CAT Explanations Week 5
• This Section
• Latest
• Web Wrap
Last Updated: Wednesday, October 02, 2013, 11:04

1. c Minimum possible value of x + y + z will be when we maximize the value of ‘x’. Maximum possible value of x will be 15 and since x, y and z are distinct positive integers, y = 1 and z = 4. So, minimum possible value of x + y + z = 15 + 1 + 4 = 20.

Maximum possible value of x + y + z will be when the value of z is maximized. Maximum possible value of z will be when y = 2 and x = 1, i.e. z = 72. Required difference is 75 – 20 = 55.

2. b Here a + (n – 1) d – a = 2 ? 2 d = (n – 1)
The sum of all the even placed terms = (a + d) + (a + 3d) +. …= 7 …(i)
The sum of all the odd placed terms = a + (a + 2d)+. …= 8.75 …(ii)
Subtracting equation (i) from (ii), we get ? - ? + ? + +… ? = ? ?
(n 1) a d d terms 1.75 2 - ? + × = (n 1) a d 1.75 2
We know that = - 2 d n 1
?a + 1 = 1.75 or a = 0.75
?The last term is 2.75
?Sum of the ‘n’ terms of the AP
= n 2 [First term + Last term] = 15.75
= n 2 [0.75 + 2.75] = 15.75 ? n = 9

3. d B A C D E H I J F G
AB2 + BC2 = AC2 ? AB = 12 cm
AH = AI, BI = BJ and CH = CJ
? 13 – AH = 5 – BJ ? AH – BJ = 8 cm
AI + BI = AH + BJ = 12 cm
AH = 10 cm and BJ = 2 cm (as BJ is equal to radius)
Let, EG = y cm, BD = z cm and GF = x cm = FJ.
DG = DI ? 7 – y = z + 2 ? y = 5 – z
In triangle DBF : DB2 + BF2 = DF2
? z2 + (2 + x)2 = (7 – x – y)2
= (2 + z – x)2 (as y = 5 – z)
2z x z 4 ? = + 2BD GF BD 4 ? = +
Therefore, k = 4.

4. d Applying distance formula, a2 + b2 = 5 So, a2 + b2 = 25
So, possible solution sets are (± 5, 0), (0, ± 5) (± 3, ± 4) and (±4, ± 3)
So there are 2 + 2 + 4 + 4 = 12 possible points.

5. d Suppose we take x 50 p. coins, y 25 p. coins and z 10 p. coins
We have x + y + z = 85 and 50 x+ 25 y + 10 z = 1875 x, y, z are all integers. ? z should be a multiple of 5.
Possible values are x y z 2 63 20 5 55 25 8 47 30
-----------------------------------
23 7 55
So, eight solutions.
You should notice here that an increase of 5 in z and 3 in x and decrease of 8 in y neither changes the number of coins nor the value.

6. d Since all of them drink three of the four beverages so a person can drop only one beverage. So one cannot drop both Vodka and Whiskey. Hence, the percentage of people taking liquor is 100.

First Published: Wednesday, October 02, 2013, 11:04