Maths-
Find the value of x:
Maths-General
- 120
- 110
- 130
- 100
Answer:The correct answer is: 120x+60=180 {linear angle sum property}
x= 180-60
x = 120
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Related Questions to study
Maths-
What is 8 x 1/32
8x 1/32 = 8/32 = ¼
What is 8 x 1/32
Maths-General
8x 1/32 = 8/32 = ¼
Maths-
iv in roman number is
V= 5 , I = 1
We subtract the smaller value in the left, therefore 5-1 = 4
We subtract the smaller value in the left, therefore 5-1 = 4
iv in roman number is
Maths-General
V= 5 , I = 1
We subtract the smaller value in the left, therefore 5-1 = 4
We subtract the smaller value in the left, therefore 5-1 = 4
Maths-
Find the area of a given triangle whose perimeter is equal to 26cm .10 cm
Perimeter= 26cm
Height=10cm
Hypotenuse=12cm
Area=?
Perimeter of triangle=sum of sides
26=10+12+ b
26=22+b
26-22=b
4cm=b
Area=1/2*base*height
=1\2 * 4*10= 20 sq. cm
Height=10cm
Hypotenuse=12cm
Area=?
Perimeter of triangle=sum of sides
26=10+12+ b
26=22+b
26-22=b
4cm=b
Area=1/2*base*height
=1\2 * 4*10= 20 sq. cm
Find the area of a given triangle whose perimeter is equal to 26cm .10 cm
Maths-General
Perimeter= 26cm
Height=10cm
Hypotenuse=12cm
Area=?
Perimeter of triangle=sum of sides
26=10+12+ b
26=22+b
26-22=b
4cm=b
Area=1/2*base*height
=1\2 * 4*10= 20 sq. cm
Height=10cm
Hypotenuse=12cm
Area=?
Perimeter of triangle=sum of sides
26=10+12+ b
26=22+b
26-22=b
4cm=b
Area=1/2*base*height
=1\2 * 4*10= 20 sq. cm
Maths-
There are two circle with same centre O. The radius of larger circle is 4cm and that of the smaller circle is 2cm.
Find area of the shaded region.
Find area of the shaded region.
Ar of shaded region =ar. of larger circle – ar. of smaller circle
{ar. of circle = 22/7*r*r}
= 22/7*4*4-22/7*2*2
=22/7(16-4)
=22/7*12
=37.7 sq cm
{ar. of circle = 22/7*r*r}
= 22/7*4*4-22/7*2*2
=22/7(16-4)
=22/7*12
=37.7 sq cm
There are two circle with same centre O. The radius of larger circle is 4cm and that of the smaller circle is 2cm.
Find area of the shaded region.
Find area of the shaded region.
Maths-General
Ar of shaded region =ar. of larger circle – ar. of smaller circle
{ar. of circle = 22/7*r*r}
= 22/7*4*4-22/7*2*2
=22/7(16-4)
=22/7*12
=37.7 sq cm
{ar. of circle = 22/7*r*r}
= 22/7*4*4-22/7*2*2
=22/7(16-4)
=22/7*12
=37.7 sq cm
Maths-
What is 8 in roman numeral
V= 5 adding 3 times I will be 8
What is 8 in roman numeral
Maths-General
V= 5 adding 3 times I will be 8
Maths-
What is 9 divided by 72
9/72, dividing Numerator and denominator by 9 we get 1/8
What is 9 divided by 72
Maths-General
9/72, dividing Numerator and denominator by 9 we get 1/8
Maths-
12 divided by 26
12/26, dividing Numerator and denominator by 2 we get 6/13
12 divided by 26
Maths-General
12/26, dividing Numerator and denominator by 2 we get 6/13
Maths-
Himani saves Rs 700 a month and pratibha saves Rs 900 a month. Find the ratio in their savings.
Himani’s ratio = Rs 700 per month
Pratibha’s ratio = Rs 900 per month
Ratio of their savings = 700/900
= (700/100) / (900/100)
=7/9
= 7:9
Pratibha’s ratio = Rs 900 per month
Ratio of their savings = 700/900
= (700/100) / (900/100)
=7/9
= 7:9
Himani saves Rs 700 a month and pratibha saves Rs 900 a month. Find the ratio in their savings.
Maths-General
Himani’s ratio = Rs 700 per month
Pratibha’s ratio = Rs 900 per month
Ratio of their savings = 700/900
= (700/100) / (900/100)
=7/9
= 7:9
Pratibha’s ratio = Rs 900 per month
Ratio of their savings = 700/900
= (700/100) / (900/100)
=7/9
= 7:9
Maths-
Aman bought an bike for Rs. 20,000 and sold it to Raman and earn a profit of 2,000. So, at what amount Aman sold the bike to Raman.
Cost price= Rs 20,000
Profit=2,000
Selling Price=?
SP=CP+profit
=20,000+2,000
=Rs 22,000
Profit=2,000
Selling Price=?
SP=CP+profit
=20,000+2,000
=Rs 22,000
Aman bought an bike for Rs. 20,000 and sold it to Raman and earn a profit of 2,000. So, at what amount Aman sold the bike to Raman.
Maths-General
Cost price= Rs 20,000
Profit=2,000
Selling Price=?
SP=CP+profit
=20,000+2,000
=Rs 22,000
Profit=2,000
Selling Price=?
SP=CP+profit
=20,000+2,000
=Rs 22,000
Maths-
11 divided by 1/3
11 ÷1/3 = 11 x 3/1 = 33
11 divided by 1/3
Maths-General
11 ÷1/3 = 11 x 3/1 = 33
Maths-
The area of a rectangular park is 1500 sq m. If the lenth is 50m, find its width. Also find the perimeter of the park.
Area of rectangular park = 1500 sq m
Length = 50m
Bredth of the park= (1500/50)
= 30m
Perimeter =2*(l+b)
= 2*(50 + 30)
=2* 80
= 160m
Length = 50m
Bredth of the park= (1500/50)
= 30m
Perimeter =2*(l+b)
= 2*(50 + 30)
=2* 80
= 160m
The area of a rectangular park is 1500 sq m. If the lenth is 50m, find its width. Also find the perimeter of the park.
Maths-General
Area of rectangular park = 1500 sq m
Length = 50m
Bredth of the park= (1500/50)
= 30m
Perimeter =2*(l+b)
= 2*(50 + 30)
=2* 80
= 160m
Length = 50m
Bredth of the park= (1500/50)
= 30m
Perimeter =2*(l+b)
= 2*(50 + 30)
=2* 80
= 160m
Maths-
Write 24 in roman numerals
XX= 20 IV = 4 , XXIV = 24
Write 24 in roman numerals
Maths-General
XX= 20 IV = 4 , XXIV = 24
Maths-
What is LCM of 4 and 6.
LCM of 4 & 6= 2*2*3 = 12
What is LCM of 4 and 6.
Maths-General
LCM of 4 & 6= 2*2*3 = 12
Maths-
What value of x satisfies 2x + 3 = 9
Substituting x = 3, 2x3 + 3 = 6+3 = 9
What value of x satisfies 2x + 3 = 9
Maths-General
Substituting x = 3, 2x3 + 3 = 6+3 = 9
Maths-
Least common multiple for 12 and 5 is
As 12 and 5 are co-prime, there product will be the LCM
Least common multiple for 12 and 5 is
Maths-General
As 12 and 5 are co-prime, there product will be the LCM