Q: When a certain number is multiplied by 13, the product consists entirely of fives. The smallest such number is?

By trial, we find that the smallest number consisting entirely of fives and exactly divisible by 13 is 555555. On dividing 555555 by 13, we get 42735 as quotient. ∴ Req. smallest number =42735.

Q: A number consist of two digit whose sum is 10 If the digit of the number are reversed then the number decreased by 36. Which of the following is correct. I. The number is divisible by a composite number. II. The number is a multiple of a prime numbe…

let the two-digit number be 10p + q Now according to the question p + q = 10 and (p +10q) + 36 = (q + 10p) ⇒ -9q + 9p = 36 ⇒ -q + p = 4 ....(ii) on adding eq (i) and (ii) we get 2p = 14 ⇒ p =7 ∴ p = 7 and q = 4 ∴ Required number is the 73 So, the number is a multiple of prime number.

Q: Every prime number of the form 3k + 1 can be represented in the form 6m + 1 (where k, m are integers), When

The only one even prime number is 2 and this is not in the form of 3k + 1. Thus any prime number of the form 3k + 1 is an odd number. This means 3k must be even number which implies that k must be even number also. Let k = 2m. Then a prime of the form 3k + 1 is of the form 3(2m) + 1 = 6m+1. Every prime number of form 3k + 1 can be represented in the form 6m + 1 only, when k is even.

Q: A number consist of two digit whose sum is 10. If the digits of the number are reversed, then the number decreased by 36. Which of the following is/are correct? I. The number is divisible by a composite number. II. The number is multiple of a prime …

Let the two digit number be 10x + y. Now, according to the question, x + y = 10............(i) if digits reversed , number get decreased by 36, and ( x +10y) = ( 10x + y ) - 36 ⇒ - 9y + 9x = 36 ⇒ x - y = 4 ..................... (ii) On adding Eqs. (i) and (ii), we get 2x = 14 ⇒ x = 7 ∴ x = 7 and y = 3 ∴ Required number is 73. So,the number is a multiple of a prime number.

Q: If the numerator of a certain fraction is increased by 2 and the denominator is increased by 1, then the resulting fraction is equal to the 1/2. If how ever the numerator is increased by 1, then denominator is decreased by 2, then the resulting frac…

Let the fraction be x / y. According to the question, ( x+2)/(y+1) = 1/2 ⇒ 2x + 4 = y+1 ⇒ 2x - y = -3 .........................(i) Again according to the question, (x+1)/(y-2) = 3/5 ⇒ 5x +5 = 3y - 6 ∴ 5x - 3y = -11..........................(ii) on multiplying Eq. (i) by 3 and then subtracting from Eq. (ii) we get, multiplying Eq. (i) by 3 6x - 3y = -9 subtracting from Eq. (ii) 5x- 3y - ( 6x - 3y ) = -11 - (-9) 5x- 3y - 6x + 3y = -11 + 9 - x = - 2 x= 2 On putting the value of x in Eq. (i) we get 2x- y = -3 2 x 2 - y = -3 ⇒ 2 x 2 -y = -3 ⇒ -y = -3 - 4 ∴ y = 7 ∴ Original fraction = x/y =2/7

Q: The sum of five consecutive odd numbers is equal to 175. What is the sum of the second largest number and the square of the smallest number among them together?

Method 1 to solve this question. Let p be the first odd number. Then next odd number is p + 2. According to question, Sum of five consecutive odd numbers 175 ⇒ p + p + 2 + p + 4 + p + 6 + p + 8 = 175 ⇒ 5p + 20 = 175 ⇒ p = (175 - 20)/5 = 31 According to question, Sum of the second largest and square of smallest one = (31 + 6) + (31)2 Sum of the second largest and square of smallest one = 37 + 961 = 998 Method 2 to solve this question. Let p be the even number. Then next odd number is p + 1. so on p + 3 , p + 5................ P + 9 According to question, Sum of five consecutive odd numbers 175 ⇒ p + 1 + p + 3 + p + 5 + p + 7 + p + 9 = 175 ⇒ 5p + 25 = 175 ⇒ p = (175 - 25)/5 ⇒ p = (150)/5 ⇒ p = 30 First odd number = p + 1 = 30 + 1 = 31 According to question, Sum of the second largest and square of smallest one = (30 + 7) + (30 + 1)2 Sum of the second largest and square of smallest one = 37 + 961 = 998

Q: Out of three given numbers, the first number is twice the second and thrice the third. If the average of these three numbers is 154, then what is the difference between the first and the third numbers?

