Question 1

Out of two sections A and B, 10 students of section B shift to A, as a result strength of A becomes 3 times the strength of B. But , if 10 students shift over from A to B, both A and B become equal in strength. Ratio of the number of students in sec…

Accepted Answer

A +10 = 3(B - 10) ⇒ A - 3B = -20 ....(i) And ( A - 10) = (B + 10) ⇒ A - B = 20 ...(ii) on subtracting Eq.(ii) from Eq.(i), we get (A - 3B) - (A - B) = -20 - 20 ⇒ - 2B = - 40 there4; B = 20 Now, by using Eq. ...(ii), there4; Ratio of the numbers of students of A and B = 40 : 20 = 2 : 1

Question 2

A sum of money is to be divided equally among P, Q and R in the respective ratio of 5 : 6 : 7 and another sum of money is to be divided between S and T equally. If S got ₹ 2100 less than P, then how much amount did Q receive?

Accepted Answer

Couldn't be determined, since the total amount of money is not given in either of the case

Question 3

Nandita scores 80% marks in five subjects together, viz., Hindi, Science, Mathematics, English and Sanskrit, where in the maximum marks of each subject were 105, How many marks did Nandita score in Science, if she scored 89 marks in Hindi, 92 marks …

Accepted Answer

Total of maximum marks of all subjects = 105 x 5 = 525 Total marks obtained by Nandita = 80% of 525 = 525 x 80/100 = 420 Obtained marks of four subjects (Hindi, Sanskrit, Mathematics and English) = 89 + 92 + 98 + 81 = 360 So, the obtained marks in Science = 420 - 360 = 60

Question 4

Number of employees in a factory decreases in the ratio of 8 : 7 and salary of employees increases in the ratio of 5 : 6 . Find whether the total salary given to the employee is increasing or decreased and in What ratio?

Accepted Answer

Let total number of employees in beginning = 8K and total number of employees in present = 7K Let salary of an employee in beginning = 5L and salary of an employee in present = 6L ∴ Total salary in beginning = 8K x 5L = 40KL and total salary in present = 7K x 6L = 42 KL Required ratio = 40KL : 42KL = 20 : 21 Clearly, salary is increased.

Question 5

Out of 120 applications for a post, 70 are males and 80 have a driver's licences . What is the ratio between the minimum to maximum number number of males having driver's license?

Accepted Answer

Since there are 70 males out of 120 applicants, there must be 50 females out must have For the minimum number of males to have a driver's licence all 50 females must have a driver's license Thus the number of males having a driver's licence will be 80 - 50 = 30. The maximum possible number of males having a driver's license is 70. The ratio between the minimum and the maximum is 30 : 70 or 3 : 7 .

Question 6

in a mixture of 60 L the ratio of acid and water is 2 : 1. If the ratio of acid and water is to be 1 : 2, then the amount of water (in liters) to be added to the mixture is?

Accepted Answer

Quantity of acid in the mixture = (2/3) x 60 = 40 L Quantity of water in the mixture = (1/3) x 60= 20 L Let the required quantity of water be x L. According to the question, 40/(20+ x) = 1/2 ⇒ 80 = 20 + x ⇒ x = 60 L

Question 7

Tea worth ₹ 126 kg and ₹ 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . If the mixture is worth ₹ 153 per kg, the price of the third variety per kg will be?

Accepted Answer

Let the cost of third variety tea will be ₹ N per kg According to the question, (126 + 135 + 2N)/4 = 153 ⇒ (261 + 2N) = 4 x 153 ⇒ 2N = 612 - 261 = 351 ∴ N = 351/2 = 175.50 ⇒ N = ₹ 175.5 Hence, the cost of third variety tea is ₹ 175.50 per kg .

Question 8

Find the compound ratio of 2 : 7, 5 : 3 and 4 : 7?

Accepted Answer

Required compound ratio = (2 x 5 x 4) / (7 x 3 x 7) = 40/147 = 40 : 147

Question 9

If A : B = 3 : 4 and B : C = 8 : 9 then find the value of A : B : C?

Accepted Answer

Given that A : B = 3 : 4 = ( 3 x 2 ) : (4 x 2 ) = 6 : 8 B : C = 8 : 9 ∴ A : B : C = 6 : 8 : 9 As consequent of the first ratio is equal to the antecedent of second ratio.

Question 10

If (a + b ) : (a - b) = 5 : 3, then find ( a 2 + b 2 ) : ( a 2 - b 2 ) .

