RATIO & PROPORTION
Note: Question numbers are numbers from the actual exam in the respective years mentioned below
- The monthly incomes of Peter and Paul are in the ratio of 4: 3. Their expenses are in the ratio of 3: 2. If each saves Rs 6,000 at the end of the month, their monthly incomes respectively are (in Rs) (2015)
(A) 24,000 and 18,000
(B) 28,000 and 21,000
(C) 32,000 and 24,000
(D) 34,000 and 26,000
Answer-A
Let's assume the monthly incomes of Peter and Paul are 4x and 3x respectively, where x is a common ratio.
According to the given information, their expenses are in the ratio of 3:2. Let's assume their expenses are 3y and 2y respectively, where y is a common ratio.
We know that their savings at the end of the month are both Rs. 6,000. So, we can set up the following equation:
4x - 3y = 6000 ---(1)
3x - 2y = 6000 ---(2)
To solve this system of equations, we can use the method of substitution.
From equation (2), we can express x in terms of y:
x = (6000 + 2y) / 3
Substituting this value of x in equation (1), we have:
4 * ((6000 + 2y) / 3) - 3y = 6000
(24000 + 8y) / 3 - 3y = 6000
24000 + 8y - 9y = 18000
24000 - y = 18000
y = 24000 - 18000
y = 6000
Substituting the value of y back into equation (1) or (2), we can find x:
3x - 2(6000) = 6000
3x - 12000 = 6000
3x = 18000
x = 18000 / 3
x = 6000
Therefore, the monthly incomes of Peter and Paul are 4x and 3x respectively:
Peter's monthly income = 4 * 6000 = 24000 Rupees
Paul's monthly income = 3 * 6000 = 18000 Rupees
So, the correct option is (a) 24000 and 18000.
- The total emoluments of two persons are the same, but one gets allowances to the extent of 65% of his basic pay and the other gets allowances to the extent of 80% of his basic pay. The ratio of the basic pay of the former to the basic pay of the latter is :(2016)
(A) 16: 13
(B) 5:4
(C) 7:5
(D) 12: 11
Answer-D
Let's assume the basic pay of the former person is x.
According to the given information, the allowances of the former person are 65% of his basic pay, which is 0.65x. So, the total emoluments of the former person would be x + 0.65x = 1.65x.
Similarly, let's assume the basic pay of the latter person is y.
The allowances of the latter person are 80% of his basic pay, which is 0.8y. So, the total emoluments of the latter person would be y + 0.8y = 1.8y.
Since the total emoluments of both persons are the same, we can equate the expressions for total emoluments:
1.65x = 1.8y
To find the ratio of their basic pay, we need to simplify this equation:
x/y = 1.8/1.65
We can multiply both the numerator and denominator by 20 to eliminate the decimals:
x/y = (1.8 * 20)/(1.65 * 20)
= 36/33
= 12/11
Therefore, the ratio of the basic pay of the former person to the basic pay of the latter person is 12:11.
The correct option is (d) 12:11.
- If there is a policy that 1/3rd of a population of a community has migrated every year from one place to some other place, what is. the leftover population of that community after the sixth year, if there is no further growth in the population during this period?(2017)
(A) 16/243rd part of the population
(B) 32/243rd part of the population
(C) 32/729th part of the population
(D) 64/729th part of the population
Answer-D
Let's assume the initial population of the community is P.
According to the given policy, 1/3rd of the population migrates every year. This means that after the first year, 1/3rd of the population remains, which is (2/3)P. Similarly, after the second year, (2/3) of the remaining population remains, which is [(2/3) * (2/3)]P = (4/9)P.
We can observe that after each year, the remaining population is multiplied by (2/3). Therefore, after the sixth year, the leftover population can be calculated as:
[(2/3) * (2/3) * (2/3) * (2/3) * (2/3) * (2/3)]P
= (2/3)^6 * P
= (64/729)P
Therefore, the leftover population of the community after the sixth year is (64/729)th part of the initial population.
The correct option is (d) 64/729th part of the population.
- P works thrice as fast as Q, whereas P and Q together can work four times as fast as R. If P, Q and R together work on a job, in what ratio should they share the earnings?(2017)
(A) 3:1:1
(B) 3:2:4
(C) 4:3:4
(D) 3:1:4
Answer-A
Given:
P works thrice as fast as Q:
P = 3Q ---(1)
P and O together can work four times as fast as R:
P + Q = 4R ---(2)
Solving equations 1 and 2 , we get Q=R—----(3)
So the answer is (a) i.e, 3:1:1
- The monthly incomes of X and Y are in the ratio of 4:3 and their monthly expenses are in the ratio of 3:2. However, each saves Rs 6,000 per month. What is their total monthly income?(2017)
(A) Rs 28,000
(B) Rs 42,000
(C) Rs 56,000
(D) Rs 84,000
Answer-B
Let's assume that X's monthly income is 4x and Y's monthly income is 3x. Also, let X's monthly expenses be 3y and Y's monthly expenses be 2y.
