Q. Find the common factors of
a) 8=2×2×2
20=2×2×5
common factor =2×2=4
b) 3×3=9,
3×5=15
3 is the common factor of 9 and 15
Q10. Find the values of the letters in each of the following and give reasons for the steps involved.
The addition of A and B is giving a number whose ones digit is 9. The sum can only be 9 not 10 as a sum of two single digits cannot exceed 18. hence there will not be any carry for the next step
2 + A = 0
implies A = 8
2 + 8 = 10 and 1 is the carry for next step.
1 + 1 + 6 = A = 8
it satisfies hence A = 8 and B = 1 is the correct answer.
View Full Answer(1)Q9. Find the values of the letters in each of the following and give reasons for the steps involved.
The additiion of B and 1 is 8 is giving a number whose ones digit is 8. this means digit B is 7 .
B + 1 = 8 and no carry for next step.
next step :
Now, A + B = 1 => A + 7
which implies A = 4
A + B = 11 and 1 is carry for next step
1 + 2 + A = B
1 + 2 + 4 = 7
Hence A = 4 and B = 7 is correct answer.
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Q8. Find the values of the letters in each of the following and give reasons for the steps involved.
The addition of 1 and B gives a number whose ones digit is 0. this is possible when digit B = 9 .
1 + B = 10 and 1 is the Carry for the next step
Now, A + 1 + 1 = B => 9
Implies A = 7.
Hence A = 7 and B = 9 is Correct answer
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Q7. Find the values of the letters in each of the following and give reasons for the steps involved.
The product of 6 & B gives a number whose unit digit is B again.
possible value of B = 0. 2 , 4, 6 or 8
If B = o then our product will be zero. hence this value of B is not possible.
If B = 2, then B x 6 = 12 . Carry for next step = 1.
6A + 1 = BB = 22
implies A = 21/6 = not any integer value hence this case is also not possible.
If B = 6 then B *6 = 36 and 3 will be carry for next step.
6A + 3 = BB = 66
implies A = 63/6 = not an integer value hence this case is also not possible.
If B = 8 then B * 6 = 48 and 4 will be carry for next step.
6A + 4 = BB = 88
implies A = 14 however A is a single digit number hence this case is also not possible.
If B = 4 then B*6 = 24 and 2 will be carry for next step.
6A + 2 = BB = 22
implies A = 7 .
Hence A = 7 and B = 4 is the correct answer.
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Q6.Find the values of the letters in each of the following and give reasons for the steps involved.
Here , multiplication of B and 5 gives a number whose ones digit is B. this is possible when B = 0 or 5
let B = 5
B * 5 = 5 * 5 = 25
Carry = 2
5*A + 2 = CA , This os possible only when A = 2 or 7
when A = 2 ,
5*2 + 2 = 12 which implies C = 1
when A = 7
5*7 + 2 = 37 which implies C = 3
now B = 0
B*5 = 0*5 = 0
Carry = 0 so
5 * A = CA which is possible when A = 0 or 5 Howerver A cannot equal to 0 since AB is a two digit number
so A = 5
5*5 = 25 which implies C = 2
hence possible values of A B and C are
A = 5, 2, 7 , B = 0, 5, 5 and C = 2, 1, 3
View Full Answer(1)Q5.Find the values of the letters in each of the following and give reasons for the steps involved. .
Here multiplication of 3 and B gives a number whose unit place digit is B .
Possible value of B = 0 and 5
let B = 5
3 * A + 1 = CA
this is not possible for any value of A.
Hence B = 0
now A * 3 = CA ( a number whose unit place digit is A itself when multiplied by 3) hence
possible value of A = 5 and 0
since AB is a two digit number A can not equal to 0.
hence A = 5
A * 3 + 1 = CA
5 * 3 + 1 = 15
hence C = 1
A = 5, B = 5 and C = 1.
View Full Answer(1)Q4.Find the values of the letters in each of the following and give reasons for the steps involved
There can be two cases
1. when the addition of unit place digit doesn't produce Carry
A + 3 = 6
A = 3
However, to get 3 in unit place of our answer our B has to be 6 and that would produce carry hence this case is not possible.
2. when the addition of unit place digit produces Carry
A + 3 +1 = 6
A = 2
for getting 2 in unit place of answer we need the sum of unit digit of numbers = 12
B + 7 = 12
B = 5
Hence A = 2 and B = 5
View Full Answer(1)Q3. Find the values of the letters in each of the following and give reasons for the steps involved.
Here
first clue :
we have A = a number which when multiplied by itself gives the same number in the unit digit.
possible numbers = 1 and 6
Second Clue:
number when multiplied with 1 and added with the reminder of previous multiplication( A*A) = 9
both first and second clue implies that A = 6.
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Q2. Find the values of the letters in each of the following and give reasons for the steps involved.
Here
answer's unit place = 3 , possible addition of unit places digit = 13
A + 8 = 13
A = 13 - 8 = 5
remainder = 1
Ten's place of answer = 4 + 9 + 1 = 14
B = 4
remainder = 1
100's place = C = 1
Hence value of A = 5 , B = 4 and C = 1.
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