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A retailer keeps Reebok shoes, which are listed at a mark up of 10% above their factory price. However on Monday he realized the shortage of demand and hence decreased the list price by 10%. On the very next day, that is, Tuesday he realized that he is making a loss so he increased yhe list price...

A retailer keeps Reebok shoes, which are listed at a mark up of 10% above their factory price. However on Monday he realized the shortage of demand and hence decreased the list price by 10%. On the very next day, that is, Tuesday he realized that he is making a loss so he increased yhe list price by 10% again. He continues this trend indefinitely. On which day will he suffer a loss of more than 10% per shoe for the first time?

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Three hundred workers are set to build a dam.In 21 weeks they have done 60% of work, but subsequent rains lasting 4 weeks washed away 25% of what they had done. The job is resumed by 200 workers. What is the additional time it will take to complete the work?
plz add detailed solution.

plz add detailed solution.

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Amity Business School, Lucknow Invites Applications for MBA 2014

Ranked 5th in India (without IIMs) and 11th in India (including IIMs) by Times B.School survey published in The Times of India on 28th Feb. '13

Applications Open!

Percentages-Foundation sum 6.

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TSD NISHIT SINHA FOUNDATION - 14

- 6Comments
- yes hw ??. 19 Feb.
- a?. 19 Feb.

TSD NISHIT SINHA FOUNDATION - 12
ship vikrant starts from point P towards a point Q at noon and at 1:00 pm ship viraat strts from Q towards P. If vikranth is expected to complete vyage in 6 hrs and viraat is movng at 2/3rd the speed of vikranth. than at what time do these meet /??
4pm...

TSD NISHIT SINHA FOUNDATION - 12

4pm

4.30 pm

3 pm

2.30 pm ?

- 6Comments
- yes. how ? i m getting 2 pm .... 18 Feb.
- 4 ?. 18 Feb.

Raghav
@rgvcat_2013
1.3
k

let d be distance btw PQ

speed of vikrant = d/6

viraat = d/9

in 1 hour vikrant moves = d/6

distance left = 5d/6

time at which they meet after 1 hour = (5d/6)/(d/6 + d/9) = 3

hence total = 1+3 = 4

speed of vikrant = d/6

viraat = d/9

in 1 hour vikrant moves = d/6

distance left = 5d/6

time at which they meet after 1 hour = (5d/6)/(d/6 + d/9) = 3

hence total = 1+3 = 4

> @ashishyog
>
> The sum of the lengths of the hypotenuse and one of the perpendicular sides of a right angled triangle is L. When the area of this triangle is maximum, the angle between the two sides is
>
>
>
> Please provide the approach as well
>
>
>
> 1) 45 degrees
>
> ...

The sum of the lengths of the hypotenuse and one of the perpendicular sides of a right angled triangle is L. When the area of this triangle is maximum, the angle between the two sides is

Please provide the approach as well

1) 45 degrees

2) 22.5 degrees

3) 60 degrees

4) None of these Skip

—

let the three sides of the triangle be P,B and H.

p^2+B^2=H^2........eq1

A/Q P+H=L

so, H=L-P

substitute H in eq1

p=(L^2-B^2)/2L

area= 1/2*B*P

for Area to be max derivative will be 0.

which gives L^2=3*(B^2)

so, B = L/sqrt(3)

p=L/3

sin X=1/2

X=30 deg

hence ANS 4 None of these

The sum of the lengths of the hypotenuse and one of the perpendicular sides of a right angled triangle is L. When the area of this triangle is maximum, the angle between the two sides is
Please provide the approach as well

The sum of the lengths of the hypotenuse and one of the perpendicular sides of a right angled triangle is L. When the area of this triangle is maximum, the angle between the two sides is

Please provide the approach as well

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Two Rabbits A and B are running a race in which they have to go upto 50 m mark and then come back to starting point. A runs in a sequence of 3 steps in which distance traveled in first jump is twice that of other two and B runs in a sequence of 3 jumps where first jump is 1.5 time the distance tr...

Two Rabbits A and B are running a race in which they have to go upto 50 m mark and then come back to starting point. A runs in a sequence of 3 steps in which distance traveled in first jump is twice that of other two and B runs in a sequence of 3 jumps where first jump is 1.5 time the distance travelled in second jump and distance traveled in 2nd jump is twice the distance traveled in 3rd jump. Distance traveled by A in 2 jumps and by B in 3 jumps is equal to 6m. Who will win the race.

Please explain the approach!

Please explain the approach!

- 5Comments
- @Oteri No options are there, i just have a list of few qu.... 02 Sep '13.
- Is it B?. 02 Sep '13.

binayak mahapatra
@binayak_89
8

@santosh.bohra n @ak.abhi03 plz explain

Ashish Yog
@ashishyog
13

@Oteri No options are there, i just have a list of few questions from Nishit Sinha...no options are given for this question.

