1) A candidate was to subtract 15 from a certain number but mistakenly added 25 and his answer was 145. Find the percentage error

2) The length and breath of a rectangle was mistakenly measured as 40m and 35m instead of 42.5m and 34.2m respectively. Find the percentage error in (a) the area (b) the perimeter.

3) in a school, 1200 of the students are boys, if 50% of the boys and 40% of the girls have paid their school fees, find the numbers of girls, given that 46% of the population has paid their school fees.

Dear nairalanders, help me solve these......

Bro na dummies full this thread oo nobody sabi maths for here

archat:
Bro na dummies full this thread oo nobody sabi maths for here

This is sad!

dejt4u and co.... Please help us out ooo

reccy:
dejt4u and co.... Please help us out ooo

I saw this post immediately you made it but it is unfortunate that I couldn't comment as I was banned by an antispam bot. The ban just expired few minutes ago.

I will attend to your questions shortly

reccy:
1) A candidate was to subtract 15 from a certain number but mistakenly added 25 and his answer was 145. Find the percentage error

1. Let the certain number be x,
Actual value = (x-15)
Measured value = (x+25)
But x+25 = 145,
x = 145 - 25,
x = 125

Therefore,
Actual value = (x-15) = 125 - 15 = 110
Measured value = (x+25) = 145
%error = (145 - 110)/110 × 100%
= (35/110) × 100%
= 31.82%

2) The length and breath of a rectangle was mistakenly measured as 40m and 35m instead of 42.5m and 34.2m respectively. Find the percentage error in (a) the area (b) the perimeter.

2.(a) Area of the Measured Value = 40m × 35m = 1400m²
Area of the Actual Value = 42.5m × 34.2m = 1453.5m²
Percentage error in the area = (1453.5 - 1400)/1453.5 × 100%
= (53.5/1453.5)×100% = 3.68%

(b)Perimeter of the Measured Value = 2(40+35) =150m
Perimeter of the Actual Value = 2(42.5+34.2) = 153.4m
Percentage error in the perimeter = (153.4 - 150)/153.4 × 100%
= (3.4/153.4) × 100%
= 2.22%

3) in a school, 1200 of the students are boys, if 50% of the boys and 40% of the girls have paid their school fees, find the numbers of girls, given that 46% of the population has paid their school fees.

Dear nairalanders, help me solve these......

3. Number of boys = 1200,
50% of boys have paid, therefore 600 boys have paid.

Let the number of girls be y
40% of girls have paid. Therefore 40y/100 girls have paid.

46% of the population have paid, this equals 46% × [1200 + y]


Total number of the POPULATION that have paid = number of BOYS that have paid + number of GIRLS that have paid, i.e.

46/100 × (1200 + y) = 600 + (40y/100)
y = 800


Therefore, the number of girls is 800

dejt4u:
I saw this post immediately you made it but it is unfortunate that I couldn't comment as I was banned by an antispam bot. The ban just expired few minutes ago.

I will attend to your questions shortly

Thanks so much sir

dejt4u:

Let the certain number be x,
Actual value = (x-15)
Measured value = (x+25)
But x+25 = 145,
x = 145 - 25,
x = 125

Therefore,
Actual value = (x-15) = 125 - 15 = 110
Measured value = (x+25) = 145
%error = (145 - 110)/110 × 100%
= (35/110) × 100%
= 31.82%


(a) Area of the Measured Value = 40m × 35m = 1400m²
Area of the Actual Value = 42.5m × 34.2m = 1453.5m²
Percentage error in the area = (1453.5 - 1400)/1453.5 × 100%
= (53.5/1453.5)×100% = 3.68%

(b)Perimeter of the Measured Value = 2(40+35) =150m
Perimeter of the Actual Value = 2(42.5+34.2) = 153.4m
Percentage error in the perimeter = (153.4 - 150)/153.4 × 100%
= (3.4/153.4) × 100%
= 2.22%


Number of boys = 1200,
50% of boys has paid, therefore 600 boys has paid.

Let the number of girls be y
40% of girls has paid. Therefore 40y/100 girls has paid.

46% of the population has paid, this equals 46% × [1200 + y]


Total number of the POPULATION that has paid = number of BOYS that has paid + number of GIRLS that has paid, i.e.

46/100 × (1200 + y) = 600 + (40y/100)
y = 800


Therefore, the number of girls is 800

Wow, this is awesome.... Thank you sir.

reccy:


Wow, this is awesome.... Thank you sir.

you're welcome