Let the third numbers = p Then, the first number = 3p and second number = 3p/2 According to question, ( p + 3p + 3p/2 )/3 = 154 ⇒ ( 2p + 6p + 3p )/6 = 154 ∴ p = 154 x 6/11 = 84 ∴ Required difference = 3p - p = 2p = 2 x 84 = 168

Q: The differences between two numbers is 18. If four times the second numbers is less than three times the first number by 18, then what is the sum of these two numbers?

Let first number = p and second number = q According to the question, p - q = 18 ................................(i) and 3p - 4q = 18.....................(ii) On multiplying Eq .(i) by 3 and subtracting Eq. (ii) from it,we get 3p - 3q = 54 3p - 4q = 18 q = 36 On putting the value of q in Eq, (i) we get p = 18 + q ⇒ p = 18 + 36 = 54 ∴ Required sum = p + q = 54 + 36 = 90

Question 1

The difference between two number is 18 if four times the second number is less then three what is the sum these two number?

Accepted Answer

Let first number = P And second number = Q According to the question P - Q = 10 ....(i) And 3P - 4Q = 18...(ii) on multiply eq (i) by 3 and then subtract eq (ii) from it 3P - 3Q = 54 3P - 4Q = 18 Q = 36 On putting the value of Q in the eq (i) P = 18 + Q = 18 +36 ⇒ P = 54 Required sum = P + Q = 54 +36 =90

Question 2

A number is greater than the square of 44 but smaller than the square of 45. If one part of the number is the square of 6 and the number is a multiple of 5, then find the number.?

Accepted Answer

Let the number be x. 442 < x < 452 ⇒ 1936 < x < 2025 ...(i) From equation (i), the required number will be any number between 1936 and 2025. Since one part of the number is the square of 6 means one factor is 36. ∴ L, C, M of 36 and 5 = 180 ∴ Number will be multiple of 180 i,e., 180 x 11 = 1980 the only value which satisfies the equation (i).

Question 3

When a certain number is multiplied by 13, the product consists entirely of fives. The smallest such number is?

Accepted Answer

By trial, we find that the smallest number consisting entirely of fives and exactly divisible by 13 is 555555. On dividing 555555 by 13, we get 42735 as quotient. ∴ Req. smallest number =42735.

Question 4

A number consist of two digit whose sum is 10 If the digit of the number are reversed then the number decreased by 36. Which of the following is correct. I. The number is divisible by a composite number. II. The number is a multiple of a prime numbe…

Accepted Answer

let the two-digit number be 10p + q Now according to the question p + q = 10 and (p +10q) + 36 = (q + 10p) ⇒ -9q + 9p = 36 ⇒ -q + p = 4 ....(ii) on adding eq (i) and (ii) we get 2p = 14 ⇒ p =7 ∴ p = 7 and q = 4 ∴ Required number is the 73 So, the number is a multiple of prime number.

Question 5

Every prime number of the form 3k + 1 can be represented in the form 6m + 1 (where k, m are integers), When

Accepted Answer

The only one even prime number is 2 and this is not in the form of 3k + 1. Thus any prime number of the form 3k + 1 is an odd number. This means 3k must be even number which implies that k must be even number also. Let k = 2m. Then a prime of the form 3k + 1 is of the form 3(2m) + 1 = 6m+1. Every prime number of form 3k + 1 can be represented in the form 6m + 1 only, when k is even.

Question 6

A number consist of two digit whose sum is 10. If the digits of the number are reversed, then the number decreased by 36. Which of the following is/are correct? I. The number is divisible by a composite number. II. The number is multiple of a prime …

Accepted Answer

Let the two digit number be 10x + y. Now, according to the question, x + y = 10............(i) if digits reversed , number get decreased by 36, and ( x +10y) = ( 10x + y ) - 36 ⇒ - 9y + 9x = 36 ⇒ x - y = 4 ..................... (ii) On adding Eqs. (i) and (ii), we get 2x = 14 ⇒ x = 7 ∴ x = 7 and y = 3 ∴ Required number is 73. So,the number is a multiple of a prime number.

