Ratio and Proportion
Exam Duration: 45 Mins Total Questions : 30
If A : B = 3 : 4 and B : C = 2 : 7, find A : B : C.
- (a)
4 : 3 : 14
- (b)
3 : 4 : 14
- (c)
14 : 4 : 3
- (d)
4 : 14 : 3
A : B = 3 : 4
B : C = 2 : 7 = (\(2\times 2:7\times 2\)) = 4 : 14
\(\therefore \) A : B : C = 3 : 4 : 14
The order of a leaf node in a B+ tree is the maximum number of (value, data recoed pointer) pairs it can hold. Given that the block size in 1 Kbytes, data record pointer is 7 nytes long, the value field is 9 byte long and a block pointer is 6 bytes long, what is the order of the leaf node?
- (a)
63
- (b)
64
- (c)
67
- (d)
68
Given B+ - tree
block size is 1 kbyte,
data record pointer is 7 bytes long
value field is 9 bytes long
a block pointer is 6 bytes long
To find The order of the leaf node
Solution Let the order of the leaf node is n
As per given, Block size = 1 k = 1024
\(\Rightarrow \) 6 + 7n + (n - 1)9 = 1024
\(\Rightarrow \) 16n = 1024
\(\Rightarrow \) n = 64
Find the duplicate ratio of \(\sqrt { 3x } :\sqrt { 5y } \)
- (a)
\({ 3x }^{ 2 }:{ 5y }^{ 2 }\)
- (b)
3x :5y
- (c)
\(\sqrt { 3x } :\sqrt { 5y } \)
- (d)
\(3\sqrt { x } :5\sqrt { x } \)
Duplicate ratio of x:y = x3 : y3
Duplicate ratio of \(\sqrt { 3x } :\sqrt { 5y } =\left( { (3x })^{ 2 }:{ (5y })^{ 2 } \right) ={ 3x }^{ 2 }:{ 5y }^{ 2 }\)
Find the triplicate ratio of 2x : 3y
- (a)
\({ 4x }^{ 2 }:{ 9y }^{ 2 }\)
- (b)
\({ 8x }^{ 3 }:{ 27y }^{ 3 }\)
- (c)
\({ 2x }^{ 2 }:{ 3y }^{ 2 }\)
- (d)
\({ 2x }^{ 3 }:{ 3y }^{ 3 }\)
Triplicate ratio of x:y = x3 : y3
Triplicate ratio of 2x : 3y = \({ (2x })^{ 3 }:({ 3y) }^{ 3 }\)
\(={ 8x }^{ 3 }:{ 27y }^{ 3 }\)
two numbers are in the ratio of 2 : 9. If 10 be added to each, they are in the ratio os 3 : 10. Find the numbers
- (a)
2 : 9
- (b)
6 : 27
- (c)
20 : 90
- (d)
8 : 36
Let the numbers be 2x and 9x. Then, \(\frac { 2x+10 }{ 9x+10 } =\frac { 3 }{ 10 } \Rightarrow 10(2x+10)=3(9x+10)\)
\(\Rightarrow \quad \quad \quad \quad \quad 7x=70\quad \Rightarrow x=10\)
If (x+y) : (y+z) : (z+x) = 4 : 6 : 5 and (x + y + z) = 15, then the value of z is
- (a)
5
- (b)
7
- (c)
3
- (d)
6
Let (x + y) = 4k, (y + z) =6k and (z + x) =5k
Then, 2(x + y + z) =15k
\(\Rightarrow \) 2\(\times \)15 = 15k\(\Rightarrow \) k =2
\(\therefore \) (x + y) =4k = 4\(\times \)2=8
\(\Rightarrow \) z = (x + y + z) - (x + y) = 15 - 8 = 7\(\Rightarrow \) z = 7
140 coins consists of 25p, 50p and re 1, their values being in the ratio of 5 : 8 : 20. Find the number of 25p coins.
