if a:b = c:d then (a+b)/(a-b) = (c+d)/(c-d) |
x ∝ y then x =
k*y where k = constant |
x ∝ 1/ y then x * y= k where k = constant |
If we multiply the numerator and the denominator of a ratio by the same number, the ratio remains unchanged. \( \frac{a}{b} = \frac{ma}{mb} \)
If we divide the numerator and the denominator of a ratio by the same number, the ratio remains unchanged. Thus \( \frac{a/d}{b/d} = \frac{a}{b} \)
The magnitudes of two ratios can be compared by equating the denominators of the two ratios and then checking for the value of the numerator. Thus, if we have to check for \( \frac{8}{3} = \frac{11}{4} \). We convert 8/3 into \( \frac{8 * 1.33}{3 * 1.33} = \frac{10.66}{4} \)Now denominators are equal and so we can compare the numerators and find which are greater and lesser.
If a/b = c/d = e/f = g/h = k then \( k = \frac{a + e + c +g}{b + d + f + h} \)
If a_{1} / b_{1} , a_{2}/b_{2}, a_{3}/b_{3}... a_{n}/b_{n} are unequal fractions then we have \( \frac{a_1 + a_2 + a_3 + .. + a_n}{b_1 + b_2 + b_3 + .. + b_n} \) lies between the lowest and the highest of these fractions.
If we have two equations containing three unknowns as a_{1}x + b_{1}y + c_{1}z = 0 and a_{2}x + b_{2}y + c_{2}z = 0 then we can find the proportion x : y : z. This will be given by \( b_1 * c_2 - b_2 * c_1 : a_2 * c_1 - c_2 * a_1 : a_1 * b_2 - a_2 * b_1 \)
If the ratio \( \frac{a}{b} > 1 \) (called a ratio of greater inequality) and if k is a positive number: \( \frac{a + k}{b + k} < \frac{a}{b} \) and \( \frac{a - k}{b - k} > \frac{a}{b} \) Similarly if \( \frac{a}{b} < 1 \) then \( \frac{a + k}{b + k} > \frac{a}{b} \) and \( \frac{a - k}{b - k} < \frac{a}{b} \)
Maintenance of equality when numbers are added in both the numerator and the denominators. This if best illustrated through an example: \( \frac{20}{30} = \frac{20 + 2}{30 + 3} \) i.e. \( \frac{a}{b} = \frac{a + c}{b + d} \) if and only if \( \frac{c}{d} = \frac{a}{b} \)
Consequently, if \( \frac{c}{d} > \frac{a}{b} \) then \( \frac{a + c}{b + d} > \frac{a}{b} \) and if \( \frac{c}{d} < \frac{a}{b} \) then \( \frac{a + c}{b + d} < \frac{a}{b} \)
To get the consolidated ratio A:B:C:D:E when individual ratios A:B, B:C, C:D, D:E are given
A:B = 2:3, B:C = 4:5, C:D = 6:11, D:E = 12:17
In order to create one consolidated ratio for this situation using the LCM process becomes too long. The short cut goes as follows:
A will correspond to the product of all numerators (2 × 4 × 6 × 12) while B will take the first denominator and the last 3 numerators (3 × 4 × 6 × 12). C on the other hand takes the first two denominators and the last 2 numerators (3 × 5 × 6 × 12), D takes the first 3 denominators and the last numerator (3 × 5 × 11 × 12) and E takes all the four denominators (3 × 5 × 11 × 17).
In mathematical terms this can be written as:
\( \frac{a}{b} = \frac{N_1}{D_1}, \frac{b}{c} = \frac{N_2}{D_2}, \frac{c}{d} = \frac{N_3}{D_3} , \frac{d}{e} = \frac{N_4}{D_4} \text{ then } a : b : c : d : e = N_1N_2N_3N_4 : D_1N_2N_3N_4 : D_1D_2N_3N_4 : D_1D_2D_3N_4 : D_1D_2D_3D_4 \)
Cross multiplication method : Compare two ratios
Two ratios can be compared using the cross multiplication method as follows. Suppose you have to compare 12/17 with 15/19. Then, to test which ratio is higher cross multiply and compare 12 × 19 and 15 × 17.
If 12 × 19 is bigger the Ratio 12/17 will be bigger. If 15 × 17 is higher, the ratio 15/19 will be higher. In this case, 15 × 17 being higher, the Ratio 15/19 is higher
Method for calculating the value of a percentage change in the ratio:
if 20/40 becomes 22/50 then Effect of numerator = 20 -> 22(10% increase) Effect of denominator = 50 -> 40(25% decrease) (reverse fashion)
Overall effect on the ratio:
100 increased by 10% gives 110 which when reduced by 25% gives 82.5.
