Mixing
Fastest way to multiply small numbers with decimals mentally
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Is there a fast way to multiply these numbers mentally? Some examples:
- $0.85 times 1.15$
- $0.5 times 1.5$
- $0.2 times 1.4$
mental-arithmetic
edited Jan 4 at 15:09
gwg
9771922
asked Apr 2 '14 at 10:59
Math is funMath is fun
5218
$endgroup$
add a comment |
$begingroup$
Is there a fast way to multiply these numbers mentally? Some examples:
- $0.85 times 1.15$
- $0.5 times 1.5$
- $0.2 times 1.4$
mental-arithmetic
edited Jan 4 at 15:09
gwg
9771922
asked Apr 2 '14 at 10:59
Math is funMath is fun
5218
$endgroup$
3
$begingroup$
Confusing to read, because of lack of line breaks and the inconsistent use of comma/period convention for the decimal.
$endgroup$
– Jason Zimba
Apr 2 '14 at 11:25
add a comment |
$begingroup$
Is there a fast way to multiply these numbers mentally? Some examples:
- $0.85 times 1.15$
- $0.5 times 1.5$
- $0.2 times 1.4$
mental-arithmetic
edited Jan 4 at 15:09
gwg
9771922
asked Apr 2 '14 at 10:59
Math is funMath is fun
5218
$endgroup$
Is there a fast way to multiply these numbers mentally? Some examples:
- $0.85 times 1.15$
- $0.5 times 1.5$
- $0.2 times 1.4$
mental-arithmetic
mental-arithmetic
edited Jan 4 at 15:09
gwg
9771922
asked Apr 2 '14 at 10:59
Math is funMath is fun
5218
edited Jan 4 at 15:09
gwg
9771922
asked Apr 2 '14 at 10:59
Math is funMath is fun
5218
edited Jan 4 at 15:09
gwg
9771922
edited Jan 4 at 15:09
gwg
9771922
edited Jan 4 at 15:09
gwg
9771922
9771922
asked Apr 2 '14 at 10:59
Math is funMath is fun
5218
asked Apr 2 '14 at 10:59
Math is funMath is fun
5218
asked Apr 2 '14 at 10:59
Math is funMath is fun
5218
5218
3
$begingroup$
Confusing to read, because of lack of line breaks and the inconsistent use of comma/period convention for the decimal.
$endgroup$
– Jason Zimba
Apr 2 '14 at 11:25
add a comment |
3
$begingroup$
Confusing to read, because of lack of line breaks and the inconsistent use of comma/period convention for the decimal.
$endgroup$
– Jason Zimba
Apr 2 '14 at 11:25
3
3
$begingroup$
Confusing to read, because of lack of line breaks and the inconsistent use of comma/period convention for the decimal.
$endgroup$
– Jason Zimba
Apr 2 '14 at 11:25
$begingroup$
Confusing to read, because of lack of line breaks and the inconsistent use of comma/period convention for the decimal.
$endgroup$
– Jason Zimba
Apr 2 '14 at 11:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
when the numbers are simple, for example your second and third example, you can multiply directly, but pay attention of the posion of the point.
$$0.5*1.5=0.75\0.2*1.4=0.28$$
when the numbers are not simple enough, you can use the formula for the difference of square: $(a+b)(a-b)=a^2-b^2$. for example,
$$0.85*1.15=(1-0.15)*(1+0.15)=1-0.15^2=0.9775\
0.5*1.5=(1-0.5)*(1+0.5)=1-0.5^2=0.75\
0.2*1.4=(0.8-0.6)*(0.8+0.6)=0.8^2-0.6^2=0.28$$
answered Apr 2 '14 at 11:11
MartialMartial
9991717
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add a comment |
$begingroup$
I move the decimal points around as needed, then move them back. I also use fraction-decimal equivalents.
For example, for $0.5times 1.5$ I think "half of 150 is 75," then put the decimal back: $0.75$.
For $0.2times 1.4$, I think "a fifth of 140, which is a fifth of 100 plus a fifth of 40, or 20 + 8 = 28." Then put the decimal back: 0.28. Or just: $2times 14 = 28$.
For $0.85times 1.15$ I would estimate "very close to 1," because $(1-x)(1+x) approx 1$ when $x$ is small.
(In the last case, I might next adjust the estimate by thinking "$15^2=225$, so come back 0.02" to give approximately 0.98.)
answered Apr 2 '14 at 11:30
Jason ZimbaJason Zimba
2,203611
$endgroup$
add a comment |
$begingroup$
Well for 0.2 x 0.14 you can convert it to 2 x 14, but remember that there were 2 numbers after the decimal point (2 and 14). So when you multiply 2 by 14 you get 28. But wait! What about the 2 spaces after the decimal point? Well that's easy! If you want to make 28, two numbers after the decimal point just make it 0.8! Therefore 0.2 x 0.14 = 0.28
Hope this worked!!!
answered Oct 8 '15 at 0:02
NaomiNaomi
1
$endgroup$
$begingroup$
We have that $0.2cdot 0.14 = 0.028$.