- (a)
40 coins
- (b)
50 coins
- (c)
45 coins
- (d)
60 coins
Duplicate ratio of x : y = x2 : y2
\(\Rightarrow \) Duplicate ratio of
\(\sqrt { 3x } :\sqrt { 5y } =(\sqrt { 3x } { ) }^{ 2 }:(\sqrt { 5y } { ) }^{ 2 }:=3{ x }^{ 2 }:{ 5y }^{ 2 }\)
Amit and Ankit have 3 and 5 pieces of bread respectively. They share their food with Arun, who pays them rs 32. If each of them got equal bread, find Amit's share.
- (a)
Rs 4
- (b)
Rs 28
- (c)
Rs 12
- (d)
Rs 24
Total number of bread pieces = ( 3 + 5) =8
Share of each = \(\frac { 8 }{ 3 } \)
Part of the food given to Arun by Amit = 3 - \(\frac { 8 }{ 3 } \) = \(\frac {1 }{ 3 } \)
Part of the food given to Arun by Anit = 5 - \(\frac { 8 }{ 3 } \) = \(\frac { 7 }{ 3 } \)
\(\therefore \) Amit and Ankit will divide Rs 32 in the ratio
= \(\frac { 8 }{ 3 } \) : \(\frac { 8 }{ 3 } \) = 1 : 7
\(\therefore \) Amit's share = \(\frac { 8 }{ 3 } \) \(\times \) 32 = Rs 4
Two vessels of equal volume contain milk and water in the ratio of 5 : 3 and x : y respectively. If the contents are mixed into a bigger vessel, then the ratio of milk to water in the resulting solution is
- (a)
\(\frac { 13x+15y }{ 3x+11y } \)
- (b)
\(\frac { 13x+15y }{ 11x+3y } \)
- (c)
\(\frac { 3x+2y }{ 5x+8y } \)
- (d)
\(\frac { 3x+2y }{ 8x+5y } \)
Quantity of milk in first vessel = \(\frac { 5 }{ 8 } \)
Quantity of milk in second vessel = \(\frac { x }{ x+y } \)
Quantity of water after mixing = \(\left( \frac { 5 }{ 8 } +\frac { x }{ x+y } \right) \)
Quantity of water in first vessel = \(\frac {3 }{ 8 } \)
Quantity of water in second vessel = \(\frac { y }{ x+y } \)
Quantity of milk after mixing = \(\left( \frac {3 }{ 8 } +\frac { y }{ x+y } \right) \)
Required ratio
= \(\left( \frac { 5 }{ 8 } +\frac { x }{ x+y } \right) \): \(\left( \frac {3 }{ 8 } +\frac { y }{ x+y } \right) \)= (13x + 5y) : (3x + 11y)
= (13x + 5y) / (3x + 11y)
A variable x varies directly as the cube of another variable y. If x = 4, y = 2, then find y, when x = 32.
- (a)
4
- (b)
2
- (c)
8
- (d)
1
\(x\infty { y }^{ 3 }\Rightarrow x={ ky }^{ 3 }\),where k is a constant
When x = 4, y = 2 \(⇒\) k(2)3 = k\(\times \) 8\(⇒\)k=\(\frac { 1 }{ 2 } \)
\(\therefore \) x = \(\frac { { y }^{ 3 } }{ 2 } \) thus where x = 32
y3 = 32 \(\times \)2 = 64 \(⇒\) y = 4
The expenses per month of a man's car are partly constant and partly vary with the nymber of kilometers he travels in that month. When he travels 100 km a month, the total expenses come to Rs 3200. If he travels 15 km, it is Rs 3800. Find the total expenses, If he travels 200 km a month.
- (a)
Rs 6400
- (b)
Rs 4400
- (c)
Rs 4800
- (d)
Rs 6200
Let the total expenses be sum of the fixed and the variable part T = F +kD
where, T = total expenses
F = fixed part
k = constant of proportoonality
D = distance travelled
\(\Rightarrow \) 3200 = F + k(100)
Also, 3800 = F +k(150)
From Eqs. (i) and (ii) we get F = 2000, k = 12
\(\therefore \) The total expenses when he travels 200km is
= 2000 + 12(200) = Rs 4400
The reststance of a wire varies directly as its length and inversely as the ares of cross-section. The resistance is 1\(\Omega \), when the length is 50 mm and the cross-sectional are 0.25mm2. Find the resistance when the cross-sectional area is 0.5 mm2 and the length of the wire is 200 mm.