Hence the overall change in the ratio is 17.5%
When two ratios are equal, the four quantities composing them are said to be proportionals. Thus if a/b = c/d, then a, b, c, d are proportionals. This is expressed by saying that a is to b as c is to d, and the proportion is written as a : b : : c : d OR a : b = c : d
The terms a and d are called the extremes while the terms b and c are called the means.
If four quantities are in proportion, the product of the extremes is equal to the product of the means. Let a, b, c, d be the proportionals. Then by definition a/b = c/d we get ad = bc Hence if any three terms of proportion are given, the fourth may be found. Thus if a, c, d are given, then b = ad/c
If three quantities a, b and c are in continued proportion, then a : b = b : c then \( ac = b^2 \) also a : c = \( a^2 : b^2 \)
If four quantities a, b, c and d form a proportion, many other proportions may be deduced by the properties of fractions. The results of these operations are very useful. These operations are
Invertendo: If a/b = c/d then b/a = d/c
Alternando: If a/b = c/d, then a/c = b/d
Componendo: If a/b = c/d, then \( \frac{a+b}{b} = \frac{c+d}{d} \)
Dividendo: If a/b = c/d, then \( \frac{a-b}{b} = \frac{c-d}{d} \)
Componendo and Dividendo: If a/b = c/d, then \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \)
Ans .
1228
The problem clearly states that when we reduce 28, 37 and 18 rupees respectively from Sherry’s, Berry’s and Cherry’s shares, the resultant ratio is: 4 : 6 : 9. Thus, if we assume the reduced values as 4x, 6x and 9x, we will have Æ Sherry’s share is 4x + 28, Berry’s share is 6x + 37 and Cherry’s share is 9x + 18 and thus we have (4x + 28) + (6x + 37) + (9x + 18) = 5783 so 19x = 5783 – 83 = 5700 Hence, x = 300. Hence, Sherry’s share is 1228.
Ans .
None of these
Solve this question using options. 1/2 of the first part should equal 1/3 rd of the second part and of the third part. This means that the first part should be divisible by 2, the second one by 3, and the third one by 6. Looking at the options, none of the first 3 options has its third number divisible by 6. Thus, option (d) is correct.
Ans .
300
(A + B) = 3 (C + D) so A + B = 375 and C + D = 125. Also, since C gets 1.5 times D we have C = 75 and D = 50, and B = 4 * C = 300.
Ans .
1/2
The given condition has a, b and c symmetrically placed. Thus, if we use a = b = c = 2 (say) we get each fraction as 1/2
Ans .
3 : 2
Solve using options. Since x > y > 0 it is clear that a ratio of x:y as 3:2 fits the equation
Ans .
ab/cd
1 : 2 = 3 : 6 So, (a^{2} + b^{2})/(c^{2} + d^{2}) = 5/45 = 1/9; From the given options, only ab/cd gives us this value.
Ans .
8
4 × 8 × 5 = 160 man-hours are required for ‘x’ no. of answer sheets. So, for ‘2x’ answer sheets we would require 320 man-hours = 2 × 20 × n Æ n = 8 Hours a day.
Ans .
40
In 40 litres, milk = 32 and water = 8. We want to create 2 : 3 milk to water mixture, for this we would need: 32 milk and 48 water. (Since milk is not increasing). Thus, we need to add 40 litres of water
Ans .
6 : 144 : 324
1 : 2 : 3 then x, 2x and 3x add up to 36. So the numbers are: 6, 12 and 18. Ratio of squares = 36 : 144 : 324
Ans .
45
The numbers would be 3x and 4x and their LCM would be 12x. This gives us the values as 45 and 60. The first number is 45.
Ans .
a + d – b – c
Assume a set of values for a, b, c, d such that they are proportional i.e. a/b = c/d. Suppose we take a:b as 1:4 and c:d as 3:12 we get the given expression: (a – b)(a – c)/a = –3 × –2/1 = 6. This value is also given by a + d – b – c and hence option (b) is correct
Ans .
7 : 5
Since equal quantities are being mixed, assume that both alloys have 18 kgs (18 being a number which is the LCM of 9 and 18). The third alloy will get, 14 kg of argentum from the first alloy and 7 kg of argentum from the second alloy. Hence, the required ratio: 21:15 = 7:5
Ans .
