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 ratio 8125 : 36 is
- (a)
6
- (b)
27
- (c)
9
- (d)
81
\(\frac { { 81 }^{ 25 } }{ { 3 }^{ 6 } } =\frac { ({ 3 }^{ 4 })^{ 25 } }{ { 3 }^{ 6 } } =\frac { { 3 }^{ 10 } }{ { 3 }^{ 6 } } { 3 }^{ 4 }=81\)
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
2 bananas and 3 apples cost as much as 1 banana and 4 apples. The ratio of the cost of 1 banana to the cost of 1 apple is
- (a)
2 : 1
- (b)
1 : 2
- (c)
1 : 1
- (d)
2 : 3
\(2b+3a=1b+4a\Rightarrow 1b=1a\Rightarrow \frac { b }{ a } =\frac { 1 }{ 1 } \)
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
\(\sqrt { 3x } :\sqrt { 5y } =(\sqrt { 3x } { ) }^{ 2 }:(\sqrt { 5y } { ) }^{ 2 }:=3{ x }^{ 2 }:{ 5y }^{ 2 }\)
In a class of 75 students the ratio of boys and girls is 2 : 3 How many more boys are to be added to make the ratio 1 : 1?
- (a)
15 boys
- (b)
30 boys
- (c)
12 boys
- (d)
24 boys
The number of boys = \(\frac { 2 }{ 5 } \times 75=30\) boys
The number of girls = \(\frac { 3 }{ 5 } \times 75=45\) girls
\(\therefore \) when 15 more boys are added the new ratio of boys and girls is 45 : 45 \(\Rightarrow \) 1 : 1
\(\Rightarrow \) 15 more boys are to added.
An amount of Rs 840 was divided between A, B and C. If each of them had received Rs 20 less, their shares would have been in the ratio 2 : 3 : 5. The money received by A was
- (a)
Rs 156
- (b)
Rs 176
- (c)
Rs 184
- (d)
Rs 193
Remainder = Rs [840 - (20\(\times \) 3)] = Rs 780
\(\therefore \) Money recived by A = \(\frac { 2 }{ 10 } \times \) 780 + 20 = Rs 176
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
A varies inversely as x3 and dirctly as y. When x = 2, y = 40 and A = 5. Find A, when x = 3 and y = 81
- (a)
6
- (b)
3
- (c)
5
- (d)
7
A\(\infty \frac { y }{ { x }^{ 3 } } A=\frac { ky }{ { x }^{ 3 } } \) where k is a constant.
Where x = 2, y = 40 and A = 5, we have
5 = \(\frac { k(40) }{ { 2 }^{ 3 } } =\frac { k\times 40 }{ 8 } \Rightarrow \) k =1
\(\therefore \) A = \(\frac { y }{ { x }^{ 3 } } \) thus, where x =3 and y = 81
A = \(\frac { 81 }{ { x }^{ 3 } } \) = \(\frac { 81 }{ { 2}{ 7 } } \)= 3 \(\Rightarrow \) A = 3
Under sonstant temperature, the pressure of a definite mass of a gas varies inversely as the volume If the volume is reduced by 50% the pressure will be
- (a)
incresae by 100%
- (b)
increase by 25%
- (c)
decrease by 25%
- (d)
decrease by 50%
\(p\infty \frac { 1 }{ v } \Rightarrow P=\frac { k }{ v } \) where k is constant
Where the value of V becomes (50% of V) ie, = \(\frac { v }{ 2 } \),
We have P1 = \(\frac { k }{ V/2 } =\frac { 2k }{ v } \) increase in pressure
= \(\left[ \frac { \left( \frac { 2k }{ V } -\frac { k }{ V } \right) }{ \frac { k }{ V } } \times 100 \right] %=100%\)
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\(\Omega \)
The ratio between the number of passengers travelling by I and II class between the two railway stations is 1 : 50 , whereas the ratio of I and II class fares between the same station is 3 : 1. If on a particular day, Rs 1325 were collected from the passengers travelling between these stations, then what was the amount collected from the II class passengers?
- (a)
Rs 750
- (b)
Rs 1000
- (c)
Rs 850
- (d)
Rs 1250
Let the number of passengers travelling by class I and class II be Rs x and Rs 50 x respectively.
Then , amount collected from Class I and Ii will be Rs 3\(\times \)x and Rs 50x respectively.
Given, 3x + 50x = 1325
53x = 1325 \(\Rightarrow \) x = 25
The numerator and denominator of a rational number are in the ratio 7 : 8. If 10 is subtracted from numerator and denominator, the resulting rational number is \(\frac { 2 }{ 3 } \). The numerator of the original number is
- (a)
12
- (b)
14
- (c)
16
- (d)
18
Suppose the numerator and denominator of the rational numbers are 7x and 8x respectively.
