Olympiad Mathematics - Direct and Inverse Proportions
Exam Duration: 45 Mins Total Questions : 25
If x : y = 2 : 3 and 2 : x = 1 : 2, then the value of y is________.
- (a)
1/3
- (b)
3/2
- (c)
6
- (d)
1/2
We have, \(\frac { x }{ y } =\frac { 2 }{ 3 } \) and \(\frac { x }{ 2 } =\frac { 1 }{ 2 } \) or x = 4.
Now putting x in \(\frac { x }{ y } =\frac { 2 }{ 3 } \), we get
\(\frac { 4 }{ y } =\frac { 2 }{ 3 } \) \(\Rightarrow\) 2y = 4 x 3 \(\Rightarrow\) y = 6
If x and y vary inversely as each other, and x = 10 when y = 6. Find y when x = 15.
- (a)
3
- (b)
4
- (c)
2
- (d)
6
As x and y vary inversely
\(\therefore\) xy = constant or x1,y1 =x2 y2
\(\Rightarrow\) 10 x 6 =15 x y2 (\(\because \) x1 = 10, y1 = 6, x2 = 15)
\(\Rightarrow\) y2 = \(\frac { 10\times 6 }{ 15 } =4\)
If x varies directly as y2 and x = 4 when y = 5, then find x when y is 15.
- (a)
4
- (b)
9
- (c)
12
- (d)
36
\(\frac { x }{ { y }^{ 2 } } =\frac { 4 }{ \left( { 5 } \right) ^{ 2 } } \frac { { x }_{ 1 } }{ \left( { 15 } \right) ^{ 2 } } \) (\(\because\) x varies directly)
or \({ x }_{ 1 }=\frac { 15\times 15\times 4 }{ 5\times 5 } \Rightarrow { x }_{ 1 }=36\)
if \(\frac { 1 }{ 5 } :\frac { 1 }{ x } =\frac { 1 }{ x } :\frac { 1 }{ 25 } \), then the value of x is ________.
- (a)
1.25
- (b)
1.5
- (c)
25
- (d)
2.25
We have , \(\frac { 1 }{ 5 } :\frac { 1 }{ x } =\frac { 1 }{ x } :\frac { 1 }{ 125 } \Rightarrow \frac { 1 }{ x } \times \frac { 1 }{ x } =\frac { 1 }{ 5 } \times \frac { 1 }{ 125 } \)
\(\Rightarrow\) \(\frac { 1 }{ { x }^{ 2 } } =\frac { 1 }{ 625 } \) \(\Rightarrow\) x2 = 625 \(\Rightarrow\) x = 25.
x and y vary in inverse proportion. When x is 12, Y is 3. Which of the following is not a possible pair of corresponding values of x and y?
- (a)
4 and 9
- (b)
10 and 3.6
- (c)
72 and 0.5
- (d)
5 and 6
For inverse proportion, xy = constant or x1Y1 = x2Y2
(A) as x1y1 = 12 x 3 = 36
x2y2 = 4 x 9 = 36
\(\therefore\) x1y1 = x2y2
(B) As, x1y1 = 36; x2y2 = 10 x 3.6 = 36
\(\therefore\) x1y1 = x2y2
(C) As x1y1 = 36; x2y2 72 x 0.5 = 36
(D) As, x1y1 = 36 x2y2 = 5x 6 = 30
\(\therefore\) x1y1 \(\neq \) x2y2
Which of the following statements is Correct?
- (a)
Length of a side of square and its area vary directly with each other.
- (b)
If one angle of a triangle is kept fixed then the measure of the remaining two angles vary inversely with each other.
- (c)
The area of circle and its diameter vary directly with each other.
- (d)
All of these
If A : B = 2 : 3 and B : C = 4 : 5, then C : A is equal to_______.
- (a)
15 : 8
- (b)
12 : 10
- (c)
8 : 5
- (d)
8 : 15
\(\frac { A }{ B } =\frac { 2 }{ 3 } \Rightarrow B=\frac { 3 }{ 2 } A\)
Also, B :C = 4 or \(\frac { B }{ C } =\frac { 4 }{ 5 } \)
\(\Rightarrow\) \(\frac { \frac { 2 }{ 3 } A }{ C } =\frac { 4 }{ 5 } or\frac { 3A }{ 2C } =\frac { 4 }{ 5 } \)
\(\Rightarrow\) \(\frac { A }{ C } =\frac { 4\times 2 }{ 5\times 3 } \Rightarrow \frac { A }{ C } =\frac { 8 }{ 15 } \Rightarrow C:A=15:8\)
Match the following.
