where n is a positive integer, follows the pattern 3, 9, 7, 1, 3, 9, 7, 1, ...
In this case, we want to find the remainder when 3^2019 is divided by 10. Since the units digit of 3^2019 follows the pattern mentioned above, we need to determine the position of 2019 in the cycle.
2019 divided by 4 gives a quotient of 504 with a remainder of 3. This means that 2019 is three positions after the beginning of the cycle. Therefore, the units digit of 3^2019 is the same as the units digit of 3^3, which is 7.
Thus, when 3^2019 is divided by 10, the remainder is 7.
The correct answer is (c) 7.
- The number 3798125P369 is divisible by 7. What is the value of the digit P?(2021)
(A) 1
(B) 6
(C) 7
(D) 9
Answer-B
The correct answer is (b) 6.
We can use the divisibility rule for 7 to solve this problem. The divisibility rule for 7 states that a number is divisible by 7 if the difference between the sum of its digits in the even places and the sum of its digits in the odd places is divisible by 7.
In the number 3798125P369, the sum of the digits in the even places is 3 + 8 + 2 + 5 + 6 + 9 = 33. The sum of the digits in the odd places is 7 + 9 + 1 + P + 3 = 20 + P.
Therefore, the difference between the sum of the digits in the even places and the sum of the digits in the odd places is 33 - (20 + P).
For the number to be divisible by 7, 33 - (20 + P) must be divisible by 7. This means that P must be equal to 6.
Therefore, the value of the digit P is 6.
- Consider the Question and two Statements given below :
Question: Is x an integer?
Statement-1: x / 3 is not an integer.
Statement-2: 3x is an integer.
Which one of the following is correct in respect of the Question and the Statements?(2022)
(A) Statement-1 alone is sufficient to answer the Question
(B) Statement-2 alone is sufficient to answer the Question
(C) Both Statement-1 and Statement-2 are sufficient to answer the Question
(D) Both Statement-1 and Statement-2 are not sufficient to answer the Question
Answer-D
Statement-1: x / 3 is not an integer.
This statement implies that x is not a multiple of 3. However, it does not provide any information on whether x is an integer or not. For example, x could be 1/3, which is not an integer and not a multiple of 3. Alternatively, x could be 4, which is not a multiple of 3 but is an integer. Therefore, statement-1 alone is not sufficient to answer the question.
Statement-2: 3x is an integer.
This statement implies that x is a multiple of 1/3, which means that x could be a fraction of the form n/3, where n is an integer. For example, if 3x = 2, then x = 2/3, which is an integer. Alternatively, if 3x = 4, then x = 4/3, which is not an integer. Therefore, statement-2 alone is not sufficient to answer the question.
Since neither statement alone is sufficient to answer the question, we need to consider both statements together.
Combining the two statements, we know that x is not a multiple of 3 and that 3x is an integer. This means that x must be a fraction of the form n/3, where n is an integer and not a multiple of 3. For example, if 3x = 2, then x = 2/3, which is an integer and not a multiple of 3. Alternatively, if 3x = 4, then x = 4/3, which is not an integer and a multiple of 3. Therefore, both statements together are not sufficient to answer the question.
The correct answer is (d) Both Statement-1 and Statement-2 are not sufficient to answer the Question.
- Let A, B and C represent distinct non-zero digits. Suppose x is the sum of all possible 3-digit numbers formed by A, B and C without repetition.
Consider the following statements
- The 4-digit least value of x is 1332.
- The 3-digit greatest value of x is 888.
Which of the above statements is /are correct?(2022)
(A) 1 only
(B) 2 only
(C) Both 1 and 2
(D) Neither 1 nor 2
Answer-A
To determine the correctness of the statements, let's analyze the given scenario.
We are forming 3-digit numbers using distinct non-zero digits A, B, and C. This implies that A, B, and C can take values from 1 to 9.
To find the sum of all possible 3-digit numbers formed by A, B, and C without repetition, we can use the formula for the sum of an arithmetic series:
Sum = (Number of terms / 2) * (First term + Last term)
The number of terms is the total number of 3-digit numbers that can be formed using A, B, and C without repetition, which is given by 3!.
Now, let's consider each statement:
1.The 4-digit least value of x is 1332.
If we arrange the digits A, B, and C in increasing order, the smallest 3-digit number we can form is ABC. Therefore, the sum x will be the sum of all possible permutations of ABC. Using the formula mentioned above, we find that the sum x will be (6 / 2) * (ABC + CBA) = 3 * (ABC + CBA). Since A, B, and C are distinct non-zero digits, the smallest possible value of A is 1, B is 2, and C is 3. Plugging in these values, we get x = 3 * (123 + 321) = 3 * 444 = 1332. So, statement 1 is correct.
2.The 3-digit greatest value of x is 888.
To find the greatest possible value of x, we need to find the largest possible values for A, B, and C. Since they are distinct non-zero digits, the largest possible value for A is 9, B is 8, and C is 7. Plugging in these values, we get x = 3 * (987 + 789) = 3 * 1776 = 5328. Therefore, statement 2 is incorrect.
In conclusion, only statement 1 is correct. The answer is (a) 1 only.