- (a)
2\(\Omega \)
- (b)
1.2\(\Omega \)
- (c)
0.2\(\Omega \)
- (d)
0.12\(\Omega \)
R\(\infty \frac { 1 }{ A } \Rightarrow R=\frac { kl }{ A } \)
Where R = 1, l = 50, A = 0.25, we have 1 = \(\frac { k\times 50 }{ 0.25 } \)
\(\Rightarrow k=\frac { 1 }{ 200 } \)
\(\therefore R=\frac { 2 }{ 200A } \) , thus when l = 200mm A = 0.5
R = \(\frac { 200 }{ 200\times 0.5 } \) = 2\(\Omega \)
In a class of 60 students, the number of boys and girls partcipating in the annuaal sports is in the ratio 3 : 2 respectively. The number of girls not participating in the sports is 5 more than the number of boys not participating in the sports. If the number of boys participating in the sports is 15, then how many girls are there in the class?
- (a)
20
- (b)
30
- (c)
25
- (d)
Data inadequate
From the information given in the question we can conclude that out of 60 students, number of boys and girls participating in sports are 15 and 10 respectively.
Suppose number of boys not participating in sports = x
Then, number not participating in sports
= 60 - 15 - 10 = 35
So, x + x + 5 = 35 \(\therefore \) x = 15
Total number of girls = 20 + 10 = 30
Three partners invested capital in the ratio 2 : 7 : 9. The time period for which each of them invested was in the ratio of the reciprocals of the amount invested. Find the share of the partner who brought in the highest capital, if profit is Rs 1080.
- (a)
Rs 120
- (b)
Rs 360
- (c)
Rs 540
- (d)
Rs 420
Ratio of capital = 2 : 7 : 9
Ratio of time = \(\frac { 1 }{ 2 } :\frac { 1 }{ 7 } :\frac { 1 }{ 9 } \)
\(\therefore \) Ratio of investment = \(2\times \frac { 1 }{ 2 } :7\times \frac { 1 }{ 7 } :9\times \frac { 1 }{ 9 } \)= 1 : 1 : 1
Each of them invested equal amount and share of each =\(\frac { 1 }{ 3 } \times \)1080 = Rs 360
The average of 11 results is 50. If the average of first 6 results is 49 and that of last 6 results is 52, find the sixth result
- (a)
64
- (b)
56
- (c)
48
- (d)
34
Sum of the first 6 results = \(49\times 6=294\)
Sum of the last 6 results = \(52\times 6=312\)
Let the sixth result be x, then
50 = \(\frac { 294+312-x }{ 11 } =\frac { 606-x }{ 11 }\)
\(\Rightarrow 606-x=50\times 11=550\Rightarrow x=606-550=56\)
The average weight of A, B, C is 48 kg. If the average weight of A and B be 45 kg and that of B and C be 50 kg. Find the weight of B.
- (a)
44 kg
- (b)
54 kg
- (c)
46 kg
- (d)
56 kg
Let the weight of A, B, and C be x, y and z.
Then ,48 = \(\frac { x+y+z }{ 3 } \Rightarrow \) x + y + z = 144 kg
Also, 45 = \(\frac { x+y}{ 2 } \Rightarrow \) = x + y = 90 kg
and 50 = \(\frac { y+z }{ 2 } \Rightarrow \) y + z = 100 kg
\(\therefore \) y = (x + y) + (y + z) - ( x + y + z)
= (90 + 100 - 144 )kg = 46 kg
Thus, the weight of B = 46 kg
Out of 10 persons, 9 person spent Rs 40 each for their meals. The 10th one spent Rs 18 more than the average expenditure of all the 10. The total money spent by all of them is
- (a)
Rs 420
- (b)
Rs 235
- (c)
Rs 540
- (d)
Rs 350
Let, the average expenditure of the 10 persons = Rs x
Then, \(\frac { 9\times 40+(x+18) }{ 10 } =x\Rightarrow 360+(x+18)=10x\\\)
\(\Rightarrow \) 9x = 378 \(\Rightarrow \) x = 42
Thus, total money spent by 10 persons = Rs (10 \(×\)42)
= Rs 420
The average weight of three brothers is 68 kg. If their weights are in the ratio 3 : 4 : 5, the weight of the heaviest of thethree brothers is
- (a)
85 kg
- (b)
70 kg
- (c)
90 kg
- (d)
82 kg
Let, the weights of the three brothers be 3x, 4x and 5x.
