Quant Boosters - Maneesh - Set 4

• 100 - (30 + 25 + 20 + 15) = 10

• Q11) N is a four-digit number in which each of the digits used appears at least two times. The number of different values that N can assume is

• let first digit is 1 then other 3 digits can be 1 itself --> 1 way
one of them can be 1 and other two any of (0,2,3....9) --> 3 * 9 = 27
total 27 + 1 = 28

similarly when first digit is 2,3,4...9 --> 9 * 28 = 252

• Q12) There are 2 bags, one consists of 3 red and 4 black balls and the other consists of 4 red and 2 black balls. One bag is selected at random and from a selected bag a ball is drawn. Let E be an event that first bag is selected, F be an event that second bag is selected, G be an event that a ball drawn is red. Find P(G/E).

• 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(A-3)
55(A-3) = 42A+31
55A-165 = 42A+31
13A = 196
A = 196/13
original expenditure = 42A = 42 * 196/13 = 633.23

• Q15) If we divide a given two-digit 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)^2-xy = 10x+y
x^2+y^2+xy = 10x+y
x^2+y^2+xy = 3xy+9
x^2+y^2-2xy = 9
(x-y)^2 = 9

x-y = +3
10y+30+y = 9+3y(y+3)
11y+30 = 9+3y^2+9y
3y^2-2y-21 = 0
y = 2+-16/6 = 18/6 = 3
x = 6

or

x - y = -3
x = y-3
11y-30 = 9+3y(y-3)
11y-30 = 9+3y^2-9y
3y^2-20y+39 = 0

b^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, 2

common 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 possibility

common ratio 1/2 ......1 possibility
common ratio 2 .....one possibility

total 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

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

61

48

61

1

61

61

61