Quant Boosters  Maneesh  Set 4

P(G/E) = P(GnE)/P(E)
P(GnE) = 1/2 * 3/7 = 3/14
P(E) = 1/2
P(G/E) = 3 * 2/14 = 3/7

Q13) The average age of A ,B ,C ,D five years ago was 45 years. By including X, the present average age of all the five is 49 years.The present age of X is
a) 64 years
b) 48 years
c) 45 years
d) 40 years

Total Age 5 years ago = 180
Total age now = 180 + 20 = 200
X included,total age = 5 * 49 = 245
Age of X = 45

Q14) There are 42 students in a hostel. Due to admission of 13 new students, the expenses of mess increase by Rs 31 per day while the average expenditure per head diminished by Rs 3. What was the original expenditure of the mess ?

Let average expenditure is A
Total expenditure is 42 A
After 13 students joined new expenditure = 42A + 31
Total expenditure 55(A3)
55(A3) = 42A+31
55A165 = 42A+31
13A = 196
A = 196/13
original expenditure = 42A = 42 * 196/13 = 633.23

Q15) If we divide a given twodigit number by the product of its digits, we obtain 3 as quotient and 9 as a remainder . If we subtract the product of the digits constituting the number, from the square of the sum of its digits, we obtain the given number. Find the number

10x+y = 3xy+9
(x+y)^2xy = 10x+y
x^2+y^2+xy = 10x+y
x^2+y^2+xy = 3xy+9
x^2+y^22xy = 9
(xy)^2 = 9xy = +3
10y+30+y = 9+3y(y+3)
11y+30 = 9+3y^2+9y
3y^22y21 = 0
y = 2+16/6 = 18/6 = 3
x = 6or
x  y = 3
x = y3
11y30 = 9+3y(y3)
11y30 = 9+3y^29y
3y^220y+39 = 0b^2 4ac = 400 468 = 68 hence no real roots
hence x= 6, y =3
number 63

Q16) How many numbers from 1000 to 10000 have their digits in AP or GP if we consider the digits from left to right?

If digits in AP ...possibilities  common difference 0, 1, 2, 1, 2, 3
if digits in GP...possibilities  common ratio 1/2, 2common difference 0....9 possibilities
common difference 1....6 possibilities
common difference 2.....3 possibilities
common difference 1.....7 possibilities
common difference 2.....4 possibilities
common difference 3......1 possibilitycommon ratio 1/2 ......1 possibility
common ratio 2 .....one possibilitytotal 32

Q17) The product of three numbers in GP is 1000 and the sum of the products of the numbers taken in pairs is 350. Find out the greatest of three numbers.

a/r, a, ar
a^3 = 1000, a = 10
a^2/r + a^2 + a^2r = 350
100/r + 100r = 250
2r^2  5r + 2 = 0
r = 1/2 , 2
numbers 5, 10, 20
Greatest is 20

Q18) How many three digit numbers increase by 495 when their digits are reversed?
a) 4
b) 20
c) 40
d) 100

100x + 10y + z 100z  10y  x = 495
99x  99z = 495
x  z = 5
(9,4)(8,3)(7,2)(6,1)
y : 0 to 9 > 10 values
4 * 10 = 40

Q19) A boy tries to climb a 15m greased pole. Each attempt, the boy climbs 3 m and slips back 2 m. How many attempts will it take to climb 10m of the pole?

in one attempt net value of distance boy climb is 3–2 = 1m
in 6 attempts boy climbs = 6 m
in the 7th attempt, he climbs = 6+3–2 = 7 m
in the eighth attempt he reaches 7+3 = 10 m
Hence answer 8 attempts

Q20) What is the remainder when (32^32)^32 is divided by 9?

32^2 mod 9 = 1024 mod 9 = 2mod 9
(32^32)^32 mod 9 = (2)^(16 * 32)mod 9 = 2^(512) mod 9
Since 2^3 mod 9 = 1
2^512 = 2^510 * 2^2
2^510 mod 9 = 2^(3 * 170) mod 9 = 1^170 = 1
2^2 mod 9 = 4 mod 9 = 4
2^512 mod 9 = 4 * 1 = 4
Hence answer 4

Q21) If (a + 1/a)^2 = 3 then (a^3 + 1/a^3) = ?

a+1/a = root3
(a+1/a)^3 = 3root3
a^3 + 1/a^3 + 3(a + 1/a) = 3 root3
a^3 + 1/a^3 + 3root3 = 3root 3
a^3 + 1/a^3 = 0

Q22) If a and b are twodigit prime numbers such that a^2  b^2 = 2a + 14b + 48. Find the largest possible value of a + b.