Question 1

The system of equations 2x + 4y = 6 and 4x + 8y = 6 has

Accepted Answer

Given equations 2x + 4y = 6 and 4x + 8y = 6 then, a1/a2 = 2/4 = 1/2; b1/b2 = 4/8 = 1/2; c1/c2 = 6/6 = 1 ∴ a1/b2 = b1/b2 ≠ c1/c2 So there is no solution for these equations.

Question 2

In a rare coin collection, there is one gold coin for every three non-gold coins. 10 more gold coins are added to the collection and the ratio of gold coins to non-gold coins would be 1 : 2, Based on the information; the total number of coins in the…

Accepted Answer

Let the number of gold coins initially be x and the number of non-gold coins be y. According to the question, 3x = y When 10 more gold coins, total number of gold coins becomes x + 10 and the number of non-gold coins remain the same at y. Now, we have 2(x + 10) = y Solving these two equations, we get x = 20 and y = 60. Total number of coins in the collection at the end is equal to x + 10 + y = 20 + 10 + 60 = 90.

Question 3

If (x + y - 8)/2 = (x + 2y - 14)/3 = (3x + y - 12)/11 then find the values of x and y, respectively.

Accepted Answer

Given, (x + y - 8)/2 = (x + 2y - 14)/3 = (3x + y - 12)/11 ⇒ (x + y - 8)/2 = (x + 2y - 14)/3 ⇒ 3x + 3y - 24 = 2x + 4y - 28 ⇒ 3x + 3y - 2x - 4y = -28 + 24 x - y = -4 ...(i) Again, (x + 2y - 14)/3 = (3x + y - 12)/11 ⇒ 11x + 22y - 154 = 9x + 3y - 36 ⇒ 2x + 19y = 118 ..(ii) On multiplying Eq. (i) by 2 and subtracting from Eq., we get 2x - 2y = -8 2x + 19y = 118 ------------------- -21y = -126 ∴ y = 6 On putting the value of y in Eq. (i), we get x - 6 = -4 ∴ x = 2 ∴ x = 2 and y = 6

Question 4

If [√ 3 + x + √ 3 - x] / [√ 3 + x - √ 3 - x] = 2, then x is equal to

Accepted Answer

[√3 + x + √3 - x] / [√3 + x - √3 - x] = 2 Let √3 + x = a and √3 - x = b Then, (a + b) / (a - b) = 2/1 ∴ a + b = 2a - 2b ⇒ a = 3b ∴ √3 + x = 3√3 - x On squaring bothv sides, we get (√3 + x)2 = (3√3 - x)2 ⇒ 3 + x = 9(3 - x) ⇒ 3 + x = 27 - 9x ⇒ 10x = 24 ∴ x = 12/5

Question 5

The graph of ax + by = c, dx + ey = f will be I. parallel, if the system has no solution. II. coincident, if the system has finite number of solutions. III. intersecting. if the system has only one solution. Which of the above statements are correct?

Accepted Answer

ax + by = c and dx + ey = f a1/a2 = a/d, b1/b2 = b/e, c1/c2 = c/f ∵ b1/b2 ≠ c1/c2 ∴ b/e ≠ c/f it represent a pair of parallel lines. ∵ a1/a2 ≠ b1/b2 ∴ a/d ≠ b/e Therefore, system has unique solutions and represents a pair of intersecting lines.

Question 6

Ten chairs and six tables together cost ₹ 6200, three chairs and two tables together cost ₹ 1900, The cost of 4 chairs and 5 tables is?

Accepted Answer

Let the cost of one chair be ₹ x and cost of one table be ₹ y. By given condition, 10x + 6y = 6200 ..(i) and 3x + 2y = 1900 ⇒ 9x + 6y = 5700 ...(ii) On subtracting Eq. (ii), we get x = ₹ 500 From Eq (i), 5000 + 6y = 6200 ⇒ 6y = 1200 ∴ y = ₹ 200 The cost of 4 chair and 5 tables = 4x + 5y = 2000 + 1000 = ₹ 3000

Question 7

If 2 a + 3 b = 17 and 2 a + 2 - 3 b +1 = 5, then

Accepted Answer

Given, 2a + 3b = 17 and 2a + 2 - 3b + 1 = 5 ⇒ 22 x 22 - 3b x 31 = 5 ⇒ 4.2a - 3.3b = 5 Let 2a = x amd 3b = y Then, x + y = 17 ...(i) 4x - 3y = 5 ...(ii) On multiplying Eq. (i) by 3 and adding to Eq (ii), we get 3x + 3y = 51 4x - 3y = 5 ------------------ 7x = 56 ⇒ x = 8 On putting the value of x in Eq. (i), we get 8 + y = 17 ∴ y = 9 Now, 2a = x ⇒ 2a = 8 (2)3 ∴ 3b = y = 9 ⇒ 3b = 32 ∴ b = 2 Hence, a = 3 and b = 2