Then, \(\frac { 3x+4x+5x }{ 3 } =68\Rightarrow 12x=68\times 3\\ \)
\(\Rightarrow x=\frac { 68\times 3 }{ 12 } =17\)
Hence, the weight of the heaviest of the brother is 5x = (5 \(\times \)17) kg = 85 kg
A man travels from destination of A to B by car to an average speed of 48 km/h and returns on his bike with an average speed of 16 km/h. Find his average speed for the entire journey.
- (a)
36 km/h
- (b)
24 km/h
- (c)
32 km/h
- (d)
21 km/h
Average speed = \(\left( \frac { 2xy }{ x+y } \right) km/h\)
= \(\\ =\left( \frac { 2\times 48\times 16 }{ 48+16 } \right) km/h=24km/h\)
The average weight of 5 persons is decreased by 2 kg when one of the men whose weight is 50 kg is replaced by a new man. The weight of thw new man ia
- (a)
40 kg
- (b)
45 kg
- (c)
42 kg
- (d)
48 kg
Let the average weight of the 5 persons = x kg
Sum of the weights of 5 persons = 5 x kg/
Now,
5x - Weight of man removed
\(\frac { +weight\quad of\quad new\quad man }{ 5 } \) = x - 2
\(\Rightarrow \frac { 5x-50+weight\quad of\quad new\quad man }{ 5} \) = x - 2
\(\Rightarrow \) Weight of new man = 40kg
12 yr ago, the average age of a husband and a wife was 20 yr. The average remains the same today, when they have two children. what is the present age of the youngest child, if they differ in age by 2 yr?
- (a)
12 yr
- (b)
9 yr
- (c)
11 yr
- (d)
7 yr
Sum of the ages of husband and wife 12 yr ago
= \(\left( 2\times 20 \right) \)yr = 40 yr
Present sum of the ages of husband and wife
= ( 40 + 2 \(\times \) 12) yr = 64 yr
Let, the age of the two children be x and x + 2.
Now, 20 = \(\frac { 64+x+x+2 }{ 4 } \Rightarrow 20=\frac { 66+2x }{ 4 } \)
\(\Rightarrow \) 66 + 2x = 80 \(\Rightarrow \)2x = 14 \(\Rightarrow \)x = 7
Hence, the age of the younger child is 7 yr.
The average income of 20 employees in an office is Rs 1800 per month. If one more employees is added the average salary becomes Rs 1810 per month. What is the monthly salary of the new employee?
- (a)
Rs 1840
- (b)
Rs 1960
- (c)
Rs 2010
- (d)
Rs 1980
Total income of 20 employees = Rs (20\(\times \)1800)
= Rs 3600
Let the salary of the new employee be Rs x.
Then, 1810 = \(\frac { 36000+x }{ 21 } \)
\(\Rightarrow \) x = 1810\(\times \)21-36000 = 2010
Hence, the monthly salary of the new employee is Rs 2010.
The average of three numbers is 52. The first is half of second is twice the third. The difference between the second and first number is
- (a)
78
- (b)
39
- (c)
72
- (d)
33
Let the third number = x
Then, the second number = 2x and the first number
= \(\frac { 2x }{ 2 } =x\)
Now, \(52=\frac { x+2x+x }{ 3 } \Rightarrow 4x=52\times 3\)
\(\Rightarrow x=\frac { 52\times 3 }{ 4 } =39\)
Difference = (2x - x) = x = 39
A batsman has a certain average of runs for 14th innings. He scores 72 runs in the 15th innings thus his average increased by 2. Find the average after 15th innings.
- (a)
44
- (b)
46
- (c)
42
- (d)
48
Let the average of 14 innings = x
Average of 15 innings = x + 2
Then, 14x + 72 = 15(x + 2)
\(\Rightarrow \) 14x + 72 = 15x + 30 \(\Rightarrow \) x = 42
\(\therefore \) Average for 15 innings = x + 2 = 42 + 2 = 44.