Column - I | Column - II |
---|---|
P. x and yare in direct proportion and x = 40 when y = 120. If x = 60 then y = | (i) 160 |
Q. x varies inversely as y and x = 12 when y = 300, if x = 24 then y = | (ii) 180 |
R. x varies directly as y and y = 50 when x = 30, if x = 96 then y = | (iii) 130 |
S. x varies inversely as y and y = 650 when x = 20 if x = 100 then y = | (iv) 150 |
- (a)
P\(\rightarrow\)(iv); Q\(\rightarrow\)(i); R\(\rightarrow\)(iii); S\(\rightarrow\)(ii)
- (b)
P\(\rightarrow\)(ii); Q\(\rightarrow\)(iv); R\(\rightarrow\)(iii); S\(\rightarrow\)(i)
- (c)
P\(\rightarrow\)(iv); Q\(\rightarrow\)(i); R\(\rightarrow\)(ii); S\(\rightarrow\)(iii)
- (d)
P\(\rightarrow\)(ii); Q\(\rightarrow\)(iv); R\(\rightarrow\)(i); S\(\rightarrow\)(iii)
P. According to Question, \(\frac { 40 }{ 120 } =\frac { 60 }{ y } \) or \(y=\frac { 60\times 120 }{ 40 } =180\)
Q. According to question, 12 x 300 = 24 x Y or \(y=\frac { 12\times 300 }{ 24 } =150\)
R. According to question,\(\frac { 30 }{ 50 } =\frac { 96 }{ y } \) or \(y=\frac { 96\times 50 }{ 30 } =160\)
S. According to question, 20 x 650 = 100 x y or \(y=\frac { 20\times 650 }{ 100 } =130\)
When x = 2, 7,11,....... , Y = 8, 28, 44, .... , then x and yare in.........
- (a)
Direct proportion
- (b)
Inverse proportion
- (c)
Neither direct nor inverse proportion
- (d)
None of these
\(\frac { x }{ y } =\frac { 2 }{ 8 } =\frac { 7 }{ 28 } =\frac { 11 }{ 44 } \) = constant.
\(\therefore\) x and y are in the direct proportion.
Two quantities x and y vary inversely with each other, then_________.
- (a)
x/y remains constant
- (b)
x - y remains constant
- (c)
x + y remains constant
- (d)
x x y remains constant
8 men can do a piece of work in 10 days. How long will 10 men take to do the same work?
- (a)
12 days
- (b)
8 days
- (c)
7 days
- (d)
6 days
Let the number of days = x
We from the table as shown.
Number of men | 8 | 10 |
Number of days | 10 | x |
It is a case of inverse proportion.
8 x 10 = 10 x x \(\Rightarrow\)x = 8 days.
An agent receives a commission of RS 73.00 on sales of RS 1000.00. The commission he will get on sales of RS 100.00 is________.
- (a)
Rs 7.30
- (b)
Rs 7.00
- (c)
Rs 6.00
- (d)
Rs 6.30
Let commission received be Rs y.
Sales | 1000 | 100 |
Commission | 73 | y |
More in sales, more will be commission.
So, it is direct proportion.
Hence, 1000 x Y = 73 x 100
y = \(\frac { 73\times 100 }{ 1000 } \Rightarrow \)y = Rs 7.30
In a birthday party, on an average, 5 bottles of coke are served for a group of 6 children. How many friends were present at Mohit's party if 15 bottles of coke were used?
- (a)
18
- (b)
24
- (c)
30
- (d)
12
Bottles | 5 | 15 |
No.of Children | 6 | ? |
NUmber of children = \(\frac { 15\times 6 }{ 5 } =18\)
In a camp there is sufficient food for 105 soldiers for 21 days. If 42 soldiers leave the camp, then how long would the food last?
- (a)
30 days
- (b)
35 days
- (c)
65 days
- (d)
45 days
Let number of days = y
Soldiers | 105 | 105-42 = 63 |
Days | 21 | y |
It is a case of inverse proportion.
Hence, 105 x 21 = 63 x Y
or y = \(\frac { 105\times 21 }{ 63 } =35\)
A shopkeeper has just enough money to buy 52 cycles worth Rs 525 each. If each cycle were to cost Rs 21 more, then number of cycles, he will be able to buy with that amount of money, is ________.
