Rene cuts out a square of the maximum possible area from a piece of paper which is in the shape
of a quadrant of a circle of radius 10 cm. She then rolls the square sheet along one of its sides to
form a cylinder. Find the radius (in cm) of the cylinder.
Root 5/2pi
root 5 Pi
1/pi
root 5/3pi
@Cat.Aspirant123 said:Rene cuts out a square of the maximum possible area from a piece of paper which is in the shapeof a quadrant of a circle of radius 10 cm. She then rolls the square sheet along one of its sides toform a cylinder. Find the radius (in cm) of the cylinder.Root 5/2pi root 5 Pi 1/pi root 5/3pi
is the ans....5/rt(2)*pi??
@Cat.Aspirant123
For Maximizing the area of the square, the length of the diagonal should be equal to the radius of the quadrant..
Thus, 10=(2)^1/2*a, or, a =5*2^(1/2)..
Now, Circumference of the Cylinder=Length of the square..
=>2*pi*r=5*_/2
=>r = 5/(_/2*pi)
PS:: Here, i think radius of the quadrant should be _/10 cm..Then, radius of the cylinder,
2*pi*r=_/5, or, r=_/5/2*pi..Option(A)
@Cat.Aspirant123 said:Rene cuts out a square of the maximum possible area from a piece of paper which is in the shapeof a quadrant of a circle of radius 10 cm. She then rolls the square sheet along one of its sides toform a cylinder. Find the radius (in cm) of the cylinder.Root 5/2pi root 5 Pi 1/pi root 5/3pi
5/rt(2)pi is the ans.
@Cat.Aspirant123 said:steps
10 is the radius of the circle. which will serve as the diagonal of the square.
so side of the square will be = 5rt(2)
wen u roll this square into a cylinder.
5rt(2) will become circumference of the cylinder.
2*pi*r= 5rt(2)
r= 5/rt(2)*pi
Let I be the set of natural numbers such that each element in I
(i) is a three digit number which leaves a remainder of 2 when divided by 4.
(ii) is divisible by 11 but not by 13.
How many elements belong to I?
(i) is a three digit number which leaves a remainder of 2 when divided by 4.
(ii) is divisible by 11 but not by 13.
How many elements belong to I?
@Cat.Aspirant123 said:AC is the Longest Chord of the circle with center O and Chord EF is parallel to AC. ADF=70* then what is EFAFILE IS ATTACHED u can check
join CF..then angle CFA= 90 degrees
angle FCA = 70 Degress..angle in the same segment on the same chord.
hence angle CAF=EFA=180-(90+70)= 20 degress
When 58123 and 59059 are divided by a three-digit number N, the same remainder is obtained. How many values can N take?
@Cat.Aspirant123 said:AC is the Longest Chord of the circle with center O and Chord EF is parallel to AC. ADF=70* then what is EFAFILE IS ATTACHED u can check
join CF..then angle CFA= 90 degrees
angle FCA = 70 Degress..angle in the same segment on the same chord.
hence angle CAF=EFA=180-(90+70)= 20 degress
@vbhvgupta said:Let I be the set of natural numbers such that each element in I(i) is a three digit number which leaves a remainder of 2 when divided by 4.(ii) is divisible by 11 but not by 13.How many elements belong to I?
16 ?
@vbhvgupta said:Let I be the set of natural numbers such that each element in I(i) is a three digit number which leaves a remainder of 2 when divided by 4.(ii) is divisible by 11 but not by 13.How many elements belong to I?
19 numbers?
Start with the first 3 digit multiple of 11..since all the numbers in the set are divisible by 11
11*10 = 110..which leaves a remainder of 2 when divided by 4
Similarly we observe a pattern of increasing multiples of 11 in the AP of 4
11*14,11*18,11*22..all leave remainders of 2 when divided by 4
But we do not need multiples of 13..so we subtract those and get the answer as 19
@vbhvgupta said:When 58123 and 59059 are divided by a three-digit number N, the same remainder is obtained. How many values can N take?
58123 = N*Q1 + R
59059 = N*Q2 + R
subtracting we get 936 =N*(Q1-Q2)
936 = 8*9*13
3 digit factors of 936 = 104,468,312,234,156,117
Hence 6 values of N possible.
And N as 936 itself is one of the factors
59059 = N*Q2 + R
subtracting we get 936 =N*(Q1-Q2)
936 = 8*9*13
3 digit factors of 936 = 104,468,312,234,156,117
Hence 6 values of N possible.
And N as 936 itself is one of the factors
Total factors = 7
@vbhvgupta said:When 58123 and 59059 are divided by a three-digit number N, the same remainder is obtained. How many values can N take?
Numbers of the form 11(2+4k) satisfy the conditions
That is, 22(1+2k)
So we find the total 3 digit numbers = 22.5, 22.7, 22.9.....22.45 = 21 numbers
Subtracting numbers like 22.13 and 22.39 = 21-2 = 19 numbers ?? ...
Farook marks up the price of an article by 50% and then offers a discount of 20% to Shahrukh. Shahrukh sells it for र20 more than the price at which he purchased it. If Shahrukh's selling price is 30% more than the original cost price of the article, then Shahrukh's profit percentage is
10%
9%
6.66%
8.33% ANS
for farook let the cp = 100 mp = 150 so sp = 120
for shahrukh cp = 120 sp = 120+20 = 140
profit = (20*100)120 = 16.66
where I am doing mistake???
10%
9%
6.66%
8.33% ANS
for farook let the cp = 100 mp = 150 so sp = 120
for shahrukh cp = 120 sp = 120+20 = 140
profit = (20*100)120 = 16.66
where I am doing mistake???
@vbhvgupta said:Farook marks up the price of an article by 50% and then offers a discount of 20% to Shahrukh. Shahrukh sells it for र20 more than the price at which he purchased it. If Shahrukh's selling price is 30% more than the original cost price of the article, then Shahrukh's profit percentage is 10%9%6.66%8.33% ANSfor farook let the cp = 100 mp = 150 so sp = 120for shahrukh cp = 120 sp = 120+20 = 140 profit = (20*100)120 = 16.66where I am doing mistake???
10%?