Question 7

If the numerator of a certain fraction is increased by 2 and the denominator is increased by 1, then the resulting fraction is equal to the 1/2. If how ever the numerator is increased by 1, then denominator is decreased by 2, then the resulting frac…

Accepted Answer

Let the fraction be x / y. According to the question, ( x+2)/(y+1) = 1/2 ⇒ 2x + 4 = y+1 ⇒ 2x - y = -3 .........................(i) Again according to the question, (x+1)/(y-2) = 3/5 ⇒ 5x +5 = 3y - 6 ∴ 5x - 3y = -11..........................(ii) on multiplying Eq. (i) by 3 and then subtracting from Eq. (ii) we get, multiplying Eq. (i) by 3 6x - 3y = -9 subtracting from Eq. (ii) 5x- 3y - ( 6x - 3y ) = -11 - (-9) 5x- 3y - 6x + 3y = -11 + 9 - x = - 2 x= 2 On putting the value of x in Eq. (i) we get 2x- y = -3 2 x 2 - y = -3 ⇒ 2 x 2 -y = -3 ⇒ -y = -3 - 4 ∴ y = 7 ∴ Original fraction = x/y =2/7

Question 8

The sum of five consecutive odd numbers is equal to 175. What is the sum of the second largest number and the square of the smallest number among them together?

Accepted Answer

Method 1 to solve this question. Let p be the first odd number. Then next odd number is p + 2. According to question, Sum of five consecutive odd numbers 175 ⇒ p + p + 2 + p + 4 + p + 6 + p + 8 = 175 ⇒ 5p + 20 = 175 ⇒ p = (175 - 20)/5 = 31 According to question, Sum of the second largest and square of smallest one = (31 + 6) + (31)2 Sum of the second largest and square of smallest one = 37 + 961 = 998 Method 2 to solve this question. Let p be the even number. Then next odd number is p + 1. so on p + 3 , p + 5................ P + 9 According to question, Sum of five consecutive odd numbers 175 ⇒ p + 1 + p + 3 + p + 5 + p + 7 + p + 9 = 175 ⇒ 5p + 25 = 175 ⇒ p = (175 - 25)/5 ⇒ p = (150)/5 ⇒ p = 30 First odd number = p + 1 = 30 + 1 = 31 According to question, Sum of the second largest and square of smallest one = (30 + 7) + (30 + 1)2 Sum of the second largest and square of smallest one = 37 + 961 = 998

Question 9

Out of three given numbers, the first number is twice the second and thrice the third. If the average of these three numbers is 154, then what is the difference between the first and the third numbers?

Accepted Answer

Let the third numbers = p Then, the first number = 3p and second number = 3p/2 According to question, ( p + 3p + 3p/2 )/3 = 154 ⇒ ( 2p + 6p + 3p )/6 = 154 ∴ p = 154 x 6/11 = 84 ∴ Required difference = 3p - p = 2p = 2 x 84 = 168

Question 10

The differences between two numbers is 18. If four times the second numbers is less than three times the first number by 18, then what is the sum of these two numbers?

Accepted Answer

Let first number = p and second number = q According to the question, p - q = 18 ................................(i) and 3p - 4q = 18.....................(ii) On multiplying Eq .(i) by 3 and subtracting Eq. (ii) from it,we get 3p - 3q = 54 3p - 4q = 18 q = 36 On putting the value of q in Eq, (i) we get p = 18 + q ⇒ p = 18 + 36 = 54 ∴ Required sum = p + q = 54 + 36 = 90

Question 11

On children's Day, sweets were to be equally distributed amongst 300 children. But on that particular day 50 children remained absent; hence each child got one extra sweet. How many sweet were distributed?