- (a)
40
- (b)
30
- (c)
50
- (d)
20
Cost of one cycle = Rs 525
\(\therefore\) Cost of 52 cycles = Rs 525 x 52 = Rs 27300
New cost of one cycle = Rs (525 + 21) = Rs 546
So, number of cycles bought with Rs 27300 = \(\frac { 27300 }{ 546 } =50\)
A labourer is paid Rs 806 for 13 days of work. If he receives Rs 1,798, then for how many days did he work?
- (a)
29
- (b)
35
- (c)
60
- (d)
40
Let number of das be y.
Rupees | 806 | 1798 |
Days | 13 | y |
More the money, more will be the number of days. So it is direct proportion.
Hence, \(\frac { 803 }{ 13 } =\frac { 1798 }{ y } \Rightarrow y=\frac { 1798\times 13 }{ 806 } =29\)
A photograph of a bacteria enlarged 50,000 times, attains a length of 5 cm What is the actual length of the bacteria?
- (a)
2 x 10-4 cm
- (b)
103 cm
- (c)
10-4 cm
- (d)
104 cm
Let actual length of bacteria be x cm.
Enlargement of the photograph | 50,000 times | 1 time |
Length of bacteria | 5 cm | x |
More the enlargement of photograph, more would be its length. So, it is a case of direct proportion.
Hence, \(\frac { 50000 }{ 5 } =\frac { 1 }{ x } \Rightarrow x=\frac { 5 }{ 50000 } ={ 10 }^{ -4 }cm\)
Raghu has enough money to buy 75 machines worth Rs 200 each. How many machines can he buy if he gets a discount of Rs 50 on each machine?
- (a)
180
- (b)
200
- (c)
150
- (d)
100
Let number of machines be y.
No.of machines | 75 | y |
Cost of each machine | 200 | (200 - 50) = 150 |
More the cost of machines, lesser the number of machines bought. So, it is a case of inverse proportion.
Hence, 75 x 200 = y x 150 or y = \(\frac { 75\times 200 }{ 150 } =100\)
A cistern has two inlets A and B which can fill it in 12 minutes and 15 minutes respectively. An outlet C can empty the full cistern in 10 minutes. If all the three pipes then are opened together in the empty cistern, time taken to fill the cistern completely is________.
- (a)
20 minutes
- (b)
10 minutes
- (c)
15 minutes
- (d)
5 minutes
Part of cistern filled by inlets A and B together in one minite = \(\frac { 1 }{ 12 } +\frac { 1 }{ 15 } =\frac { 5+4 }{ 60 } =\frac { 9 }{ 60 } \)
and part of cistern emptied by outlet C in 1 minute = \(\frac { 1 }{ 10 } \)
So, part of the cistern filled in one minute when all three pipes are opened simultaneously,
= \(\frac { 9 }{ 60 } -\frac { 1 }{ 10 } =\frac { 9-6 }{ 60 } =\frac { 3 }{ 60 } =\frac { 1 }{ 20 } \)
So, time required to fill the cistern = 20 mins.
In 15 days, the earth picks up 1.2 x 108 kg of dust from the atmosphere. In how many days it will pick up 4 8 x 108 kg of dust?
- (a)
50
- (b)
40
- (c)
60
- (d)
100
Number of days | 15 | x |
Weight of dust (in kg) | 1.2 x 108 | 4.8 x108 |
More the dust, more the number of days it will take to pick it up. So, it is a case of direct proportion.
\(\therefore\) x = \(\frac { 15 }{ 1.2\times { 10 }^{ 8 } } \times 4.8\times { 10 }^{ 8 }\) days \(\Rightarrow\)x = \(\frac { 15\times 4.8 }{ 1.2 } =60\) days
State 'T' for true and 'F' for false.
(i) If x and yare in direct proportion, then (x - 1) and (y - 1) are also in direct proportion.
(ii) If x and yare in inverse proportion then (x + 2) and (y + 2) are also in inverse proportion.
(iii) Two quantities x and yare said to vary directly with each other if where \(\frac { x }{ y } =k\) is a positive constant.
(iv) When distance is kept fixed, then speed and time vary inversely with each other.
- (a)
(i) (ii) (iii) (iv) T F F T - (b)
(i) (ii) (iii) (iv) T T F T - (c)
(i) (ii) (iii) (iv) F F T T - (d)
(i) (ii) (iii) (iv) T F T F
Fill In the blanks.
(i) The perimeter of circle and its diameter vary P with each other.
(ii) (ii) If two quantities p and q vary inversely with each other then Q of their corresponding values remain constant.
(iii) When x and yare in indirect proportion and if y doubles then x becomes R.
