### CAT 2019 – Slot 1 – Quantitative Ability – The number of integers x such that 0.25 <2x < 200

Q. 1: The number of integers x such that 0.25 <2x < 200, and 2x +2 is perfectly divisible by either 3 or 4, is?

as given 0.25 < 2x < 200

Or ¼ < 2x < 200

So x = { -2, -1, 0,1,2,3,4,5,6,7}

Now 2x +2 will be perfectly divisible by 3 if x is even non-negative integer and will be divisible by 4 if x = 1

So number of possible solution = 5 { x = 0, 1, 2, 4 ,6}

##### Q. 2: If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals

Given , f(x + 2) = f(x) + f(x + 1)

f(15) = f(13) + f(14)

f(13) + f(14) = 617 —————1)

f(12) + f(13) = f(14) ————-2)

f(11) + f(12) = f(13)————-3)

from eq 1) , 2) & 3)

2f(11) + 3f(12) = 617

Putting f(11) = 91

f(12) = (617 – 2*91)/3 = 145

f(12) = f(11) + f(10)

so f(10) = f(12) – f(11) = 145 – 91 =54

Q. 3: In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?
1. N ≤ 200
2. 243 ≤ N ≤ 252
3. N ≥ 253
4. 201 ≤ N ≤ 242

From the question we can say ( 100% – 68%) of ( 45% of N ) = 36

Or 32/100×45/100 N=36

N=360000/(32×45 )=250

Q. 4: Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio
1. 18 : 25
2. 19 : 24
3. 21 : 25
4. 17 : 25

When A and B are mixed in the ratio 3:2, CP = (3A+2B)/(5 )=40/1.1

3A +2B = 2000/11 ——-1)

When A and B are mixed in the ratio 2:3, CP = (2A+3B)/(5 )=40/1.05

2A +3B = 20000/105 = 4000/21 ———–2)

By solving eq 1) & 2)

A = 7600/21*11 and B = 9600/21*11

Thus A:B = 19:24

Q. 5:John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment, in Rs., is

Let the amount repaid in each installment was x. So

210000×(1.1)^2=x ×1.1+x

210000×1.21=2.1x

Or x=121000

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