Quantifying Quant

gaurav200x

gaurav200x

Gaurav Mittal
I have some test papers which i had taken from some quant text books and PaGaLGuY.com.... I shall be posting some quant questions regularly from it. Try to keep solving them. Correct answers with solutions shall be available soon.

Q1 A man puts in a tray letters to be typed by his secretary one at a time during the day at various times in the order 1, 2, 3, 4, 5, 6. The secretary takes the letters at the top for typing when he feels free to type it. What is the order in which the letters will definitely not get typed ?

1] 1, 2, 3, 4, 5, 6
2] 6, 5, 4, 3, 2, 1
3] 3, 2, 5, 4, 6, 1
4] 4, 5, 6, 2, 3, 1



Q2 A and B run in opposite directions from point P on the circumference of a circle with different speeds. A runs in clockwise direction and meets B for the first time after running 700 m. For the second time, A meets B, at a distance of 300m, from P, in the clockwise direction. The circumference of the circle, if only one of them has crossed P once, is

1] Can’t say
2] 900 m
3] 1000 m
4] 1100 m


Q3 A cube of length 5 cm has its faces painted and is cut into cubes of length 1cm each and one cube is picked at random. The probability of getting a cube with two faces painted is

1] 36/125
2] 27/125
3] 8/125
4] 54/125


Q5 The number of diagonals that can be drawn by joining the angular points of an Octagon is

1] 28
2] 20
3] 22
4] 21


Q5 If 17th December 1982 was a Saturday, then what will be the day on 22nd September 1984?

1] Saturday
2] Wednesday
3] Friday
4] Sunday
 
need some more enthusiasm here guys....... keep coming with ur responses... I have got many more questions to post.
 
we can turn this into some sort of online competition... the fastest person with correct ans gets some points... should make ppl more enthu
 
bips said:
we can turn this into some sort of online competition... the fastest person with correct ans gets some points... should make ppl more enthu
Click to expand...
yes, we're working on such a system and would let u know soon.

neeraj_uchila said:
the ans to the question no 5 is 20...hey gaurav if there is a direct formula for the soln please post it
Click to expand...

Were u able to find out how to calculate the dates? Post ur solutions to the others and let me know whether u know the others or not, so that i will post my answers.
 
Resuming our Quant marathon. This time, the questions are taken from a different source.

Answer the questions based on the following data:

A salesman enters the quantity sold and the price into the computer. Both the numbers are two-digit numbers. Once, by mistake, both the numbers were entered with their digits interchanged. The total sales value remained the same, i.e. Rs. 1148, but the inventory reduced by 54.

1. What is the actual price per piece?
1. 82
2. 41
3. 56
4. 28

2. What is the actual quantity sold?
1. 28
2. 14
3. 82
4. 41


Answer the questions based on the following data:

A thief, after committing a burglary, started fleeing at 12:00 noon, at a speed of 60 kmph. He was then chased by a policeman X. X started the chase 15 minutes after the thief had started, at a speed of 65 kmph.

3. At what time did X catch the thief?
1. 3:30 p.m.
2. 3:00 p.m.
3. 3:15 p.m.
4. None of these

4. If another policeman had started the same chase along with X, but at a speed of 60 kmph, then how far behind was he when X caught the thief?
1. 18.75 km
2. 15 km
3. 21 km
4. 37.5 km

5. The cost of a diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs. 70,000 less. Find the original price of the diamond.
1. Rs. 1.4 lakh
2. Rs. 2.0 lakh
3. Rs. 1.0 lakh
4. Rs. 2.1 lakh

6. A cube of side 12 cm is painted red on all the faces and then cut into smaller cubes, each of side 3 cm. What is the total number of smaller cubes having none of their faces painted?
1. 16
2. 8
3. 12
4. 24

7. The points of intersection of three lines, 2X + 3Y - 5 = 0, 5X - 7Y + 2 = 0, and 9X - 5Y - 4 = 0:
1. form a triangle.
2. are on lines perpendicular to each other.
3. are on lines parallel to each other.
4. are coincident.

8. If n is any odd number greater than 1, then n (n² - 1) is
1. divisible by 48 always
2. divisible by 24 always
3. divisible by 60 always
4. None of these
 
1)41
2)28
The inventory was reduced by 54 so the difference between the interchanged no and original no is 54.So 41 and 14 are eliminated.The remaining nos are 82 and 28.Since the inventory was reduced by 54 the orginal no is smaller no among the 2 so its 28i.e 82-28=54
Since the no of items is 28 the price per item is 1148/28=41

3)3:15pm
the thief had a headstart of 15 mins so at the speed of 60 kmph he covers 15 kms.Now the cop X start his chase so the time is 12:15.He is running at 65kmph so the relative speed between him and the thief is 65-60=5 kmph.So at 5 kmph it takes him 3 more hrs to cover the 15kms.

