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The ______ terms of four different numbers are in proportion are called Means.
The middle terms of four different numbers are in proportion are called Means.
In a cricket coaching camp, 1500 children are trained out of which 300 are selected for various matches.
Ratio of nonselected children to the total number of children is___.
The total number of children are 1500.
The total number of selected children are 300
therefore,
the number of nonselected children=1500−300=1200
Ratio
A bus travels 126 km in 2 hours and a train travels 320 km in 5 hours. The ratio of their speeds is _____.
A bus travels 126 Km in 2 hours
A train travels 320Km in 5 hours
We know that, Speed =
Speed of bus = Km/ hr
Speed of train = Km/ hr
Ratio of speeds = =63:64
There are 360 boys and 240 girls in a certain school. What is the ratio of the number of boys to the number of girls?
Number of boys = 240
Number of girls = 360
Ratio = =3:2
A sack contains 150 oranges and 50 mangoes. Find the ratio of the number of mangoes to the number of oranges.
Number of oranges = 150
Number of mangoes = 50
Ratio
A man had 15 hectares of land to share between his two children, Akash and Rahul in the ratio 2: 3. How many hectares of land did Akash and Rahul respectively get?
Total land = 15 hectares
Let the share of land Akash and Rahul get be 2x,3x respectively.
So,
2x+3x=15
5x=15
x
Therefore,
Share of land Akash got
And Share of land Rahul got
Thus,
Akash got 6 hectares of land and Rahul got 9 hectares of land.
Raj and Ram shared ₹ 750 in the ratio of 7: 8 respectively. What was Raj’s share?
Total amount = ₹ 750
Let the share of money Raj and Ram shared be 7x,8x respectively.
So,
7x+8x= 750
15x = 750
X=
Therefore,
Share of money Raj got = 7 x =₹
And share of money Ram got
Thus, Raj’s share of money is 350₹
Fill in the blanks \(\dfrac{3}{8} = \dfrac{{}}{{24}}\)
Find the ratio of 75 cm to 1.5 m.
The given numbers are not in the same units. So, converting them into same units.
1.5 m = 1.5 100 cm = 150 cm
[∵ 1 m = 100 cm]
∴ The required ratio is 75 cm: 150 cm.
∴ Required ratio = 1: 2
39 packets of 12 pens each cost ₹ 374.40. Find the cost of 52 packets of 10 pens each.
Number of pens in 1 packet = 12
Number of pens in 39 packets = 12 39 = 468
Now if,
Number of pens in 1 packet = 10
Number of pens in 52 packets = 10 x 52 = 520
Now cost of 468 pen = ₹ 374.40
Cost of 1 pan = ₹
Cost of 520 pens = ₹ ₹
Find the value of x, if 4, x, x, 100 are in proportion.
Since 4, x, x, 50, are in proportion.
∴ x x = 4 100
⇒ = 400
∴ x = 200
Are 10, 15, 12, 25 in proportion?
We have 10, 25, 10, 25
Product of extremes = 10 25 = 250
Product of middles = 25 10 = 300
Since both the products are same.
∴ The four numbers 10, 25, 10, 25 are in proportion
The first, second and fourth terms in a proportion are 32, 112, 217 respectively. Find the third term.
Let the third term be x.
∴ 32, 112, x and 217 are in proportion.
∴ 32: 112:: x : 217
Or
Thus, the third term = 62.
Find the ratio of the diagonal of a square of side 10 cm , to its side
Diagonal of a square = side
Ratio =
56 to 63 =?
The cost of a pen is ₹ 5. The cost of a pencil is ₹ 2. How many times of the cost of a pencil is the cost of a pen?
The ratio of cost of pen to the cost of pencil is
So, the cost of pen is equal to times the cost of pencil.
Two distances are in the ratio 14 : 9. The longer distance is 28 km. What is the length of the shorter distance?
The monthly salary of Hari Kishan is ₹120000. The monthly salary of Manish is ₹ 20000. How many times of the salary of Manish is the salary of Hari Kishan?
So, salary of Hari Krishan is 6 times the salary of Manish.
Ram spent 54 s to complete a task while Priya took 0.6 min to complete the same task. Identify simplest form of the ratio of the time taken by Ram to the time taken by Priya.
Time taken by Ram to complete the task = 54 sec
Time taken by Priya to complete the task = 0.6 min =36 sec
Ratio =
Anisha is 9 months old and Nishith is 1.5 years old. Find the simplest form of the ratio of Nishith’s age to Anisha’s age.
Age of Anisha = 9 months
Age of Nishith = 1.5 yrs. = 12+5=17 months
Ratio =
If the ratio between the length and the perimeter of a rectangular plot is 2: 5, then the ratio between the length and breadth of the plot is
So, we get
By cross multiplication, we get:
On further calculation
We get,
Hence, the ratio of the length and the breadth is 4: 1.