- (a)
P Q R Inversely Ratio Double - (b)
P Q R Directly Product Half - (c)
P Q R Inversely Ratio Half - (d)
P Q R Directly Product Double
Match the following.
Column - I | Column - II |
(P) If the cost of 93 m of a certain kind of plastic sheet is Rs 1395, then what would it cost (in Rs) to buy 105 m of such plastic sheet? | (1) 42 |
(Q) 55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days? | (2) 18 |
(R) 18 men can reap a field in 35 days. For reaping the same field in 15 days; how many men are required? | (3) 88 |
(S) Suneeta types 1080 words in one hour. What is her GWPM (gross words per minute)? | (4) 1575 |
- (a)
P Q R S 4 3 1 2 - (b)
P Q R S 2 1 4 3 - (c)
P Q R S 3 2 1 4 - (d)
P Q R S 1 3 2 4
(P) Let the cost be Rs x
Cost of Sheet (In Rs) | 1395 | x |
Length of sheet | 93 | 105 |
It is the case of direct variation.
\(\frac { 1395 }{ 93 } =\frac { x }{ 105 } \Rightarrow x=\frac { 1395\times 105 }{ 93 } =1575\)
So, cost of plastic sheet is Rs 1575
(Q) Let the number of cows be x.
No.of coves | 55 | x |
No. of days | 16 | 10 |
It is the case of inverse variation.
\(\Rightarrow\) 55 x 16 = x x 10 \(\Rightarrow\) x = \(\frac { 55\times 16 }{ 10 } =88\)
So, 88 cows will graze the field in 10 days.
(R) Let the number of men be x.
No.of men | 18 | x |
No. of days | 35 | 15 |
It is the case of inverse variation.
18 x 35 = x x 15 \(\Rightarrow\) x = \(\frac { 18\times 35 }{ 15 } =42\)
So, 42 men can reap a field in 15 days.
(S) Let the number of words = x
No.of words | 1080 | x |
Time (in minutes) | 60 | 1 |
It is the case of direct variation.
\(\Rightarrow\) x = \(\frac { 1080 }{ 60 } =18\)
So, Sunita types 18 words per minute.
A worker is paid Rs 139.20 for 3 days.
(i) What will he get in the month of June (In Rs)?
(ii) For how many days will he be working for Rs 696?
- (a)
(i) (ii) 1392 15 - (b)
(i) (ii) 15 1392 - (c)
(i) (ii) 1382 20 - (d)
(i) (ii) 1392 20
Let Rs x2 be paid in the month of June.
Amount paid (x) (in Rs) | x1 = 139.20 | x2 |
No.of days (y) | y1 = 3 | y2 = 30 |
Since, it is a case of direct proportion.
\(\therefore\) \(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \Rightarrow \frac { 139.20 }{ 3 } =\frac { { x }_{ 2 } }{ 30 } \)
\(\Rightarrow\) \({ x }_{ 2 }=\frac { 139.20\times 30 }{ 3 } \Rightarrow { x }_{ 2 }=1392\)
Thus, Rs 1392 will be paid in the month of June.
(ii) Let Y2 be the number of days he will work for Rs 696.
Amount paid (x) (in Rs) | x1 = 139.20 | x2 = 696 |
No.of days (y) | y1 = 3 | y2 |
It is a case of direct proportion.
\(\therefore\) \(\frac { 139.20 }{ 3 } =\frac { 696 }{ { y }_{ 2 } } \Rightarrow { y }_{ 2 }=\frac { 696\times 3\times 10 }{ 1392 } \Rightarrow { y }_{ 2 }=15\)
Thus, he will work 15 days for Rs 696.
Which of the following tables shows the inverse proportion?
(i)
x | 6 | 12 | 30 | 48 |
y | 250 | 125 | 50 | 31.25 |
(ii)
x | 115 | 130 | 145 | 160 |
y | 615 | 600 | 585 | 570 |
(iii)
x | 50 | 100 | 300 | 120 |
y | 300 | 150 | 100 | 75 |
- (a)
Only (i)
- (b)
Both (i) and (ii)
- (c)
Both (ii) and (iii)
- (d)
None of these
(i) For inverse proportion, xy = constant
Now, 6 x 250 = 1500,
12 x 125 = 1500, 30 x 50 = 1500
And 48 x 31.25 = 1500
\(\therefore\) The data is in inverse proportion
(ii) 115x615=70725,
130 x 600 = 78000
\(\therefore\) The data is not in inverse proportion
(iii) 50 x 300 = 15000,
100 x 150 = 15000,
and 300 x 100 = 30000
\(\therefore\) The data is not in inverse proportion.