Quant Boosters  Vikas Saini  Set 3

8x + y + 8y + x = abc.
9x + 9y = abc.
9(x+y) = abc.
x = 5, y = 7 then abc = 108.
x = 6, y =7 then abc = 117.
but abc are distinct digits.
so only 1 value 108.

Q2) A two digit number in base 7 is equal to the two digit number formed by reversing it digits but in base 13. How many such numbers are possible ?

(xy)_7 = (yx)_13
7x+y = 13y+x
6x = 12y
x = 2y.
x = 1,2,3.
y = 2,4,6.
three numbers 12, 24 and 36

Q3) A 99 digit number is formed by writing first 54 natural numbers in front of each other as 123 ... 54. Find the remainder when this number is divided by 8.
(a) 4
(b) 7
(c) 2
(d) 0 (CAT 1998)

For the divisibility by 8, we check only for the last three digits (because 8 = 2^3) of the number. If the last three digits of a number is divisible by 8, then the number will be divisible otherwise not.
So, here the number is 1234567891011121314…….525354.
Last three digits of the number = 354
Remainder when the number is divided by 8 = Rem(354/8) = 2

Q4) If log_10(x) – log_10(x^1/3) = 6log_x (10) then find the value of x
(a) 10
(b) 30
(c) 100
(d) 1000. (IIFT 2013)

log_10(x) – (1/3)log_10(x) = 6log_x(10).
= > (2/3)log_10(x)= 6 / log_10 (x)
Put log_10 (x) = m.
= > (2/3) m = 6 / m
= > m^2 = 9.
= > m = 3, 3.
We need to remove m = 3.
m = 3
log_10 (x) = 3.
x = 1000.
Hence option d.

Q5) The sum of the possible values of x in the equation x + 7 + x8 = 16.
(a) 0
(b) 1
(c) 2
(d) 3
(e) none of the above. (XAT 2014)

First put extreme value of x = 8.
At x = 8, equation = 15.
At x = 8.5, equation = 16.
Now put extreme negative value x = 7.
At x = 7, equation = 15.
At x = 7.5, equation = 16.
Sum = 8.5 – 7.5 = 1.

Q6) Sum of 6 terms of an AP is 6. If 3rd term of AP is 0 then find the 6th term of it.

Suppose 3rd term of AP is a.
2nd & 1st term are ad & a2d respectively.
4th, 5th & 6th terms are a+d,a+2d,a+3d respectively.
So (ad) + (a2d) + a + (a+d) + (a+2d) + (a+3d) = 6.
= > 6a + 3d = 6.
= > 0 + 3d = 6.
= > d = 2.
So a+3d = 6.
6th term is 6.

Q7) Sum of 3 terms of an AP is 30. If we add 6 in product of these 3 terms then it becomes a perfect square. Find all 3 terms of AP.

Let three terms are ad, a, a+d.
= > (ad)+a+(a+d) = 30.
= > 3a = 30.
= > a = 10.
(10 – d) x 10 x (10+d) + 6 = k^2.
= > 10(100 – d^2)+6 = k^2.
d = 1,2,3,4,5,6,7,8,9.
d^2 = 1,4,9,16,25,36,49,64,81.
100 – d^2 = 99,96,91,84,75,64,51,36,19.
10(100 – d^2) + 6 = 996,966,916,846,756,646,516,366,196.
196 is the only number here which is perfect square.
So d = 9.
3 terms of AP are = (109), 10, (10+9)
= 1, 9, 19.

Q8) Sum of first 4 terms of an AP is 42. Sum of square of 4 terms are 566. Find all the 4 terms

Let terms are a2d, ad , a+d , a+2d.
(a3d)+(ad)+(a+d)+(a+3d) = 42.
= > 4 a = 42.
a = 10.5.
(a3d)^2 + (ad)^2 + (a+d)^2 + (a+3d)^2 = 566.
= > a^2 + 9d^2 – 6ad + a^2 + d^2 – 2ad + a^2 + d^2 + 2ad + a^2 + 9d^2 + 6ad = 566.
= > 4a^2 + 20 d^2 = 566.
= > 4 x (10.5)^2 + 20 d^2 = 566.
= > 441 + 20d^2 = 566.
= > 20 d^2 = 125.
= > d^2 = 6.25
= > d = 2.5 .
Numbers are 3, 8, 13, 18.

Q9) From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formed such that the common difference is a factor of the fifth term ?

For d=1, total = 16.
(1,2,3,4,5), (2,3,4,5,6) ... (16,17,18,19,20)For d=2, total = 6.
(2,4,6,8,10), (4,6,8,10,12) ... (12,14,16,18,20).For d=3, total = 2.
(3,6,9,12,15), (6,9,12,15,18 )For d=4, total = 1.
(4,8,12,16,20)Total = 16 + 6 + 2 + 1 = 25.

Q10) Ratio of sum of two APs is (n + 1) : 2n. Find the ratio of the 6th terms of both series.

To find ratio of any term m.
We need to put n = 2m – 1.
m = middle term.
Here m = 6.
n = 11.
So ratio of 6th terms = 11 + 1 / 2 x 11 = 12 / 2 x 11
= 6 / 11.

Q11) Find total no of APs with 5 distinct terms that can be formed from the first 50 natural numbers.