If the ratio of perimeter of two squares is 15: 16, then the ratio of their areas is
Consider ABCD and PQRS as the two squares.
Let the lengths of each side of ABCD and PQRS be x and y.
We know that
So we get
Or
By taking square on both sides,
By taking the ratio of their areas, we get
Hence, the ratio of their perimeters = 225: 256
If the ratio between the length and the perimeter of a rectangular plot is 1: 4, then the ratio between the length and breadth of the plot is
Given that Length of rectangle/Perimeter of rectangle
So, we get
The sides of a rectangle are in the ratio 5: 4. If its perimeter is 72 cm, then its length is
Consider the sides of the rectangle as 5x and 4x.
We know that, perimeter of rectangle = 2 (Length + Breadth)
By substituting the values
72 = 2 (5x + 4x)
On further calculation
72 = 2 9x
So, we get
72 = 18x
By division
x =
Hence, the length of the rectangle = 5x = 5 4 = 20 cm
One side of a square plot
Convert the ratio 66 : 18 in its simplest form
HCF of 66 and 18 is 6
Dividing both 66 and 18 by their HCF =
Hence, the simplest form of 66: 18 is 11: 3
Find the ratio of 48 min to 4 hours
Taking both the quantities in same unit, we have
4 hours = (4 x 60) = 240 min
The equation now becomes 48 min: 240 min
Or
Dividing both the numbers by their HCF, i.e., 48
Hence, the required ratio is 1: 5
Find the Equivalent ratio of 6: 7?
On multiplying or dividing each term of a ratio by the same non zero number, we get a ratio equivalent to the given ratio
For,
Both numerator and denominator of given fraction is multiplied by same non zero number i.e., 4
is an equivalent ratio of .
Two numbers are in the ratio 5: 7 and their sum is 120. Find the numbers?
Let the required number be 5a and 7a
Since the sum of these two numbers is given, we can say that
5a + 7a = 120
12a = 120
a =
a = 10
So, the first number is 5a = 5 x 10
= 50
Second number is 7a = 7 x 10
= 70
Hence, two numbers are 50 and 70
Divide ₹ 2000 between X and Y in the ratio 5 : 3
Ratio between x and y is 5:3
The sum of the ratios = 5+3=8
So, x= and y
Divide ₹ 5000 among X, Y and Z in the ratio 1 : 2 : 7
Total money = ₹ 5000
Given ratio = 1 : 2 : 7
Sum of ratio terms = ( 1 + 2 + 7 )
= 10
Share of X = ₹ ₹
Share of Y = ₹ ₹
Share of Z = ₹ ₹
15 m: 45 m and 30 km: 90 km in proportion?
We have 15 m: 45 m
HCF of 15 and 45 is 15
30 km: 90 km
HCF of 30 and 90 is 30
= 30: 90
Since, the ratios 15 m: 45 m and 30 km: 90 km are equal to . So, they are in proportion.
Are 3, 6, 5, 15 in proportion?
Product of means = Product of extremes
Here, Means are 6 and 5
Extremes are 3 and 15
Product of extremes = 3 x 15 = 45
Product of means = 6 x 5 = 30
Since, Product of extremes ≠ Product of means
Hence, 3 , 6 , 5 , 15 are not in Proportion
If 45 : y : : y : 5, find the value of y?
Clearly, Product of means = Product of extremes
y x y = 45 x 5
y² = 45 x 5
y² = 225
Hence, y = 15
If 3A = 5B = 4C, find A : B : C?
Let,
3A = 5B = 4C = k
This implies that 3A = k
A =
Also, if 5B = k
B =
Further, if 4C = k
C =
A : B : C = : :
LCM of 3 , 5 , 4 is 60
Multiplying each of the ratio by we get the ratios as
=20:12:15
Hence, A : B : C = 20 : 12 : 15
What must be added to each term of the ratio 7 : 9 so that the new ratio becomes 7 : 8 ?
Let the required number to be added be ‘a ‘
then, (7 + a): (9 + a) = 7: 8
Therefore,
8 (7 + a) = 7 (9 + a)
56 + 8a = 63 + 7a
8a – 7a = 63 – 56
1a = 7
Hence, the required number is 7
Present age of Sanjay is 52 years and the age of his Daughter is 28 years. Find the ratio of present age of Daughter to the present age of Sanjay?
Present age of Sanjay = 52 years
Present age of Daughter = 28 years
To find the ratio of present age of Daughter to the present age of Sanjay
On simplifying,
HCF of 28 and 52 is 4
Hence, the ratio of the present age of Daughter to Sanjay is 7 : 13
Present age of Seema is 22 years and the age of her Brother is 26 years. Find the ratio of Seema’s age to her Brother’s age after 10 years.