Accepted Answer

(a + b) / (a - b) = 5/3 ⇒ 3a + 3b = 5a - 5b ⇒ 2a = 8b ⇒ a = 4b, ⇒ a/b = 4/1 Now, ( a2 + b2 ) / (a2 - b2 ) = [a2 / (b2+1)] / [a2 / (b2 -1)] = [( a/b )2 + 1] / [( a/b )2 - 1] = [( 4/1 )2 + 1] / [( 4/1 )2 - 1] = [16 + 1] / [16 - 1] = 17/15 ∴ ( a2 + b2 ) : ( a2 - b2 ) = 17 : 15

Question 11

If x/2y = 6/7, then find the value of (x - y) / (x + y) + 14/19?

Accepted Answer

Given that, x/2y = 6/7 ⇒ x/y = 12/7 By componenod dividendo, (x - y) / (x + y) = (12 - 7) / (12 + 7) = 5/19 ∴ (x - y) / (x + y) + 14 /19 = 5/19 + 14/19 = 19/19 = 1

Question 12

The total weight of sanjay and suresh is 120 kg. If sanjay weighs 30 kg more than Suresh, then what is the ratio of the weight of suresh to that of sanjay?

Accepted Answer

Given that Sanjay + Suresh = 120 ...(i) Sanjay - Suresh = 30 ...(ii) By adding Eqs (I) and (ii) we get 2 x sanjay = 150 ∴ sanjay = 150/2 = 75 kg ∴ Weight of Suresh = 75 - 35 = 45 kg ∴ Suresh : Sanjay = 45 : 75 = 3 : 5 = 3/5 = 0.6

Question 13

In a school the ratio of boy and girls is 4 : 5 When 100 girls leave the school the ratio become 6 : 7. How many boys are there in the school?

Accepted Answer

Let number of boys = 4N and number of girls = 5N Accounting to the question 4N / (5N - 100) = 6/7 ⇒ 28N = 6 ( 5N - 100 ) ⇒ 28N = 30N - 600 ⇒ 2N = 600 ∴ N = 600/2 = 300 ∴ Number of boys = 4N = 4 x 300 = 1200

Question 14

In a class the number of boys and girls is in the ratio of 4 : 5. If 10 more boys join the class, the ratio of numbers of boys and girls become 6 : 5. How many girls there in the class?

Accepted Answer

Let original number of boys = 4N and number of girls = 5N According to the question (4N + 10)/5N = 6/5 ⇒ 30N = 20N + 50 ⇒ 10N = 50 ∴ N = 50/10 = 5 ∴ Number of girls = 5N = 5 x 5 = 25

Question 15

The total number of student in a school is 8670. If the number of boys in the school is 4545, then what will be the ratio of the total number of boys to the total number of girls in the school?

Accepted Answer

Total number of student = 8670 Total number of boys = 4545 ∴ Total number of girls = 8670 - 4545 = 4125 ∴ Required ratio = 4545 : 4125 = 303 : 275

Question 16

A truck covers a distance 640 km in 10 h. A car covers the same distance in 8 h. What is the respective ratio of speed of the truck to that of car?

Accepted Answer

Speed of truck 640/10 = 64 km/h Speed of car = 640/8 = 80 km/h So, the required ratio = 64/80 = 4/5 = 4 : 5

Question 17

A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speeds is

Accepted Answer

Case I Speed = Distance/ Time Case II Speed = (Distance/2) / (2 x Time) = Distance / (4 x Time) ∴ Required ratio = 1/4 = 1:4

Question 18

A person distributes his pens among four friends A, B, C and D in the ratio 1/3 : 1 / 4 : 1/5 : 1/6. What is the minimum number of pens that the person should have?

Accepted Answer

The ratio among A, B, C and D = 1/3 : 1 / 4 : 1/5 : 1/6 On rearranging the ratio = 60/3 : 60/4: 60/5 : 60/6 = 20 : 15 : 12 : 10 So minimum number of pens can be when the common ration is 1. So minimum number of pen = 20 + 15 + 12 + 10 = 57

Question 19

Divide ₋ 990 into 3 parts in such a way that half of the first part, one - third of the second part and one -fifth of the third part are equal .

Accepted Answer

If three parts be P, Q and R, then, P/2 = Q/3 = R/5 ⇒ P : Q : R = 2 : 3 : 5 ∴ P = (2/10) x 990 = 198 Q = (3/10) x 990 = 297 and R = (5/10) x 990 = 495

Question 20

The quantity that must be added to each term of a : b, so as to make it c : d, is

Accepted Answer

Let the quantity be k. Then, (a + k) / (b + k) = c/d ⇒ (a + k)d = (b + k)c ⇒ ad + dk = bc + ck ⇒ ad - bc = ck - dk ⇒ ad - bc = k(c - d) ∴ k = (ad - bc) / (c - d)

Ratio and Proportion Questions