If is+this here then value of numeric value of t*e+i*r*h-s
Answers
Answer:
157
Step-by-step explanation:
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If “ IS + THIS = HERE ”.
Then, the numerical value of: T x E + I x R x H-S = 72
Given:
IS + THIS = HERE
To Find:
T x E + I x R x H-S = ?
Solution:
Given that IS + THIS = HERE
This can be re-written as T H I S
+ I S
= H E R E
- T + nothing or (0) is not equal to H.
[Explanation: T + 0 = T
(any number added with zero will remain same)]
Thus 1 should have been carried forward from the previous digit or number (H) i.e., T + 1 = H.
Therefore, T = H – 1 --------------- Step 1
- Similarly H + nothing or (0) is not equal to E.
And 1 should have been carried forward from the previous digit or number.
[Explanation: I + I = R
(Where R value would have been greater than or equal to 10 Thus 1 being carry forwarded to H.)]
i.e., H + 1 >/= 10. -------------- Step 2
Therefore, H = 10 – 1
= 9
H = 9
So, Step 1: T = H – 1
= 9 – 1
= 8
T = 8
- From Step 2: H + 1 = 10 Then the value of E will be the One’s digit.
i.e., E = 0
- S + S = E And E = 0
S + S can be equal to 0 only if the addition of both equal to 10.
Therefore, S = 5
[Explanation: 5 + 5 = 10]
- Since S + S = 10 ; 1 will be carried to the next digit.
i.e., I + I [+1] = R
= 2I + 1 = R
[Based on the explanation of step 1 R value will be greater than or equal to 10.]
- Assumptions
Assuming I value as 7; 2 x 7 + 1 = 15 where, R cannot be 5 as S = 5
Assuming I value as 6; 2 x 6 + 1 = 13 where. R is 3 which is possible,
[because I + I = R greater than or equal to 10 and the when assumed I value as 6 the addition of the I + I + 1 = 13 which is greater than 10 also the digit 3 does not collide with any other values.
Therefore, T = 8; H = 9; I = 6; S = 5; E = 0; and R = 3
Thus T x E + I x R x H-S = (T x E) + (I) (R) (H-S)
= (8 x 0) + 6 x 3 x (9 – 5)
= 0 + 18 x 4
= 72
Therefore, If “ IS + THIS = HERE ”.
Then, the numerical value of: T x E + I x R x H-S = 72
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