\(\frac { 7x-10 }{ 8x-10 } =\frac { 2 }{ 3 } \Rightarrow \) 21x - 30 = 16x - 20
5x = 10 \(\therefore \) x = 2
\(\therefore \) Numerator 7x = 14
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 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 = \(\left( \frac { 2xy }{ x+y } \right) km/h\)
= \(\\ =\left( \frac { 2\times 48\times 16 }{ 48+16 } \right) km/h=24km/h\)
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 of 100 numbers is 50. It is found that while calculating the average, two numbers, namely 81 and 66 were wrongly read as 18 and 6. The correct average is
- (a)
51.23
- (b)
41.45
- (c)
36.42
- (d)
52.46
Sum of the numbers = (100\(\times \)50-18-6+81+66)
= 5123
\(\therefore \) Correct average = \(\frac { 5123 }{ 100 } \) = 51.23
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.
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.
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
A company employed 600 men and 400 women. Their average wage was Rs 68 per day. If a woman gets Rs 30 less than a man, what is the average wage of a woman?
- (a)
Rs 50
Let the average wage of a man = Rs x
Then, the average wage of a woman = Rs (x - 30)
Then, 68 = \(\frac { 600\times x+400\times (x-30) }{ 600+400 } \)
\(\Rightarrow 600x+400x-12000=68\times 1000=68000\\ \Rightarrow 1000x=68000+12000=8000\Rightarrow x=80\)
Thus, the average wage of woman = Rs (x - 30)
= Rs (80 - 30) = Rs 50
If the total sales for a business in a certain year were Rs 132000. What were the sales in June, if june sales were half the monthly average?
- (a)
Rs 8400
- (b)
Rs 7200
- (c)
Rs 5500
- (d)
Rs 6300
Monthly sale = Rs \(\left( \frac { 132000 }{ 12 } \right) \)= Rs 11000
June sales = Rs \(\left( \frac { 11000 }{ 12 } \right) \) = Rs 5500
The average monthly salary of 11 workers and one officer in an organisation is Rs 600. When the officer whose salary was Rs 1600 per month retired, a new officer was appointed and the average salary of the 12 employees is Rs 570 per month. The salary of the new officer is
- (a)
Rs 1260
- (b)
Rs 1240
- (c)
Rs 1220
- (d)
Rs 1280
Let, the average month salary of each worker = Rs x
Total salry of 11 workers and 1 officer = 11x + 1600
Then, 600 = \(\frac { 11x+1600 }{ 12 } \)
\(\Rightarrow 11x=600\times 12-1600=5600\)
Let, the salary of the new officer by Rs y.
Then, 570 = \(\frac { 11x+y }{ 12 } \) \(\Rightarrow\) 11x + y = 570 \(\times \) 12
\(\Rightarrow y=570\times 12-11x\\ \Rightarrow y=570\times 12-5600=1240\)
Hence, the salary of the new officer = RS 1240
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
There were 35 students in a hostel. If the number of students increase by 7, the expenses of the mess increase by Rs 42 per day while the average expenditure per head diminishes by 1. Find the original expenditur of the mess.
- (a)
Rs 480
- (b)
Rs 520
- (c)
Rs 420
- (d)
Rs 460
Let the average expenditure per student per day be Rs x.
Then, total expenditure per day = Rs 35x
New total expenditure per day = Rs (35x + 42)
New average expenditure = (x - 1)
Given, \(\frac { 35x+42 }{ 42 } \) = x - 1 \(\therefore \) 35x + 42 = 42x - 42
7x = 42 + 42 or x = \(\frac { 2\times 42 }{ 7 } \) = Rs 12
\(\therefore \) Original expenditure of mess = 35 \(\times \)12 = Rs 420
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
If 6 yr are subtracted from the present age of Randheer and the remainder is divided by 18, then the present age of his grandson Anup is obtained. If Anup is 2 yr younger to Mahesh whose age is 5 Yr, whai is the age of Randheer?
- (a)
84 yr
- (b)
48 yr
- (c)
60 yr
- (d)
96 yr
\(\frac { R-6 }{ 18 } =A\)
Given, Mahesh = 5 yr \(\therefore \) Anup = 3yr
\(\therefore \) R = 18\(\times \)3 + 6 = 60yr
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
The total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expenses per boarder is Rs 700 when there are 25 boarders and 600 when there are 50 boarders. What is the average expense per boarders when there are 100 boarders?
- (a)
Rs 550
- (b)
Rs 580
- (c)
Rs 540
- (d)
Rs 570
Let fixed expense be a and variable expenses be b.
\(\Rightarrow 700\times 28=a+25b\quad \\ \Rightarrow 17500=a+25b\quad \quad ..(i)\\ \Rightarrow 30000=a+50b\quad \quad ..(ii)\\ \)
on solving Eqs. (i) and (ii)
a = 5000, b = 500
When there are 100 boarders
E = 5000 + 500(100) = 5500
Average expenses = Rs 550