$endgroup$
– Winther
Oct 8 '15 at 0:34
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
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$begingroup$
when the numbers are simple, for example your second and third example, you can multiply directly, but pay attention of the posion of the point.
$$0.5*1.5=0.75\0.2*1.4=0.28$$
when the numbers are not simple enough, you can use the formula for the difference of square: $(a+b)(a-b)=a^2-b^2$. for example,
$$0.85*1.15=(1-0.15)*(1+0.15)=1-0.15^2=0.9775\
0.5*1.5=(1-0.5)*(1+0.5)=1-0.5^2=0.75\
0.2*1.4=(0.8-0.6)*(0.8+0.6)=0.8^2-0.6^2=0.28$$
answered Apr 2 '14 at 11:11
MartialMartial
9991717
$endgroup$
add a comment |
$begingroup$
when the numbers are simple, for example your second and third example, you can multiply directly, but pay attention of the posion of the point.
$$0.5*1.5=0.75\0.2*1.4=0.28$$
when the numbers are not simple enough, you can use the formula for the difference of square: $(a+b)(a-b)=a^2-b^2$. for example,
$$0.85*1.15=(1-0.15)*(1+0.15)=1-0.15^2=0.9775\
0.5*1.5=(1-0.5)*(1+0.5)=1-0.5^2=0.75\
0.2*1.4=(0.8-0.6)*(0.8+0.6)=0.8^2-0.6^2=0.28$$
answered Apr 2 '14 at 11:11
MartialMartial
9991717
$endgroup$
add a comment |
$begingroup$
when the numbers are simple, for example your second and third example, you can multiply directly, but pay attention of the posion of the point.
$$0.5*1.5=0.75\0.2*1.4=0.28$$
when the numbers are not simple enough, you can use the formula for the difference of square: $(a+b)(a-b)=a^2-b^2$. for example,
$$0.85*1.15=(1-0.15)*(1+0.15)=1-0.15^2=0.9775\
0.5*1.5=(1-0.5)*(1+0.5)=1-0.5^2=0.75\
0.2*1.4=(0.8-0.6)*(0.8+0.6)=0.8^2-0.6^2=0.28$$
answered Apr 2 '14 at 11:11
MartialMartial
9991717
$endgroup$
when the numbers are simple, for example your second and third example, you can multiply directly, but pay attention of the posion of the point.
$$0.5*1.5=0.75\0.2*1.4=0.28$$
when the numbers are not simple enough, you can use the formula for the difference of square: $(a+b)(a-b)=a^2-b^2$. for example,
$$0.85*1.15=(1-0.15)*(1+0.15)=1-0.15^2=0.9775\
0.5*1.5=(1-0.5)*(1+0.5)=1-0.5^2=0.75\
0.2*1.4=(0.8-0.6)*(0.8+0.6)=0.8^2-0.6^2=0.28$$
answered Apr 2 '14 at 11:11
MartialMartial
9991717
answered Apr 2 '14 at 11:11
MartialMartial
9991717
answered Apr 2 '14 at 11:11
MartialMartial
9991717
answered Apr 2 '14 at 11:11
MartialMartial
9991717
9991717
add a comment |
add a comment |
$begingroup$
I move the decimal points around as needed, then move them back. I also use fraction-decimal equivalents.
For example, for $0.5times 1.5$ I think "half of 150 is 75," then put the decimal back: $0.75$.
For $0.2times 1.4$, I think "a fifth of 140, which is a fifth of 100 plus a fifth of 40, or 20 + 8 = 28." Then put the decimal back: 0.28. Or just: $2times 14 = 28$.
For $0.85times 1.15$ I would estimate "very close to 1," because $(1-x)(1+x) approx 1$ when $x$ is small.
(In the last case, I might next adjust the estimate by thinking "$15^2=225$, so come back 0.02" to give approximately 0.98.)
answered Apr 2 '14 at 11:30
Jason ZimbaJason Zimba
2,203611
$endgroup$
add a comment |
$begingroup$
I move the decimal points around as needed, then move them back. I also use fraction-decimal equivalents.
For example, for $0.5times 1.5$ I think "half of 150 is 75," then put the decimal back: $0.75$.
For $0.2times 1.4$, I think "a fifth of 140, which is a fifth of 100 plus a fifth of 40, or 20 + 8 = 28." Then put the decimal back: 0.28. Or just: $2times 14 = 28$.
For $0.85times 1.15$ I would estimate "very close to 1," because $(1-x)(1+x) approx 1$ when $x$ is small.
(In the last case, I might next adjust the estimate by thinking "$15^2=225$, so come back 0.02" to give approximately 0.98.)
answered Apr 2 '14 at 11:30
Jason ZimbaJason Zimba
2,203611
$endgroup$
add a comment |
$begingroup$
I move the decimal points around as needed, then move them back. I also use fraction-decimal equivalents.
For example, for $0.5times 1.5$ I think "half of 150 is 75," then put the decimal back: $0.75$.