Question 8

The solution of the equations (p/x) + (q/y) = m and (q/x) + (p/y) = n is

Accepted Answer

(p/x) + (q/y) = m ..(i) (q/x) + (p/y) = n ...(ii) On multiply Eq (i) by q and Eq. (ii) by p and subtracting, we get (pq/x) + (q2)/y = mq (pq/x) + (p2)/y = np ----------------------------------------- (q2/y) - (p2/y) = mq - np ∴ (q2 - p2) = y(mq - np) ∴ y = (q2 - p2)/mp - np = (p2 - q2)/np - mq Again, on multiplying Eq. (i) by p and Eq. (ii) by q and subtracting, we get (p2/x) + (pq/y) = mp (q2/x) + (pq/y) = nq ----------------------------------------- (p2/x) - (q2/x) = mp - nq ⇒ (p2 - q2) = x (mp - nq) ⇒ x = (p2 - q2)/(mp - nq) ∴ x = (p2 - q2)/(mp - nq) and y = (p2 - q2) / (np - mq)

Question 9

One says "Give me a hundred friend ! I shall then become twice as rich as you ". The other replies, "If you give me ten, I shall be six times as rich as you." Find the amount of their capitals.

Accepted Answer

Let the capital of one is x and that of another is y . According to the question. x + 100 = 2(y - 100) x + 100 = 2y - 200 or x - 2y = -300 ...(i) Again, according to the question y + 10 = 6(x - 10) ⇒ y + 10 = 6x - 10 ⇒ 6x - y = 70 ..(ii) On multiplying Eq. (ii) by 2 and subtracting from Eq. (i) , we get x - 2y = -300 12x - 2y = 140 ---------------- -11x = -440 ∴ x = 40 On putting the value of x in Eq. (i) , we get 40 - 2y = -300 ⇒ 2y = 340 ∴ y = 170 So, their capital are ₹ 40 and ₹ 170.

Question 10

The ratio of incomes of two persons is 8 : 5 and the ratio of their expenditure i 2 : 1. If each of them manages to save ₹ 1000 per month, find the difference of their monthly income

Accepted Answer

Let the incomes of two persons be 8x and 5x and their expenditure be 2y and y , respectively. ∵ Saving = Income - Expenditure ∴ 1000 = 8x - 2y ...(i) and 1000 = 5x - y ...(ii) On multiplying Eq. (ii) by 2 and subtracting from Eq. . (i) , we get 8x - 2y = 1000 10x - 2y = 2000 ---------------------- -2x = -1000 ∴ x = 500 ∴ Monthly incomes are 8x = ₹ 4000 and 5x = ₹ 2500 ∴ Difference = ₹ 4000 - 2500 = ₹ 1500

Question 11

A fraction becomes 7/8, if 5 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 6/7. Find the fraction.

Accepted Answer

Let the fraction be x/y, According to the question, (x + 5) / (y + 5) = 7/8 ⇒ 8x + 40 = 7x + 35 ∴ 8x - 7y = -5 ..(i) Again, according to the question, (x + 3)/(y + 3) = 6/7 ⇒ 7x + 21 = 6y + 18 ∴ 7x - 6y = -3 ..(ii) On multiplying Eq. (i) by 6 and Eq. (ii) by 7 and subtracting, we get 48x - 42y = -30 49x - 42y = -21 --------------------- -x = -9 ∴ x = 9 On putting te value of x in Eq. (i) , we get 72 - 7y = -5 ⇒ -7y = -5 - 72 ⇒ y = -5 - 72 ⇒ y = (-77) / (-7) = 11 ∴ Required fraction = 9/11

Question 12

If the number obtained on the interchanging the digits of two-digits number is 18 more than the original number and the sum of the digits is 8, then what is the original number?

Accepted Answer

Method 1 to solve the given question. Lets assume the unit's digit is y and the ten's digit is x. then, the number is 10 x + y. According to question, after interchanging the digits, the new number is 10y + x. Then, New Number - 18 = Original number 10y + x = 10x + y + 18 ⇒ 9y - 9x = 18 ⇒ y - x = 2 .....................(1) Again according to question, Sum of digits of Original Number = 8 x + y = 8 ..............................(2) Add the equation (1) and (2) , we will get y - x + x + y = 2 + 8 ⇒ 2y = 10 ⇒ y = 5 Put the value of Y in equation (2) , we will get x + 5 = 8 ⇒ x = 8 - 5 = 3 Then , the original number = 10x + y Put the value of x and y and get the original number. The original number = 10x + y = 10 x 3 + 5 = 30 + 5 = 35 Method 2 to solve the given question. Given that sum of digits of original number is 8. Let us assume that unit digit is x , then ten digit will be 8 - x. (since sum of digits is 8.) Now Original number = 10( 8 - x) + x After interchanging the digits , the number = 10x + (8 - x) = 9x + 8 According to question. New number = original number + 18 ⇒ 9x + 8 = 10(8 - x) + x + 18 ⇒ 9x + 8 = 80 - 10x + x + 18 ⇒ 9x + 8 = 98 - 9x ⇒ 9x + 9x = 98 - 8 ⇒ 18x = 90 ⇒ x = 90/18 ⇒ x = 5 The original number = 10(8 - x) + x = 10(8 - 5) + 5 = 10(3) + 5 = 30 + 5 = 35

Question 13

There are two numbers such that the sum of twice the first number and thrice the second number is 100 and the sum of thrice the first number and twice the second number is 120. Which is the larger number?