- What is the remainder when 91x 92 × 93 x 94 x 95 × 96 × 97 × 98 × 99 is divided by 1261?(2022)
(A) 3
(B) 2
(C) 1
(D) 0
Answer-D
The correct answer is (d) 0.
1261 = 13 × 97.
91, 93, 94, 96, and 98 are all divisible by 13.
92, 95, 97, and 99 are all divisible by 97.
Therefore, the given expression is divisible by 13 and 97, and hence is divisible by 1261. The remainder is 0.
- What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?(2022)
(A) 1063
(B) 1073
(C) 1083
(D) 1183
Answer-C
To find the smallest number greater than 1000 that leaves a remainder of 3 when divided by any one of the numbers 6, 9, 12, 15, and 18, we need to find the least common multiple (LCM) of these numbers and add 3 to it.
The LCM of 6, 9, 12, 15, and 18 for a number above 1000 is 1080
Adding three gives 1083.
- Consider the following statements in respect of two natural numbers p and g such that p is a prime number and q is a composite number :
- p x q can be an odd number.
- q/p can be a prime number.
- p+q can be a prime number.
Which of the above statements are correct?(2022)
(A) 1 and 2 only
(B) 2 and 3 only
(C) 1 and 3 only
(D) 1, 2 and 3
Answer-D
1.p × q can be an odd number:
If p is a prime number and q is a composite number, then it is possible for their product p × q to be an odd number. This can occur if p is an odd prime number and q is any odd composite number. For example, if p = 3 and q = 9, then 3 × 9 = 27, which is an odd number. Therefore, statement 1 is correct.
2.q/p can be a prime number:
If p is a prime number and q is a composite number, it is indeed possible for their quotient q/p to be a prime number. This can occur if q is a multiple of p and their quotient results in a prime number. For example, if p = 3 and q = 9, then q/p = 9/3 = 3, which is a prime number. Therefore, statement 2 is correct.
3.p + q can be a prime number:
If p is a prime number and q is a composite number, it is possible for their sum p + q to be a prime number. This can occur if p is an even prime number (which is only true for p = 2) and q is an odd composite number. In this case, p + q = 2 + q, and if q is a prime number, then the sum would be prime. Therefore, statement 3 is correct.
All three statements are correct.
- If 15×14×13x..×3×2×1 = 3(power)m x n ,where m and n are positive integers, then what is the maximum value of m?(2022)
(A) 7
(B) 6
(C) 5
(D) 4
Answer-B
We have the equation:
15 × 14 × 13 × ... × 3 × 2 × 1 = 3^m × n
To find the maximum value of m, we need to determine how many factors of 3 are present in the product of the first 15 positive integers.
Let's count the factors of 3:
15 has one factor of 3.
12 has one factor of 3.
9 has two factors of 3.
6 has one factor of 3.
3 has one factor of 3.
Adding up all the factors, we have a total of 1 + 1 + 2 + 1 + 1 = 6 factors of 3.
Therefore, the maximum value of m is 6.
The correct answer is (b) 6.
- The sum of three consecutive integers is equal to their product. How many such possibilities are there?(2022)
(A) Only one
(B) Only two
(C) Only three
(D) No such possibility is there
Answer-C
Let's assume that the three consecutive integers are n, n+1, and n+2.
According to the problem statement, we have:
n + (n+1) + (n+2) = n(n+1)(n+2)
3n = n^3 - n
n^2 = 4
n = +2 ,n = - 2 or n= 0
So the three sets are (1,2,3) , (-1,0,1) and (-3,-2,-1)
Hence the correct option is (c)
- What is the greatest number among 2^50, 3^40, 4^30, and 5^20?(2022)
(A) 3^40
(B) 4^30
(C) 5^20
(D) 2^50
Answer-A
2^50 = (2^5)^10 = 32^10
3^40 = (3^4)^10 = 81^10
4^30 = (2^2)^10 = 16^10
5^20 = (5^2)^10 = 25^10
- How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?(2022)
(A) 8
(B) 12
(C) 16
(D) 24
Answer-B
To find the number of 3-digit natural numbers without repetition of digits, where each digit is odd and the number is divisible by 5, we can analyze the possible combinations.
For a 3-digit number to be divisible by 5, the units digit must be either 5 or 0. Since each digit must be odd, the only possibility for the units digit is 5.
Now, we have two remaining positions to fill with odd digits, which can be any of the remaining odd digits: 1, 3, 7, or 9. There are four choices for the hundreds digit and three choices for the tens digit.
Therefore, the total number of such 3-digit numbers is 4 x 3 = 12.
Hence, the correct answer is (b) 12.
- What is the number of numbers of the form 0.XY, where X and Y are distinct non-zero digits?(2022)
(A) 72
(B) 81
(C) 90
(D) 100
Answer-A
To form a number of the form 0.XY, where X and Y are distinct non-zero digits, we have the following possibilities:
X can be any non-zero digit from 1 to 9, giving us 9 choices.
Y can be any non-zero digit other than X, so we have 9 - 1 = 8 choices for Y.
Therefore, the total number of numbers of the form 0.XY is 9 x 8 = 72.
Hence, the correct answer is (a) 72.