What is the remainder when (1^1+2^2+3^3+4^4+.....+100^100) is divided by 4? Please share the approach...
a) 0
b) 1
c) 2
d) 3
:angry: :angry: :angry:

What is the remainder when (1^1+2^2+3^3+4^4+.....+100^100) is divided by 4? Please share the approach...

a) 0

a) 0

b) 1

c) 2

d) 3

- 6Comments
- @kingleon Please explain the approch dude....!!. 02 Sep '13.
- 0. 31 Aug '13.

swapnil more
@kingleon
141

all even nos and powers will be divisible by 4....like 2^2 4^4 and so on....bache all odd nos and theirs powers like 1^1 3^3 5^5 will give remainders in form of +1 -1 +1 -1 and so on....since we have even no of nos all get cancelled leaving us with remainder 0

D Som
@D.Som
71

Just break it down in 4 consecutive terms in a block. So the nth block is (4n+1)^(4n+1)+(4n+2)^(4n+2)+(4n+3)^(4n+3)+(4n+4)^(4n+4). So the the sequence given is the sum of 25 blocks from n=0 to n=24. Now (4n+1)^(4n+1)= sum all multiples of 4 +1^(4n+1) {By binomial theorem}. The remainder when divided by 4 is 1. The second term (4n+2)^(4n+2) is clearly divisible by 4 {take 2 common out from and the term becomes a multiple of 4} and hence remainder is 0. The same happens for the last term (4n+4)^(4n+4). The third term we can write as (4n+4-1)^(4n+3)=(4m-1)^(4n+3) [m=n+1]. Now again (4m-1)^(4n+3)= sum of multiples of 4 +(-1)^(4n+3). When divided by 4 the remainder is (-1)^(4n+3)=(-1)^3=-1. So the sum of the remainder from four terms is 1+0+(-1)+0=0 and when summed fro all 25 blocks the final remainder is also 0.

Ashish Yog
@ashishyog
13

@kingleon Please explain the approch dude....!!

> @Oteri
>
> the average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and fro journey is 17 hours, covering a distance of 800km.speed of the train in onward j...

the average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and fro journey is 17 hours, covering a distance of 800km.speed of the train in onward journey is:

A 45kmph

B 47.5kmph

C 52kmph

D 56.25kmph

45kmph

Hi,
Could anyone share some insight on the approach to solve this problem
Q> Which of the following is smallest?
a) 5^(1/2) b)6^(1/3) c)8^(1/4) d)12^(1/5)
:embarrased:

Hi,

Could anyone share some insight on the approach to solve this problem

Q> Which of the following is smallest?

a) 5^(1/2) b)6^(1/3) c)8^(1/4) d)12^(1/5)

Q> Which of the following is smallest?

a) 5^(1/2) b)6^(1/3) c)8^(1/4) d)12^(1/5)

- 12^1/5. 04 Mar.
- LCM(2,3,4,5)= 60 5^(1/2)= (5^30)^(1/60) 6^(1/3)= (.... 28 Aug '13.

LCM(2,3,4,5)= 60

5^(1/2)= (5^30)^(1/60)

6^(1/3)= (6^20)^(1/60)

8^(1/4)= (8^15)^(1/60)

12^(1/5)=(12^12)^(1/60)

Now since the power over all the numbers is (1/60), the ans will be smallest of all the numbers, i.e (12^12) in this case.

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the average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and fro journey is 17 hours, covering a distance of 800km.speed of the train in onward journey is:
A ...

the average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and fro journey is 17 hours, covering a distance of 800km.speed of the train in onward journey is:

A 45kmph

B 47.5kmph

C 52kmph

D 56.25kmph

- D 56.25kmph. 28 Aug '13.

i am getting stuck with this particular type of question : ( 1!^1! + 2!^2! + 3!^3! ....... 1000!^1000! )/ (2^3 x 5). what will be the remainder ?? this is question no. 34 of number systems (page 1.13)
:rolleyes:

i am getting stuck with this particular type of question : ( 1!^1! + 2!^2! + 3!^3! ....... 1000!^1000! )/ (2^3 x 5). what will be the remainder ?? this is question no. 34 of number systems (page 1.13)

- 13Comments
- 17?. 02 Aug '13.
- 5???. 02 Aug '13.

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aspire 2shine
@aspire2shine
95

2^3*5=40..all the terms from 5!^5! will have (2^3 and 5) as its factors..So all of them will be divisible by (2^3*5).So the remaining expression is (1!^1!+2!^2!+3!^3!+4!^4!)/(40)...which should give 17 as the remainder..according to me

*7\. **Remainder when 22^**33+10^**35 divided by 45? **a. **2*
*b. **8*
*c. **11*
*d. **None*

**7. ****Remainder when 22^****33+10^****35 divided by 45? ****a. ****2**

**b. ****8**

**c. ****11**

**d. ****None**

- 15Comments
- 32. 28 Aug '13.
- me gettin 32. 04 Aug '13.

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- Showing 11-15 of 15
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AYUSH AGARWAL
@ayush1511
16

@chillfactor 5 gives remainder 2 and 9 gives remainder 2 then waht should i do ..

Maverickster
@AshishDaryani
99

What's the correct answer.. 2 or 32..

6Commentsso 1 hour = 7x2x60/13x3 =280/13 metric ton

let it fill 70 in x hrs so

(280/13-15)*x=70

x=14*13/17

so speed = 156/x

x=14.42 = 14.5

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