22.5
The total number of manhours required = 30 × 7 × 18 = 3780 21 × 8 × no. of days Æ 3780/168 = 22.5 days Note, you could have done this directly by: (30 × 7 × 18)/(21 × 8)
Ans .
6000
Solve using options. Option (c) fits the situation as if you take A’s income as t` 6,000, B’s income will become t` 4,000 and if they each save t` 1,000, their expenditures would be t` 5,000 , t` 3,000 respectively. This gives the required 5: 3 ratio.
Ans .
480
Solve by options. Option (a) C = 480 fits perfectly because if C = 480, B = 120 and A = 80
Ans .
28,000, 42,000, 14,000
A’s contribution = 33.33% B’s contribution = 50% C’s contribution = 16.66% Ratio of profit sharing = Ratio of contribution = 2 : 3 : 1 Thus, profit would be shared as : 28000 : 42000 : 14000
Ans .
100
2x + 20 : 3x + 20 : 5x + 20 = 4 : 5 : 7 so x = 10 and initially the number of students would be 20, 30 and 50 and a total of 100.
Ans .
6 : 4 : 3
The ratio of time would be such that speed × time would be constant for all three. Thus if you take the speeds as 2x, 3x and 4x respectively, the times would be 6y, 4y and 3y respectively
Ans .
ab + cd
Let us say we assume a = 1, b = 4, c = 3 and d = 12 we get: a^{2} + c^{2} = 10 and b^{2} + d^{2} = 160. The mean proportional between 10 and 160 is 40. ab + cd gives us this value and can be checked by taking another set of values to see that it still works.
Ans .
1/z > \( \sqrt{z} \)
Option (d) is true since 1/z will be greater than 1 and square root of z would be less then 1.
Ans .
1080
Amar’s share should be divisible by 6. Option d gets rejected by this logic. Further: A + B + C = 2250. If Amar’s share is 720 (acc. To option a) Bijoy’s share should be 480 & Chandra’s share should be 300. (Gives us a total of 720 + 480 + 300 = 1500). But the required total is 2250 (50% more than 1500). Since all relationships are linear, 1500 will increase to 2250 if we increase all values by 50%. Hence, Amar’s share should be 1080.
Ans .
2/5
Trial and error would give us 2/5 as the original fraction
Ans .
375
Their ratio being 5:3, the difference according to the ratio is 2. But this difference is 10. To get the values, expand the ratio 5 times. This gives 25 and 15 as the required values. Hence, the product is 375.
Ans .
20
60 oxen days = 1/7 of the field so 420 oxen days are required to plough the field. Thus, the remaining work would be 360 oxen days. With 18 oxen, it would take 20 days
Ans .
15 : 16
Assume that 1 cat leap is equal to 3 metres and 1 dog leap is equal to 4 metres. Then the speed of the cat in one unit time = 3 × 5 = 15 meters. Also, the speed of the dog in one unit time = 4 × 4 = 16 meters. The required ratio is 15:16
Ans .
90
4x, and 5x are their current ages. According to the problem, 4x – 18 : 5x – 18 = 11:16 so x = 10 and hence the sum total of their present ages is 90 years (40 + 50).
Ans .
49
x + 3x + 4x + 7x = 105 Æ x = 7 Thus, 7x = 49
Ans .
14.68
The share of the rent is on the basis of the ratio of the number of cow months. A uses 330 cow months (110 × 3), B uses 660(110 × 6) and C uses 1320 cow months (440 × 3) Hence, the required ratio is: 330:660:1320 = 1:2:4
Ans .
1/4
10 × 8 = 80 man days is required for the job. If only 8 students turn up, they would require 10 days to complete the task. The number of days is increasing by 1/4
Ans .
6
Initial wine = 35 litres Initial water = 14 litres Since, we want to create 7 : 4 mixture of wine and water by adding only water, it mean that the amount of wine is constant at 35 litres. Thus 7 : 4 = 35 : 20. So, we need 6 litres of water
Ans .
none
If we were to draw out 4 litres of wine and substitute it with plain water, the ratio of wine to water would become 1:1. Hence, option (d) is correct.
Ans .
165
The overall ratio is: 21:35:55. Dividing 333 in 111 parts (21 + 35 + 55) each part will be 3 and Class III will have the highest number of pupils are 55 × 3 = 165
Ans .