Given that X and Y each save Rs 6,000 per month, we can set up the following equations:
4x - 3y = 6000 (X's monthly savings)
3x - 2y = 6000 (Y's monthly savings)
We can solve these equations to find the values of x and y:
Multiplying the first equation by 2 and the second equation by 3, we get:
8x - 6y = 12000
9x - 6y = 18000
Subtracting the second equation from the first, we get:
-x = -6000
x = 6000*1 = 6000
Substituting the value of x in either of the two equations, we can find the value of y:
4x - 3y = 6000
4(6000) - 3y = 6000
24000 - 6000 = 3y
3y = 18000
y = 18000/3 = 6000*2/3 = 4000
Therefore, X's monthly income is 4x = 4(6000) = Rs. 24,000 and Y's monthly income is 3x = 3(6000) = Rs. 18,000.
Their total monthly income is Rs. 24,000 + Rs. 18,000 = Rs. 42,000.
Hence, option (b) Rs 42,000 is the correct answer.
- The figure drawn below gives the velocity graphs of two vehicles A and B. The straight line OKP represents the velocity of vehicle A at any instant, whereas the horizontal straight line CKD represents the velocity of vehicle B at any instant. In the figure, D is the point where perpendicular from P meets the horizontal line CKD such that PD = 1/2 LD :
What is the ratio between the distances covered by vehicles A and B in the time interval OL?(2018)
(A) 1:2
(B) 2: 3
(C) 3:4
(D) 1:1
Answer-C
Displacement of vehicle A = Area of triangle OPL
1/2 * QL * LP
1/2 * QL * (LD + PD)
1/2 * QL * (LD + LD/2)
1/2 * QL * 3LD/2
3/4 * OL * LD
Displacement of vehicleB = Area of rectangle OLDC = OL * LD
The ratio between the distances covered by vehicles A and B in the time interval OL is
= Displacement of vehicle A /Displacement of vehicle B
=3/4 * OL * LD /OL * LD
= 3/4
So option c is the right answer
- A sum of Rs 2,500 is distributed among X, Y and Z in the ratio 1/2: 3/4: 5/6.
What is the difference between the maximum share and the minimum share?(2020)
(A) Rs 300
(B) Rs 350
(C) Rs 400
(D) Rs 450
Answer-C
X, Y and Z in the ratio 1/2: 3/4: 5/6.
LCM of 2, 4, 6 is 12.
[(1/2) × 12] : [(3/4) × 12] : [(5/6) × 12] = 6 : 9 : 10
According to the question: 6x + 9x + 10x = 2500
⇒ 25x = 2500
⇒ x = 100
X gets = Rs. 6x = 6 × 100 = Rs. 600
Y gets = Rs. 9x = 9 × 100 = Rs. 900
Z gets = Rs.10x = 10 x 100 = Rs. 1000
Difference between the maximum share and the minimum share = Rs. (1000 – 600) = Rs. 400
- A student appeared in 6 papers. The maximum marks are the same for each paper. His marks in these papers are in the proportion of 5: 6 : 7: 8 : 9 : 10. Overall he scored 60%. In how many number of papers did he score less than 60% of the maximum marks?(2021)
(A) 2
(B) 3
(C) 4
(D) 5
Answer-B
Given the ratio of marks: 5:6:7:8:9:10
Assuming the maximum marks for each paper is 100, the total marks obtained in all six papers would be 600.
Now, let's find the actual marks obtained in each paper:
Paper 1: 5x
Paper 2: 6x
Paper 3: 7x
Paper 4: 8x
Paper 5: 9x
Paper 6: 10x
The total marks obtained would be:
Total marks obtained = 5x + 6x + 7x + 8x + 9x + 10x = 45x
The overall score is given as 60%, which means the total marks obtained as a percentage of the maximum marks would be:
(45x / 600) * 100 = 60
Simplifying the equation, we get:
x = 8
Now, let's calculate the actual marks obtained in each paper:
Paper 1: 5x = 5 * 8 = 40
Paper 2: 6x = 6 * 8 = 48
Paper 3: 7x = 7 * 8 = 56
Paper 4: 8x = 8 * 8 = 64
Paper 5: 9x = 9 * 8 = 72
Paper 6: 10x = 10 * 8 = 80
From the above calculations, we can see that the student scored less than 60% of the maximum marks in the first three papers.
Therefore, the student scored less than 60% of the maximum marks in 3 papers.
Hence, the correct answer is (b) 3.