Accepted Answer

Let total number of sweets distributed on children's day = x According to the question, x/250 - x/300 = 1 ⇒ ( 6x - 5x )/1500 = 1 ∴ x = 1500 ∴ Required number of sweets distributed on children's day = 1500

Question 12

There are two examination halls P and Q. If 10 students shifted P to Q, then the number of students will be equal in both the examination halls. If 20 students shifted from Q to P, then the students of P would be doubled to the students of Q. The nu…

Accepted Answer

Let number of students in examination halls P and Q is x and y, respectively Then as per the first condition, x - 10 = y + 10 ⇒ x - y = 20 ..............(i) As per the second condition, ⇒ x + 20 = 2( y - 20) ⇒ x - 2y = - 60 ...............(ii) On subtraction Eq. (ii) from Eq. (i), we get - y + 2y = 20 + 60 ⇒ y = 80 Putting the value of y Eq. (i) we get, x - 80= 20 ⇒ x = 100 Hence, number of students in examination halls P and Q is 100 and 80, respectively.

Question 13

The sum of the square of two numbers is 97 and the square of their difference is 25. The product of the two number is

Accepted Answer

Let the two number are a and b According to the question sum of squares of two numbers = 97 i.e.., a 2 + b 2 = 97 .....................(i) and square of their difference = 25 i.e., (a - b) 2 = 25 .................(ii) ⇒ a - b = 5...............(iii) from Eq. (ii) a2 + b2 - 2ab = 25 (a2 + b2) - 2ab = 25 Now put the value of a 2 + b 2 from Eq. (i) ⇒ 97 - 2ab = 25 ⇒ 2ab = 72 ⇒ ab = 36.................(iv) Now, we have (a + b)2 = ( a2 + b2) + 2ab Now put the value of a 2 + b 2 from Eq. (i) and ab from Eq. (iv) (a + b)2 = 97 + 72 = 169 a + b = 13.............(v) Now add the Eqs. (iii) and (v), we get a - b + a + b = 5 +13 2a = 18 ⇒ a = 9 now put the value of a in Eq. (v) a + b = 13 9 + b = 13 b = 13 - 9 b = 4 ∴ Product of both the numbers = ab = 9 x 4 = 36

Question 14

There are 200 question in a 3 hour examination. Among 200 question, 50 are Mathematics, 100 are from GK and 50 are from Sciences. Ram spent twice as much time on each Mathematics question for each other question. How many minute did he spend on Math…

Accepted Answer

Let Ram spends p minute on each Mathematics question. According to the question. 50 x p + 100 x p/2 + 50 x p/2 = 3 x 60 p(50+ 50 + 25) = 180 ⇒ p = 180/125 ∴ Required time = 50 x 180/125 ∴ Required time = 2 x 180/5 ∴ Required time = 2 x 36 = 72 min Method 2 Let Ram spends a minute on each sciences and Gk questions. According to the question, 50 x 2a + 150 x a = 3 x 60 100a + 150a = 180 250a = 180 a = 180/250 So total time spend on Mathematics question = 2a x 50 question So total time spend on Mathematics question = 2 x 50 x 180/250 = 72 minutes

Question 15

The taxi charges in a city contain fixed charges and additional charge per kilometer. The fixed charge is for a distance of up to 5 km and additional charge per kilometer thereafter. The charge for a distance of 10 km is ₹ 350 and for 25 km is ₹ 800…

Accepted Answer

let the fixed charges = ₹ p for first 5 km and the additional charges = ₹ q per km according to the question p + 5q = 350........... (1) p + 20q = 800...........(2) on subtracting Eq. 1 from Eq. 2 we get 15q = 450 q = 30 On putting the value of q in Eq. (1) we get p = 200 ∴ charge for a distance of 30 km = p + 25q = 200 + 30 x 25 = ₹ 950

Question 16

X, Y and Z had have dinner together. The cost of the meal of Z was 20% more then that of Y and the cost of the meal of X was 5/6 as much as the cost of the meal of Z. If Y paid ₹ 100, then what was the total amount that all the three of them had pai…

Accepted Answer

Given that, The cost of meal of Y = ₹ 100 Now, according to the question, The cost of the meal of Z = 20% more than that of Y The cost of the meal of Z = (100 + 100 x 20 %) = (100 + 100 x 20/100 ) = (100 + 20) = ₹ 120 and the cost of the meal of X = 5/6 as much as the cost of the meal of Z = 5/6 x 120 = ₹ 100 ∴ Total amount that all the three has to be paid = 100 + 120 + 100 Total amount that all the three has to be paid = ₹ 320