15
Solve using options. 15/24 becomes 24/33 gives 8/11
Ans .
none
Ratio of no. of coins = 12 : 10 : 7 Ratio of individual values of coins = 1 : 0.5 : 0.25 Ratio of gross value of coins = 12 : 5 : 1.75 = 48 : 20 : 7 so 75 Thus, he has rs 7 in 25 paisa coins. Which means that he would have 28 such coins.
Ans .
85
The ratio of the values of the three coins are: 10 × 10 : 17 × 20 : 7 × 100 = 100:340:700 = 5:17:35 is the ratio of division of value of coins. Thus, 20 paise coins correspond to t` 17. Hence, there will be 85 coins.
Ans .
25
Let the values of milk and water be 5x and x respectively. Then when we add 5 liters of water to this mixture, water would become x + 5. Now: 5x/(x + 5) = 5:2 so x = 5. Thus, 5x is 25.
Ans .
102
The ratio : 1/2 : 2/3 : 3/4 Converts to 6 : 8 : 9 (on multiplying by 12) Thus, the first monkey would get (391/23) × 6 = 102 bananas
Ans .
For option (c), 68 girls. Hence, 82 boys Amount with Girls = 68 × 0.25 = 17 Amount with Boys = 82 × 0.5 = 41. Total of t` 58. Thus, option (c) fits the conditions.
Ans .
+25%
Since pressure and volume are inversely proportional, we get that if one is reduced by 20% the other would grow by 25%.
Ans .
30
6x + 15 : 5x + 15 = 9 : 8 so 45x + 135 = 48x + 120 3x = 15 so x = 5 Maya’s present age = 6x = 30
Ans .
1000
From the first statement A = 1200 and B + C = 1800. From the second statement C = 1000 and A + B = 2000
Ans .
14 : 3
The initial amount of water is 9 liters and milk is 27 liters. By adding 15 liters of milk the mixture becomes 42 milk and 9 water Æ 14:3 the required ratio.
Ans .
54
x = ky gives 18 = 7k and k = 18/7 Hence, x = 18/7 × y When y = 21, x = 54.
Ans .
35
A = K × B × C so It is known that when A = 6, B = 3 and C = 2. Thus we get 6 = 6K gives K = 1. Thus, our relationship between A, B and C becomes A = B × C. Thus, when B = 5 and C = 7 we get A = 35
Ans .
Cannot be determined
x = ky/z We cannot determine the value of k from the given information and hence cannot answer the question
Q. If a:b = 3:4 and b:c is 5:4 then find a:b:c
A. The value common to both is b so b has to made same in both ratios. The LCM of 4,5 is 20 so we convert the ratios by multiplication
a:b = (3*5):(4*5) and b:c = (5*4)/(4*4)
a:b = 15:20 and b:c = 20:16 now b value is same in both sides and so we can merge on basis of 'b'.
a:b:c = 15:20:16
Q.Find fourth proportional between 4,9,12
A. Assume fourth proportional as 'x'. so we get 4:9::12:x .
As we know from above formula: 4*x = 9*12 so we can get value of x.
Q. Third proportional to 16, 36.
A. Assume third proportional is 'x'. so we get 16:36::36:x
As we know from formula that 16*x = 36*36 we can get 'x'.
Q. Mean proportional between 0.08 and 0.18
A. mean proportional = square-root(0.08*0.18)
= square-root(8/100 * 18/100)
= square-root(144/10000)
=12/100
=0.12
Q. If x : y = 3 : 4, find (4x + 5y) : (5x - 2y)
A. 4x+5y : 5x-2y = 4(x/y)+5 / 5(x/y) - 2 we get this by dividing by y to both numerator and denominator.
substituting x/y in equation we can solve it.
Q. Divide Rs. 600 in the ratio 4:2
A. the ratio 4:2 means first value shall be 4/4+2 and second value shall be 2/4+2 i.e. 600 has to be divided into 4/6 and 2/6 parts. so we get 400,200.
Q. A bag contains 50 p, 25 P and 10 p coins in the ratio 5: 9: 4, amounting to Rs. 206. Find the number of coins of each type.
A. Since ratio is 5:9:4 this means actual number of coins can be 5x, 9x and 4x.
We know that, 5x*(1/2) + 9x*(1/4) + 4x*(1/10) = 206 as we converted the paisa values to rupee.
Q. A mixture contains alcohol and water in the ratio 4 : 3. If 5 litres of water is added to the mixture, the ratio becomes 4: 5. Find the quantity of alcohol in the given mixture.
A. Alcohol and water were 4x and 3x and
now 4x / 3x+5 = 4 / 5.
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