The average weight 55 students in a class is 30kg. If 15 of the students whose average weight is 28 kg. leave the class. Find the average weight of the remaining students
- (a)
31.25 kg
- (b)
29.45 kg
- (c)
32.50 kg
- (d)
30.75 kg
Total weight of 55 students = (55\(\times \)30)kg = 1650 kg
Total weight of 15 students
= (15\(\times \)28) kg = 420 kg
Total weight of 40 of the students = (1650 - 420)kg
= 1230 kg
Average weight of 40 students = \(\frac { 1230 }{ 40 } \) = 30.75 kg
One half- of a certain distance is covered at 40 km/h, one-third of it at 80 km/h and the rest at 12 km/h. Find the average speed for the whole journey.
- (a)
\(51\frac { 1 }{ 13 } \) km/h
- (b)
\(55\frac { 1 }{ 13 } \) km/h
- (c)
\(52\frac { 1 }{ 13 } \) km/h
- (d)
\(56\frac { 2}{ 13 } \)km/h
Let total journey = x km.
Total time taken = \(\left( \frac { x }{ 2\times 40 } +\frac { x }{ 3\times 80 } +\frac { x }{ 6\times 120 } \right) h\)
= \(\left( \frac { x }{ 40 } +\frac { x }{ 240 } +\frac { x }{ 720 } \right) h\)
\(\therefore \) Average speed = x \(\times \) \(\frac { 720 }{ 13x } =\frac { 720 }{ 13 } =55\frac { 5 }{ 13 } \)km/h
What is the average of all numbers from 1 to 100 that end in 3?
- (a)
46
- (b)
46
- (c)
48
- (d)
47
Numbers between 1 and 100 which end in 3 are 3,
13 ,23, 33, 43, 53, 63, 73, 83, and 93 .
Sum of the numbers = 3 + 13 + 23 +........ + 93
This is an AP with a = 3, l = 93 and x = 10
\({ S }_{ n }=\frac { n }{ 2 } (a+l)=\frac { 10 }{ 2 } [3+93]=480\)
Hence, required average = \(\frac { 480 }{ 10 } \) = 48
The mean daily profit made by a shopkeeper in a month of 30 days was Rs 350. If the mean profit for the fifteen days was Rs 275, then the mean profit for the last fifteen days would be
- (a)
Rs 200
- (b)
Rs 350
- (c)
Rs 275
- (d)
Rs 475
Total profit by the shopkeeper for 30 days
= 350 \(\times 30=\) Rs 10500
Total profit by the shopkeeper for remaining for first 15 days
= 275\(\times 15=\) Rs 4125
\(\therefore \) Profit of the shopkeeper for remaining 15 days
= (10500 - 4125) = Rs 6375
Hence, mean profit for the last 15 days = \(\frac { 6375 }{ 15 } \) = Rs 425
The average marks of a student in 8 subjects are 7. Of these, the heighest marks are 2 more than the one next in value. If these two subjects are eliminated, the average marks of the remaining subjects are 85. What are the highest marks now obtained by him?
- (a)
89
- (b)
94
- (c)
91
- (d)
96
Sum of marks of 8 subjects = \(87\times 8=696\)
Sum of marks of 6 subjects = \(85\times 6=510\)
\(\therefore \) The total of second highest and highest marks = (696 - 510) = 186
Given, x + (x + 2) = 186 \(\therefore \) 2x = 184 and x = 92
\(\therefore \) Heighest marks = (92 + 2) = 94
The captian of a football team of 11 players is 25 yr old and the goalkeeper is 3 yr older to him. If the ages of these two are excluded, the average age of the remainning players is 1 yr less than the average age of the whole team. what is the average age of the whole team?
- (a)
22.5 yr
- (b)
23.5 yr
- (c)
22 yr
- (d)
25 yr
Let the average age of 11 players be x.
Then, sum of age of 11 players = 11 x
Average age of 9 players = \(\frac { 11x-25-28 }{ 9 } \)
Given, \(\frac { 11x-53 }{ 9 } =x-1\Rightarrow \) 11x - 53 = 9x - 9
2x = 44 \(\Rightarrow \) x = 22 yr