Quants 4 (tough ones)
Q1. If K = 231 x 319 , how many positive divisors of k2 are less than k but do not divide K?
a. 1228 b. 1028 c. 639 d. 589
Q2. A triangle of side 20cm, 48cm and 52cm is cut into two pieces of equal area by a straight cut. What is the maximum possible sum of the perimeters of the two pieces (approximately)
a. 178cm b. 190cm c. 198cm d. 218cm
Q3. How many integral solutions exist for the equation |x –105| + |y-303| <= 3?
a. 14 b. 25 c. 63 d. None of these
Q4. A right circular cone is cut, parallel to its base into 5 slices. Heights of all slices are equal. By what percentage is the volume of the middle most slice lower than the volume of the largest slice?
a. 82.4% b. 41.3% c. 68.9% d. 55.3 %
Q5. Find the number of solutions for positive integers a, b and c such that a + b + c = 38 and a 2 + b 2 + c 2 = 722
a. 1 b. 3 c. 6 d. None of these
Q6. A dice is rolled three times. Find the probability of getting a larger number than previous number each time.
a. 5/74 b. 7/54 c. 11/54 d. 13/54
n/a
The perimeter‘ll be maximum when we cut the triangle with the base as the smallest side.
i.e we ‘ll cut the triangle at base 20 (BC) with a median drawn from point A.
Now the length of the medians
M20 = sqrt [ {2( 48)^2 + 2 (52)^2 – 20^2}/4]
=49
So the sum of the perimeters of the newly formed two triangles
= 48 + 49 + 10 +
52 + 49 + 10 = 218 cm
n/a
Q4. A right circular cone is cut, parallel to its base into 5 slices. Heights of all slices are equal. By what percentage is the volume of the middle most slice lower than the volume of the largest slice?
a. 82.4% b. 41.3% c. 68.9% d. 55.3 %
ans is 68.9%
Little Star
n/a
|x –105| + |y-303| <= 3?
Above is same as |A| + |B| <= 3
For |A| + |B| = 3 number of vals of A, B = 12
For |A| + |B| = 2 number of vals of A, B = 8
For |A| + |B| = 1 number of vals of A, B = 4
For |A| + |B| = 0 number of vals of A, B = 1
So total number of solutions = 25
btw do you see any pattern??
is there any easier method to solve this sort of question ???
Q6. A dice is rolled three times. Find the probability of getting a larger number than previous number each time.
a. 5/74 b. 7/54 c. 11/54 d. 13/54
dice rolled three times
if getting 1 on first dice then
first dice second third
1 2 3/4/5/6
3 4/5/6/
4 5/6
5 6
2 3 4/5/6
4 5/6
5 6
3 4 5/6
5 6
4 5 6
so total 20 case
so probability=20/216 =5/54
Little Star
n/a
n/a