For $0.2times 1.4$, I think "a fifth of 140, which is a fifth of 100 plus a fifth of 40, or 20 + 8 = 28." Then put the decimal back: 0.28. Or just: $2times 14 = 28$.
For $0.85times 1.15$ I would estimate "very close to 1," because $(1-x)(1+x) approx 1$ when $x$ is small.
(In the last case, I might next adjust the estimate by thinking "$15^2=225$, so come back 0.02" to give approximately 0.98.)
answered Apr 2 '14 at 11:30
Jason ZimbaJason Zimba
2,203611
$endgroup$
I move the decimal points around as needed, then move them back. I also use fraction-decimal equivalents.
For example, for $0.5times 1.5$ I think "half of 150 is 75," then put the decimal back: $0.75$.
For $0.2times 1.4$, I think "a fifth of 140, which is a fifth of 100 plus a fifth of 40, or 20 + 8 = 28." Then put the decimal back: 0.28. Or just: $2times 14 = 28$.
For $0.85times 1.15$ I would estimate "very close to 1," because $(1-x)(1+x) approx 1$ when $x$ is small.
(In the last case, I might next adjust the estimate by thinking "$15^2=225$, so come back 0.02" to give approximately 0.98.)
answered Apr 2 '14 at 11:30
Jason ZimbaJason Zimba
2,203611
edited Apr 2 '14 at 20:19
answered Apr 2 '14 at 11:30
Jason ZimbaJason Zimba
2,203611
answered Apr 2 '14 at 11:30
Jason ZimbaJason Zimba
2,203611
answered Apr 2 '14 at 11:30
Jason ZimbaJason Zimba
2,203611
2,203611
add a comment |
add a comment |
$begingroup$
Well for 0.2 x 0.14 you can convert it to 2 x 14, but remember that there were 2 numbers after the decimal point (2 and 14). So when you multiply 2 by 14 you get 28. But wait! What about the 2 spaces after the decimal point? Well that's easy! If you want to make 28, two numbers after the decimal point just make it 0.8! Therefore 0.2 x 0.14 = 0.28
Hope this worked!!!
answered Oct 8 '15 at 0:02
NaomiNaomi
1
$endgroup$
$begingroup$
We have that $0.2cdot 0.14 = 0.028$.
$endgroup$
– Winther
Oct 8 '15 at 0:34
add a comment |
$begingroup$
Well for 0.2 x 0.14 you can convert it to 2 x 14, but remember that there were 2 numbers after the decimal point (2 and 14). So when you multiply 2 by 14 you get 28. But wait! What about the 2 spaces after the decimal point? Well that's easy! If you want to make 28, two numbers after the decimal point just make it 0.8! Therefore 0.2 x 0.14 = 0.28
Hope this worked!!!
answered Oct 8 '15 at 0:02
NaomiNaomi
1
$endgroup$
$begingroup$
We have that $0.2cdot 0.14 = 0.028$.
$endgroup$
– Winther
Oct 8 '15 at 0:34
add a comment |
$begingroup$
Well for 0.2 x 0.14 you can convert it to 2 x 14, but remember that there were 2 numbers after the decimal point (2 and 14). So when you multiply 2 by 14 you get 28. But wait! What about the 2 spaces after the decimal point? Well that's easy! If you want to make 28, two numbers after the decimal point just make it 0.8! Therefore 0.2 x 0.14 = 0.28
Hope this worked!!!
answered Oct 8 '15 at 0:02
NaomiNaomi
1
$endgroup$
Well for 0.2 x 0.14 you can convert it to 2 x 14, but remember that there were 2 numbers after the decimal point (2 and 14). So when you multiply 2 by 14 you get 28. But wait! What about the 2 spaces after the decimal point? Well that's easy! If you want to make 28, two numbers after the decimal point just make it 0.8! Therefore 0.2 x 0.14 = 0.28
Hope this worked!!!
answered Oct 8 '15 at 0:02
NaomiNaomi
1
answered Oct 8 '15 at 0:02
NaomiNaomi
1
answered Oct 8 '15 at 0:02
NaomiNaomi
1
answered Oct 8 '15 at 0:02
NaomiNaomi
1
1
$begingroup$
We have that $0.2cdot 0.14 = 0.028$.
$endgroup$
– Winther
Oct 8 '15 at 0:34
add a comment |
$begingroup$
We have that $0.2cdot 0.14 = 0.028$.
$endgroup$
– Winther
Oct 8 '15 at 0:34
$begingroup$
We have that $0.2cdot 0.14 = 0.028$.
$endgroup$
– Winther
Oct 8 '15 at 0:34
$begingroup$
We have that $0.2cdot 0.14 = 0.028$.
$endgroup$
– Winther
Oct 8 '15 at 0:34
add a comment |
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$begingroup$
Confusing to read, because of lack of line breaks and the inconsistent use of comma/period convention for the decimal.
$endgroup$
– Jason Zimba
Apr 2 '14 at 11:25