1303 : Arithmetic _CAT 2007

A confused bank teller transposed the rupees and paise when he cashed a cheque for Shailaja.
giving her rupees instead of paise and paise instead of rupees. After buying a toffee for 50 paise,
Shailaja noticed that she was left with exactly three times as much as the amount on the cheque.
Which of the following is a valid statement about the cheque amount?
(1) Over Rupees 13 but less than Rupees 14
(2) Over Rupees 7 but less than Rupees 8
(3) Over Rupees 22 but less than Rupees 23
(4) Over Rupees 18 but less than Rupees 19
(5) Over Rupees 4 but less than Rupees 5


Soln :

Suppose the cheque for Shailaja is of Rs. X and Y paise
As per the question: 3 × (100X + Y) = (100Y + X) – 50
⇒ 299X = 97Y – 50
⇒ Y = 299X + 50 /97

Now the value of Y should be a integer. Checking by options only for X = 18, Y is a integer and
the value of Y = 56

From Puneet Singh (www.bsaitmfbd.com) Eduventures

Alzebra : 1021

Problem 1

The equations x² + ax + 1 = 0 and x² + bx + c = 0 have a common real root, and the equations x² + x + a = 0 and x² + cx + b = 0 have a common real root. Find a + b + c.

 

Solution

The common root of x² + ax + 1 = 0 and x² + bx + c = 0 must also satisfy (a-b)x + (1-c) = 0, 

so it must be (c-1)/(a-b). 

Note that the other root of x² + ax + 1 = 0 must be (a-b)/(c-1), since the product of the roots is 1. 

Similarly the common root of x² + x + a = 0 and x² + cx + b = 0 must satisfy (c-1)x + (b-a) = 0, 

so it is x = (a-b)/(c-1). 

Hence x² + x + a = 0 and x² + ax + 1 = 0 have a common root. 

Hence it satisfies (a-1)x + (1-a) = 0. Now we cannot have a = 1, for then x² + ax + 1 has no real roots. Hence the common root must be 1. 

Hence both roots of x² + ax + 1 = 0 are 1 and so a = -2.

So x² + bx + c = 0 has one root 1. 

Then its other root must be c/1 = c. Hence -b = 1 + c, or b + c = -1. Hence a + b + c = -3. 

Answer

-3