4)15
Since the other cop runs at 60 kmph ans the thief has already a headstart of 15 kms at 60 kmph... both will cover the same distance in the next 3hrs...thus the distance of 15 kms is still maintained between them
 
Last edited by a moderator:
neeraj_uchila said:
1)41
2)28
The inventory was reduced by 54 so the difference between the interchanged no and original no is 54.So 41 and 14 are eliminated.The remaining nos are 82 and 28.Since the inventory was reduced by 54 the orginal no is smaller no among the 2 so its 28i.e 82-28=54
Since the no of items is 28 the price per item is 1148/28=41

3)3:15pm
the thief had a headstart of 15 mins so at the speed of 60 kmph he covers 15 kms.Now the cop X start his chase so the time is 12:15.He is running at 65kmph so the relative speed between him and the thief is 65-60=5 kmph.So at 5 kmph it takes him 3 more hrs to cover the 15kms.

4)15
Since the other cop runs at 60 kmph ans the thief has already a headstart of 15 kms at 60 kmph... both will cover the same distance in the next 3hrs...thus the distance of 15 kms is still maintained between them
Click to expand...
Well done neeraj, all ur answers so far are correct. Post the rest of the answers too and if u can answer all of them correctly u can win 10 points. :thumb: I shall post the solutions tomorrow.


p.s. Post all ur answers in 1 post only.
 
5) the answer is Rs. 1.0 lakhs
ok let the original diamond be of weight 10(1+2+3+4)
then the square of weight is 100 and price say X*100 , where X is a constant
now for the fractions we have price accordingly as 16+9+4+1=30
there fore price is X*30
the difference is X*70 and that is equal to 70000
therefore the price of the diamond is 100000

6) cant really explain in words but i suppose the answer mut be 9

7) comparing the coefficients of the variables the options 2,3,4 are ruled out...

8)putting value of n as 3,5,7,9,11,13 etc wecan say teh answert is option 2 i.e. 24
 
Set 1 :
1. Option 1 is possible if the man puts one letter at a time and immediately the secretary types it out.
Option 2 is possible if the man puts all the letters in order at the same time and then the secretary starts typing
Option 4 is possible if the secretary starts typing after the man has put 4 letters in the tray and before she can finish typing the fourth letter he puts in letter #5 and #6
Only option 3 is not possible
Ans : option 3

2. when A and B meet for the first time, let x be the distance traveled by B and speed of B be SB and that of A be SA
therefore x/SB = 700/SA – (i)
(x+300)/SA = 400/SB – (ii)
Solving the two equations, we get x = 400
Hence circumference of circle = 1100
Ans : option 4

3. If a cube of length 5 is cut into smaller cubes of length 1, then there are 125 smaller cubes of which there are 3 cubes on each edge with 2 sides painted. Therefore there are a total of 12*3=36 such cubes
hence the probability is 36/125
Ans : option 1

4. the number of lines that can be drawn by joining any 2 of the available 8 points is 8C2. Out of these 8 lines form the edges of the octagon. Hence there are 8C2–8 = 20 diagonals
Ans : option 2

5. Using the below formula
First figure out the values for a, y, and m -- variables to be plugged into a formula.
a = (14 - month)/12 (month = # of month, 1 for Jan, 2 for Feb, etc)
y = year - a (year = the 4 digit year)
m = month + 12a - 2
Next, plug the values of y and m into the following formula to calculate the day:
d = (day + y + y/4 - y/100 + y/400 + 31m/12) mod 7
We get the value of d as 0 (Sunday)
Ans : option 4


NOTE : Since nobody answered this set before, do i get my points?? :big_grin:
 
Set 2
1-2. Since its given that the inventory value reduces by 54, we can find that the diff between the digits is 6. Therefore options 2,4 for q2 are eliminated.
Since the inventory reduced by 54, the original value must be smaller of the two ie 28.
If qty=28 then it can be found that price=41
1. Ans : option 2
2. Ans : option 1
3. Thief traveled a distance of 15 km before the X started. If X catches thief at a dist of x km then,
(x+15)/65 = x/60 implies x=180
therefore X catches thief after 3 hours of starting
Ans : option 2

4. Since second policeman and thief run at same speeds, they both will cover the same distance in 3 hours and hence the second policeman will be lagging back by a dist of 5 km.
Ans : option 2

5. let w be the weight and p be the price. Therefore p = kw^2
let the weight of the 4 pieces be w,2w,3w,4w therefore the total weight = 10w
therefore, k(10w)^2 - kw^2(1+4+9+16) = 70000
kw^2 = 1000 hence actual price = 10*1000 = 100000
Ans : option 3

6. In a cube of length 12 cm cut into 3 cm cubes, there are a total of 8 cubes that are not colored
Ans : option 2

7. it can be seen that for (x,y) = (1,1) satisfy all the 3 equations. Hence they are co-incident at that point.
Ans : option 4

8. Taking few random odd values, it can be seen that n(n^2 – 1) is always divisible by 24
Ans : option 2
 
hi gaurav
i have found out the answer
tell me whether they r correct or not
Q1)is 1 i.e.1, 2, 3, 4, 5, 6
Q2)is 3 i.e. 1000 m
Q3)is 1 i.e. 36/125
Q4)is 2 i.e. 20
Q5)is 3 i.e. friday
 
Rahul, now u got what u wanted.... more and more are coming forward to participate. So i expect u to handle the sections and keep posting the questions for all sections.
 
general formula to find no. of diagonals in a n-sided polygon: nC2 - n
[working : nC2 gives no. of lines joining any 2 points of the polygon but we need to subtract n from it corresponding to the n sides of the polygon ]
 
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