Present age of Seema = 22 years
After 10 years Seema’s age = 22 + 10 = 32 years
Present age of Brother = 26 years
After 10 years Brother’s age = 26 + 10 = 36 years
Ratio of age of Seema and Brother after 10 years
Taking, HCF of 32 and 36 is 4
Hence, the ratio of Seema’s age to her Brother’s age after 10 years is 8 : 9
Find the third proportion to 6 and 12?
Let, the third proportion to 6 and 12 be a
6: 12 :: 12 : a
(Product of Extremes= Product of Means)
Here, Extremes are = 6 and a
Means are = 12 and 12
Hence, the value of ‘a’ is 24
Find the mean proportional between 10 and 40?
Let, the mean proportional between 10 and 40 be ‘ a ‘
10: a :: a : 40
(Product of extremes = Product of means)
Here, Extremes are 10 and 40
Means are a and a
Hence, the value of ‘a’ is 20
If 12, 42, a are in continued proportion, find the value of a?
Given:12, 42, a are in continued proportion.
Then,
12: 42 :: 42 : a
(Product of extremes = Product of means)
Here, Extremes are 12 and a
Means are 42 and 42
Hence, the value of ‘a’ is 147
Find the mean proportional between 9 and 16.
Let x be the mean proportional between 9 and 16.
9: x :: x : 16
⇒ x × x = 9 × 16
⇒
⇒
Hence, the required mean proportional = 12.
Divide Rs 351 into two parts such that one may be to the other as 2: 7.
Given total amount is to be divided = 351
Ratio 2: 7
The sum of terms = 2 + 7= 9
First ratio of amount = = 2 × 39 = ₹78
Second ratio of amount = = 7 × 39
= ₹273
The ratio of zinc and copper in an alloy is 7: 9. It the weight of the copper in the alloy is 11.7 kg, find the weight of the zinc in the alloy.
Given that ratio of zinc and copper in an alloy is 7: 9
Let their ratio = 7x: 9x
Weight of copper = 11.7kg
9x = 11.7
x =
x = 1.3
Weight of the zinc in the alloy = 1.3 × 7
= 9.10kg
The sides of a triangle are in the ratio 1: 2: 3. If the perimeter is 36 cm, find its sides.
Given sides of a triangle are in the ratio 1: 2: 3
Perimeter = 36cm
Sum of the terms of the ratio = 1 + 2 + 3 = 6
First side = = 6cm
Second side = = 2 × 6 = 12cm
Third side = = 6 × 3= 18cm
The boys and the girls in a school are in the ratio 7:4. If total strength of the school be 550, find the number of boys and girls.
Given that boys and the girls in a school are in the ratio 7:4
Sum of the terms of the ratio = 7 + 4 = 11
Total strength = 550
Boys’ strength = = 7 × 50 = 350
Girls’ strength = = 4 × 50 = 200
Two numbers are in the ratio 7: 11. If 7 is added to each of the numbers, the ratio becomes 2: 3. Find the numbers.
Let the required numbers be 7x and 11x
If 7 is added to each of them then
Thus, the numbers are 7x
And 11x
If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers.
Given two numbers are in the ratio 6: 13
Let the required number be 6x and 13x
The LCM of 6x and 13x is 78x
78x = 312
Thus, the numbers are 6x =
13x =
If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y
Given x: y = 3: 5
We can write above equation as
By substituting the value of x in given equation 3x + 4y: 8x + 5y we get,
The sides of a rectangle are in the ratio 7: 8. If its perimeter is 150 cm, then its bigger side is?
Consider the sides of the rectangle as 7x and 8x.
We know that, perimeter of rectangle = 2 (Length + Breadth)
By substituting the values
150 = 2 (7x + 8x)
On further calculation
150 = 2 15x
So, we get
150 = 30x
By division
x = = 5
Hence, the bigger side of the rectangle = 8x = 8 4 = 32 cm
The ratio of the perimeters of two squares, one having its diagonal double than the other, is___
Consider ABCD and PQRS as the two squares. We know that, the diagonal of square PQRS is twice the diagonal of square ABCD.
PR = 2 AC
Side of the square =
Side of PQRS =
Perimeter of PQRS =
In the same way,
Side of ABCD =
Perimeter =
From the question, we know that:
If AC = x units, we get PR = 2x units
Perimeter of PQRS/Perimeter of ABCD = 2√2PR/ 2√2AC
= PR/AC
By substituting the value
Hence, the ratio of the perimeters of squares PQRS and ABCD is 2: 1.