M4maths Previous Puzzles

22 Dec 2011 percent

k % of 90 = 80 % of y
relation between k and y will be

OPtion

1) k=y
2) 3k=2y
3) 3k=8y
4) 9k=4y
5) 9k=8y
6) 8k=9y
7) 9k=10y
8) 10k=9y
9) 2k=9y
10) none of these

Solution

90k/100 = 80y/100
or 9k = 8y

Correct Option: 5 Best Solution (0)

21 Dec 2011 AP and GP

There are 4 positive numbers k, k^2, k^3, k^4. If we subtract 12 by first number and 36 by last number, we will get an AP of 4 numbers. Calculate k?

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10) none of these

Solution

According to the given condition k-12, k^2, k^3, k^4-36 are in AP.
so k^2-(k-12) = k^3-k^2
so k^3-2k^2+k-12 = 0 (i)
equation (i) has 3 as its real root
on taking the value of k, given 4 numbers will be -9,9,27,45

Correct Option: 3 Best Solution (0)

20 Dec 2011 probability

What will be the probability that k1*k2*k3 is an even number, if k1, k2, k3 are 3 positive integers taken randomly.

OPtion

1) 1/8
2) 1/4
3) 3/8
4) 1/2
5) 5/8
6) 3/4
7) 7/8
8) 1
9) 1/3
10) none of these

Solution

if you take all odd numbers, then product will be odd. In all other 7 cases, it will be even so probability = 7/8

Correct Option: 7 Best Solution (0)

19 Dec 2011 integer

Two positive integers x and y are possible for which sum of x,x^2,x^3 and similarly y,y^2,y^3 are nearer to 100 than the sum of number, its square and cube for other integers. What will be the positive difference between the sum of x,x^2,x^3 and y,y^2,y^3

OPtion

1) 45
2) 54
3) 58
4) 63
5) 68
6) 70
7) 71
8) 75
9) 87
10) none of these

Solution

The value of sum (x+x^2+x^3) for 3, 4, 5 and 6 are 39, 84, 155 and 258 respectively. Nearer to 100 are 84 and 155, so difference = 155-84 = 71

Correct Option: 7 Best Solution (2)

16 Dec 2011 age

Sum of the ages of a boy and his father is 70 years. Before 20 years Father's age was 14 times the age of the boy. What was the age of father at the time of birth of the child?

OPtion

1) 22
2) 25
3) 26
4) 27
5) 29
6) 31
7) 34
8) 36
9) 38
10) none of these

Solution

Let present age of father and the son are x and y
then according to the given condition
x+y = 70
before 20 years
(x-20) = 14(y-20)
so 70-y-20 = 14(y-20)
or 50-y = 14y-280
or 15y = 330
or y = 22 and x = 48
so at the time of birth of child, age of father = 48-22 = 26

Correct Option: 3 Best Solution (25)

15 Dec 2011 number game

If x = 3, y = 4, z = 5,
then sum of the modulus of the numbers obtained by differences in the manner E1 - E2 and E2 - E3
Meaning of operations are
E1 : sum of x,x^2,x^3;
E2 : sum of y,y^2,y^3 and
E3 : sum of z,z^2,z^3

OPtion

1) 96
2) 104
3) 110
4) 116
5) 129
6) 134
7) 140
8) 148
9) 152
10) none of these

Solution

E1 = 39
E2 = 84
E3 = 155
now E1 - E2 = 39 - 84 = -45 its modulus means numerical value = 45
and E2 - E3 = 84 - 155 = -71 its modulus means numerical value = 71
so sum = 45+71 = 116

Correct Option: 4 Best Solution (12)

14 Dec 2011 lines

There are 5 lines in a plane. Difference between maximum and minimum number of intersecting points. No two lines are parallel

OPtion

1) 2
2) 3
3) 4
4) 5
5) 8
6) 9
7) 10
8) 15
9) 19
10) none of these

Solution

Maximum no. of intersecting points = 5*4/2 = 10
minimum no. of intersecting points = 1

Correct Option: 6 Best Solution (10)

13 Dec 2011 AP and GP

A number a, its square and its cube are in GP. What will be the third term of AP, for which first two terms are obtained by taking the difference of terms of GP in 2nd-1st and 3rd-2nd manner?

OPtion

1) 6*sum of all numbers upto (a-1)
2) 6*sum of squares of all numbers upto (a-1)
3) a+a^2+a^3
4) 1+a+a^2+a^3
5) 6*sum of all numbers upto a
6) 6*sum of squares of all numbers upto a
7) 6*sum of cubes of all numbers upto (a-1)
8) 6*sum of cubes of all numbers upto a
9) always a fixed number
10) none of these

Solution

Let the 3 numbers are a, a^2 and a^3
so differences = a^2-a and a^3-a^2
third term of the AP, for which first 2 terms are a^2-a and a^3-a^2 will be 2(a^3-a^2)-(a^2-a) = 6*[a(a-1)(2a-1)/6]
= 6*sum of squares of all numbers upto (a-1)

Correct Option: 2 Best Solution (7)

12 Dec 2011 sequence

next number of the sequence
12, 20, 30, 42, 56, ..... will be

OPtion

1) 58
2) 60
3) 64
4) 66
5) 68
6) 70
7) 72
8) 76
9) 78
10) none of these

Solution

Difference of the sequence is 8, 10, 12, 14, 16
so 12, 12+8, 20+10, 30+12, 42+14, 56+16,
56+16 = 72

Correct Option: 7 Best Solution (40)

09 Dec 2011 powers

What will be the power of 10 in the product 1*2*3*4*5......*40

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10) none of these

Solution

10 = 5*2, so we should consider either the number of 5 or number of 2, whichever is less,
Total no of fives = 8+1 = 9

Correct Option: 9 Best Solution (9)