CAT Previous Year Questions: Progressions (June 4) - CAT MCQ

x is the Arithmetic Mean of z & y, therefore, z, x, y form an A.P.
It goes without saying that y, x, z also forms an A.P.

Let the alloy contain x Kg silver and y kg copper 
Now when mixed with 3Kg Pure silver 
we get 
we get 10x + 30 = 9x + 9y + 27
9y - x = 3    (1)
Now as per condition 2
silver in 2^nd alloy = 2(0.9) = 1.8
So we get 
we get 21y -4x =3   (2)
solving (1) and (2) we get y= 0.6 and x = 2.4
so x + y = 3

Let price of smallest cup be 2x and medium be 5x and large be y 
Now by condition  1 
2x ×  5x × y =800
we get x²y =80    (1)
Now as per second condition ;

Now dividing (2) and (1)
we get 
we get  
we get 30x² − 42x − 36=0
5x² −7x − 6 = 0
we get x = 2
So 2x = 4 and 5x = 10
Now substituting in (1) we get y = 20
Now therefore sum = 4 + 10 + 20 = 34

Let us solve this question by assuming values(multiples of 100) and not variables(x).
Since we know that the female population was twice the male population in 1990, let us assume their respective values as 200 and 100.
Note that while assuming numbers, some of the population values might come out as a fraction(which is not possible, since the population needs to be a natural number). However, this would not affect our answer, since the calculations are in ratios and percentages and not real values of the population in any given year.

Now, we know that the female population became 1.25 times itself in 1990 from what it was in 1980.
Hence, the female population in 1980 = 200/1.25 = 160
Also, the female population became 1.2 times itself in 1980 from what it was in 1970.
Hence, the female population in 1970 = 160/1.2 = 1600/12 = 400/3
Let the male population in 1970 be x. Hence, the male population in 1980 is 1.4x.
Now, the total population in 1980 = 1.25 times the total population in 1970.
Hence, 1.25 (x + 400/3) = 1.4x + 160
Hence, x = 400/9.
Population change = 300 - 400/9 - 400/3 = 300 - 1600/9 = 1100/9
Percentage change
= 

Initially number of matches = 40 
Now matches won = 12 
Now let remaining matches be x 
Now number of matches won = 0.6x
Now as per the condition :

24 +1.2x = 40 + x
0.2x = 16
x = 80
Now when they won 90% of remaining = 80(0.9) =72
So total won = 84

Let the number of large shirts be l and the number of small shirts be s.
Let the price of a small shirt be x and that of a large shirt be x + 50.
Now, s + l = 64l (x + 50) = 5000
sx = 1800
Adding them, we get,
lx + sx + 50l = 6800
64x + 50l = 6800
Substituting l = (6800 - 64x) / 50, in the original equation, we get

So, x = 75
x + 50 = 125
Answer = 75 + 125 = 200.

Let Total matches played be n and in initial n-10 matches his goals be x 
so we get 
we get x+1 =0.15n                (1)
From condition (2) we get :

we get x + 2 = 0.2n                  (2)
Subtracting (1) and (2)
we get 1 = 0.05n
n = 20
So initially he played n - 10 =10 matches

Let the number of white balls be x and black balls be y
So we get x + y = 450 (1)
Now metallic black balls = 0.5y
Metallic white balls = 0.4x
From condition 0.4x= 0.5y
we get 4x - 5y = 0 (2)
Solving (1) and (2) we get
x = 250 and y = 200
Now number of Non Metallic balls = 0.6x + 0.5y = 150 + 100 = 250

T_n = 12
Tm = 3 / 4


To get the minimum value for r + n - m, r should be minimum. 
∴ r = - 4
n - m = 2
∴ Required answer =-2 

x₁ = −1
 
......
If we observe the series, it is a series that has a difference between the consecutive terms in an AP.
Such series are represented as t(n) = a + bn + cn²
We need to find t(100).
t(1) = -1
a + b + c = -1
t(2) = -1
a + 2b + 4c = -1
t(3) = 0
a + 3b + 9c = 0
Solving we get,
b + 3c = 0
b + 5c = 1
c = 0.5
b = -1.5
a = 0
Now,
t(100) = (−1.5)100 + (0.5)100²  = −150 + 5000 = 4850