# Quant Boosters - Kamal Lohia - Set 2

• Regular Sol: Let salary of new one is x, so we have
12n - 350 + x = 12(n - 5) where n is the initial average salary.
i.e. x = 350 - 60 = 290. (d)

Alt Sol: As average salary dropped by 5 i.e. salary of new clerk is less by 5*12 = 60 than the leaving clerk's 350 i.e. 350 - 60 = 290. (d)

• Q26) Ram prepares solutions of alcohol in water according to customers' needs. This morning Ram has prepared 27 liters of a 12% alcohol solution and kept it ready in a 27 liter delivery container to be shipped to the customer. Just before delivery, he finds that the customer had asked for 27 liters of 21% alcohol solution. To prepare what the customer wants, Ram replaces a portion of 12% solution by 39% solution. How many liters of 12% solution is replaced? [XAT2012]
a. 5
b. 9
c. 10
d. 12
e. 15

• Regular Sol: Let x liters of solution has been replaced, so
39x + 12(27 - x) = 21(27)
i.e. 27x = 9(27)
i.e. x = 9 liters (b)

Alt Sol: 12%---(9) --> 21% < -- (18)--- 39%
So the quantities of 12% and 39% solutions are mixed in the INVERSE ratio of gaps i.e. 2 : 1.
As total solution is 27 liters, so the replaced mixture's quantity is (1/3) * 27 = 9 liters (b)

• Q27) The owner of an art shop conducts his business in following manner, Every once a while he raises his prices by X% and then a while later he reduces all the new prices by X%. After one such up-down cycle, the price of a painting decreased by Rs.441. After a second up-down cycle the painting was sold for Rs. 1944.81. What was the original price of the painting? [CAT2001]
a. 2756.25
b. 2256.25
c. 2500
d. 2000

• Regular Sol(with least calculation): Px²/100² = 441 = 21²
P(1 - x²/100²)² = 1944.81 = (44.1)²
i.e. (100² - x²)/100x = 44.1/21 = 2.1
i.e. x² + 210x - 10000 = 0
i.e. (x + 250)(x - 40) = 0
i.e. x = 40
And P = 441*10000/1600 = 2756.25 (a)
Alt Sol: As in first up-down cycle price reduced by 441, so in second such cycle the reduction in price will be lesser than 441 but it'll not be much lesser than 441. So taking it to be near 400, you can easily eliminate options and zero down to 2756.25 (a)

• Q28) Find the remainder when 10³ + 11³ + 12³ + 13³ + 14³ + 15³ is divided by 75.
a. 0
b. 15
c. 25
d. 50

• Sol: Just remember the property of numbers that (aⁿ + bⁿ + cⁿ + ...) is always divisible by (a + b + c + ...)
where
(i) all alphabets represent positive integers,
(ii) a, b, c, ... are in arithmetic progression and
(iii) n is odd.
So here (10³ + 11³ + 12³ + 13³ + 14³ + 15³) is divisible by (10 + 11 + 12 + 13 + 14 + 15) i.e. 75 as all the above mentioned three conditions are being satisfied.

• Q29) N⁴ is a positive integer. Which of the following statement/s is/are certainly correct?
I. N is a positive integer.
II. N² is a positive integer.
III. N³ is a positive integer.
A. I only
B. I and III only
C. II only
D. none of these

• Mind that information provided is "N⁴ is a positive integer", but nothing has been said about N. So it can be an imaginary number also. One possible set of values, which refutes all the three statements, is:
N = i
N² = -1
N³ = -i
N⁴ = 1
Answer is D

• Q30) TG² = GOG where T, G, O are distinct single digit positive integers. Find O.
A. 0
B. 2
C. 7
D. 8

• Sol: Remember that: If square of a number ends in same unit digit as that of number, then that unit digit can be 0, 1, 5 or 6 only. So possible values for G are 0, 1, 5 or 6. Now by checking you can easily negate that G can't be 0 as GOG will not remain a three digit number. It can't be 1 either as in this case the three digit number will be 1O1 and it'll be perfect square only when O is 2 i.e. T is 1 which is not possible as T and G are "DISTINCT" single digit positive integers. Similarly you can check for 5 also that no number of the form 5O5 is a perfect square.
Only possible case is 26² = 676 i.e. O = 7.

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