*CHALLENGE FOR MATHEMATICIANS*
Here is a mathematical quiz to solve.
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A man gave a cheque to his servant and asked to get it encashed from bank. Cheque was for few rupees and paisas ( Rs.ab.xy).
बैंक कै्शियर ने गलती से चैक में जितने रुपए थे, उतने पैसे और जितने पैसे थे, उसकी जगह उतने रुपए नौकर को दे दिए। (Means, cashier gave give Rs. xy.ab. in place of Rs. ab.xy).
On the way to home, नौकर ने उसी पैसों मे से 20 पैसे की बीङी पी ली। और बाकी सारे पैसे मालिक को लाकर दे दिए।
मालिक ने पैसे गिने तो देखा कि
Amount brought by servant was exactly double the amount written in the cheque.
Now question is *WHAT WAS THE AMOUNT OF THE CHEQUE*
*लगाओ दिमाग*
Answers
Let us take the number of rupee is x and that of paise is y
Then written on cheque = 100x + y paise
Amount given by cashier = 100y + x paise
Servant spent = 20 paise
Amount left = 100y + x - 20 paise
By the given conditions,
100y + x - 20 = 2 (100x + y)
or, 100y + x - 20 = 200x + 2y
or, 199x = 98y - 20
or, x = (98y - 20)/199
We need to find a 2 digit integral value of x and 2 digit integral value of y.
Let us put y = 10, 11, 12, 13, ... and calculate the value of x.
When y = 10, x = 4.824...
When y = 11, x = 5.317...
When y = 12, x = 5.809...
..... ..... ..... .....
..... ..... ..... .....
When y = 53, x = 26.
Here both x = 26 and y = 53 are integral values in two digits.
So it was written Rs. 26.53 on cheque.
Confirmation:
- Written >> Rs. 26.53
- Cash given to servant >> Rs. 53.26
- Spent by servant >> Rs. 0.20
- Cash given to the man >> Rs. 53.06
- Rs. 53.06 = 2 * Rs. 26.53
Read more on Brainly:
- https://brainly.in/question/16318498
Answer:
26.53
Step-by-step explanation:
Let ab = A(two digit number) &
xy = X(two digit number)
As given in the question:
⇒ Amount brought by servant was exactly double the amount written in the cheque
⇒ 2(A.X) = X.A - 0.2
⇒ 2(A + .X) = X + .A - 0.2
As X is a 2 digit number, in terms of face value
⇒ 2(A + X/100) = X + A/100 - 0.2
⇒ 199A = 98X - 20
⇒ A = (98X - 20)/199 ...(1)
As A is a two digit number, (98X - 20)/199 must be greater than 10 or 10.
To do so, X must be at least (199*10-20)/98 = 20.1. (atleast 21).
Now we have A ≥ 10, X ≥ 21
On observing (1),
For every change(by 1) in value of A, X must be increased by 2. (not precised). So we would prefer to go for X = 21, 21 + 2, 21 + 4 , etc
Every time, A increases by 2*96/199 ≈ 0.9849
For X = 21 , A = 10.2412
For X = 23, A = 10.2412 + (196/199)
For X = 25, A = 10.2412 + 2(0.9849)
For X = 27, A = 10.2412 +3(0.9849)
We need to need such an integral value of (0.9849) which has (1 - 0.2412) as its first decimal values for natural value of A.
We need to keep on adding,
We get that 16(0.9849) ends with 0.758 as required.
Hence,
A = (98*21-20)/199 + 16(196/199)
= 26
Hence, X = 53 [substituting in (1)]
∴ number is 26.53
Method 2
⇒ 2(A.X) = (X.A) - 0.2
Possible only when X is actually 2A or 2A + 1 (which might had been decreased by 0.2)
⇒ 199A = 98X - 20
⇒ 199A = 196A - 20 or 199A = 196A + 98 -20
⇒ A ≠ -ve or A = 26
Therefore, X = 53
Number is 26.53