Accepted Answer

Let us assume the first number is x and second number is y. According to question, sum of twice the first number and thrice the second number is 100. 2x + 3y = 100 ...................(1) and sum of thrice the first number and twice the second number is 120. 3x + 2y = 120...................(2) Subtract equation (1) from Equation (2) after multiply 2 with Equation (1) and 3 with Equation (2), we will get. 3(3x + 2y) - 2(2x + 3y) = 3 x 120 - 2 x 100 9x + 6y - 4x - 6y = 360 - 200 5x = 160 x = 32 put the value of x in equation (1) 2 x 32 +3y = 100 3y = 100 - 64 3y = 36 y = 12 The larger Number is 32 .

Question 14

A series of books was published at seven years interval. When the seventh book was issued, the sum of the publication year was 13,524. When was the first book published?

Accepted Answer

Let us assume the first book published in year x. According to question, Books are published at seven years interval, First edition book published in year = x Second edition book published in year = x + 7 Third edition book published in year = x + 14 Four edition book published in year = x + 21 Five edition book published in year = x + 28 Six edition book published in year = x + 35 Seven edition book published in year = x + 42 When the seventh book was issued, the sum of the publication year was 13,524. x + x + 7 + x +14 + x +21 + x + 28 + x + 35 + x + 42 =13524 147 + 7x = 13524 7x = 13524 -147 ⇒ x = 13377/7 = 1911 The first book published in year x = 1911

Question 15

The difference between a two- digit number and the number obtained by interchanging the position of its digits is 36.What is the difference between the two digits of that number?

Accepted Answer

Let the ten's digits be x and unit's digit be y. Then,(10x + y) - (10y + x)=36 ⇔9(x-y)=36 ⇔ x - y=4

Question 16

If the digits of a two-digit number are interchanged , the number formed is greater than the the original number by 45. If the difference between the digits is 5, then what is the original number?

Accepted Answer

Let us assume the two digits number of the number is a and b. According to given question, The original number is 10a + b. After interchanging the digits , the Number is 10b + a. The number formed is greater than the the original number by 45, We will get, New number = Original number + 45 ⇒ 10b + a = 10a + b + 45 ⇒ 10b + a - 10a - b = 45 ⇒ 9b - 9a = 45 ⇒ b - a = 5 According the question the difference between the digits is 5 which will not help us to solve this question because both condition given in question are same. So can't be determined the answer. Now try Hit and Trail method Try one by one all option. All 3 Options are true. But we cannot choose all 3 option, but as per question we cannot find the answer.

Question 17

In a two-digit number, the digit in the unit's place is three times the digit in tenth's place. The sum of the digits is equal to 8. What is the number?

Accepted Answer

Let us assume the ten's digit of the number is n. Then the unit's digit will be 3n. According to question, The sum of the digits is equal to 8. sum of the digit = 8 n + 3n = 8 ⇒ 4n = 8 ⇒ n = 2 So, ten's digit is 2 and unit's digit is 6. So the number = 10n + 3n = 10 x 2 + 3 x 2 = 20 + 6 = 26.

Question 18

In a group of equal number of cows and herdsmen, the number of legs was 28 less than four times the number of heads . The number of herdsmen was

Accepted Answer

Suppose the number of cows = a therefore , the number of herdsmen = a The total number of legs = Legs of cows + Legs of herdsmen ( Cow has 4 legs and herdsmen has 2 legs ) The total number of legs = a x 4 + a x 2 = 4a + 2a = 6a The total number of heads = Heads of cows + Heads of herdsmen ( Cow has 1 head and herdsmen has also 1 head ) The total number of heads = a + a = 2a According to question, The total number of legs was 28 less than four times the number of heads, 6a = 4 x 2a - 28 ⇒ 8a - 28 = 6a ⇒ 8a - 6a = 28 ∴ a = 28/2 = 14

Question 19

A number when subtracted by 1/7 of itself gives the same value as the sum of all the angles of a triangle. What is the number?

Accepted Answer

Let us assume the number be N. According to question, A number when subtracted by 1/7 of itself gives the same value as the sum of all the angles of a triangle, N - N x 1/7 = 180 ( As we know sum of all angels of a triangle is 180 .) ⇒ N - N/7 = 180 ⇒ (7N - N)/7 = 180 ⇒ (7N - N) = 180 x 7 ⇒ 6N = 180 x 7 ⇒ N = 180 x 7/6 ⇒ N = 30 x 7 ⇒ N = 210

Question 20

A number of two digits has 3 for its unit's place,and the sum of digits is 1/7 of the number itself, The number is

Accepted Answer

Let us assume the ten's place digit be x. Then, the number = 10x + 3 and sum of digits = x + 3 According to question, So, ( x + 3 ) = 1/7( 10x + 3 ) ⇔7x + 21 =10x +3 ⇔ 3x = 18 ⇔ x = 6 The number = 10x + 3 = 10 x 6 + 3 = 60 + 3 = 63

Linear Equation Questions