Question 17

In an examination paper of five questions 5% of the candidates answered all of them and 5% answered none. Of the rest 25% candidates answered only one question and 20% answered 4 question. If 396 candidates answered either 2 question or 3 question. …

Accepted Answer

Let total number of candidates in Exam = a Number of candidates answered 5 questions = a x 5% = a x 5/100 = 5a/100 = a/20 Number of candidates answered not any questions = a x 5% = a x 5/100 = 5a/100 = a/20 ∴ Remaining students = a - ( a/20 + a/20) = a - ( 2a/20 ) = a - ( a/10 ) = (10a - a)/10 = 9a/10 Number of candidates answered only 1 question = ( 9a/10 ) x 25% = ( 9a/10 ) x 25/100 = 9a/40 Number of candidates answered 4 questions = ( 9a/10 ) x 20% = ( 9a/10 ) x 20/100 = 9a/50 Given number of candidates awarded either 2 questions or 3 questions = 396 ⇒ a - ( a/20+ a/20 + 9a/40 + 9a/50 ) = 396 ⇒ a - ( a/10 + 9a/40 + 9a/50 ) = 396 ⇒ a - ( ( a x 20 + 9a x 5 + 9a x 4 )/200) = 396 ⇒ a - ( ( 20a + 45a + 36a )/200) = 396 ⇒ a - ( 101a/200) = 396 ⇒ ( 200a - 101a)/200 = 396 ⇒ ( 99a)/200 = 396 ∴ a = 396 x 200/99 ∴ a = 4 x 200 = 800 ⇒ a = 800 Hence, number of candidates = 800

Question 18

In a three-digit number, the digit in the unit's place is four times the digit in the hundred's place. If the digit in the unit's place and the ten's places are interchanged, the new number so formed is 18 more than the original number. If the digit…

Accepted Answer

Let hundred's place digit = p Then according to question, unit's digit = 4 x hundred's place digit = 4p and Ten's place digit = 3 x hundred's place digit = 3p So Number = 100 x p + 10 x 3p + 1 x 4p= 134p If the digit in the unit's place and the ten's places are interchanged according to question, unit's digit = 3p and Ten's place digit = 4p Now Number = 100 x p + 10 x 4p + 1 x 3p = 143p According to the question, after interchanging, 143p - 134p = 18 ⇒ 9p = 18 ∴ p = 2 Original number = 134p = 134 x 2 = 268 25% of original number = 268 x 25/100 = 67

Question 19

In a two digit positive number the unit digit is equal to the square of ten's place digit. The difference between the original number and the number formed by interchanging the digit is 54. What of 40% of the original number?

Accepted Answer

Let ten's place digit be a and unit's place digit be a2 Original number = 10 x a + 1 x a2 = 10a + a2 The number formed by interchanging the digits, New number = 10 x a 2 + 1 x a = 10a 2 + a According to the question (10a2 + a) - ( 10a + a2) = 54 ⇒ 10a2 + a - 10a - a 2 = 54 ⇒ 9a2 - 9a = 54 ⇒ 9( a2 - a) = 54 ⇒ ( a2 - a) = 54/9 ⇒ ( a2 - a) = 6 ⇒ a2 - a - 6 = 0 ⇒ a2 - 3a + 2a - 6 = 0 ⇒ a (a - 3) + 2 (a - 3) = 0 ∴ (a - 3) (a + 2) = 0 ∴ a = 3, - 2 ∴ Ten,s digit = a = 3 Unit's digit = a2 = 32 = 9 Original number = 39 ∴ Required number = 39 x 40/100 = 15.6

Question 20

Find the maximum number of trees which can be planted 20 m apart on the two sides of a straight road 1760 m long.

Accepted Answer

Number of trees that can be planted on one side of road = (Length of road/gap between two trees ) + 1 Number of trees that can be planted on one side of road = ( 1760/20 )+ 1 Number of trees that can be planted on one side of road = 88 + 1 = 89 ∴ Threes on the both sides = 2 x 89 = 178

Number System Questions