I am unsure if it is inverse or direct variation. Please correct me if I am wrong
" A shopkeepr sold certain number of toys. The no. of toys as well as the price of each toy (in Rs) was two digit number. By mistake he reversd the digits of both the no. of toys he sold and the price of each toy.As a result he found that his stock account at the end of the day showed 81 items more than it actually was. "
Find the actual number of toys sold..
1) 92
2)81
3) 90
4) 29
5) cannot be determined.
can someone help me in solving this que...
pls explain the approach tooo...Many thanks!!!
" A shopkeepr sold certain number of toys. The no. of toys as well as the price of each toy (in Rs) was two digit number. By mistake he reversd the digits of both the no. of toys he sold and the price of each toy.As a result he found that his stock account at the end of the day showed 81 items more than it actually was. "
Find the actual number of toys sold..
1) 92
2)81
3) 90
4) 29
5) cannot be determined.
can someone help me in solving this que...
pls explain the approach tooo...Many thanks!!!
90 and 09 are the 2 numbers that differ by 81 when reversed...so number of toys sold must be 90, which he writes as 09 sold..therefore his inventory, which should be reduced by 90, is reduced only by 09...hence it reflects 81 more toys than are actually in stock 😁
r11gupta Says90 and 09 are the 2 numbers that differ by 81 when reversed...so number of toys sold must be 90, which he writes as 09 sold..therefore his inventory, which should be reduced by 90, is reduced only by 09...hence it reflects 81 more toys than are actually in stock :D
Hi r11 thanks..but was wondering is there any other way to solve this problem?
" A shopkeepr sold certain number of toys. The no. of toys as well as the price of each toy (in Rs) was two digit number. By mistake he reversd the digits of both the no. of toys he sold and the price of each toy.As a result he found that his stock account at the end of the day showed 81 items more than it actually was. "
Find the actual number of toys sold..
1) 92
2)81
3) 90
4) 29
5) cannot be determined.
can someone help me in solving this que...
pls explain the approach tooo...Many thanks!!!
It can be solved using 10x + y and reversing that ... but that approach will take a lot for you... the best way is to go thorough the option in this case and 90 then tends out to be the right answer.
X and Y are playing a game.There are eleven 50p coins on the table and each player must pick up at least one coin but not more than 5.The person picking up the last coin loses.X starts.How many shud he pick to ensure a win no matter what strategy Y employes?
Options:
4,3,2,5
The Question is of Pre-assessment of Block 1,Arun Sharma
Plz Help.The Explanation is insufficient...
X and Y are playing a game.There are eleven 50p coins on the table and each player must pick up at least one coin but not more than 5.The person picking up the last coin loses.X starts.How many shud he pick to ensure a win no matter what strategy Y employes?
Options:
4,3,2,5
The Question is of Pre-assessment of Block 1,Arun Sharma
Plz Help.The Explanation is insufficient...
see,the one who picks the last coin will lose.similarly,the one who picks a coin when there are 7 remaining is bound to lose.so,X should pick up 4 coins to ensure his victory.
hope that helps.
People,,
How effective do you think is doin the LOD3 of Quants by Arun Sharma for CAT?
see,the one who picks the last coin will lose.similarly,the one who picks a coin when there are 7 remaining is bound to lose.so,X should pick up 4 coins to ensure his victory.
hope that helps.
Yup,got it...
was easy,i m cursing myself,i didnt get...
Here's on question from LOD2 Time n Work...
According to a plan, a drilling team had to drill to a depth of 270m.
For first 3 days they drilled according to the plan.After that they started
drilling 8m more than the plan every day.Therefore, a day before the planned date they had drilled a depth of 280m. What was the actual planned drilling per day?
a)38m b)30m c)27m d)none of these 
Here's on question from LOD2 Time n Work...
According to a plan, a drilling team had to drill to a depth of 270m.
For first 3 days they drilled according to the plan.After that they started
drilling 8m more than the plan every day.Therefore, a day before the planned date they had drilled a depth of 280m. What was the actual planned drilling per day?
a)38m b)30m c)27m d)none of these
the answer is b)30 m
here's how it is done:
let rate of work=M
let no of days = k
M.K=270
as per qn: 3.M + ((k-1)-3).M + ((k-1)-3).8 =280
simplifying we get....8.k - M =42
k=(42+M)/8...
subs in M.(42 + M)/8 = 270..
checking from options..
M=30
Here's on question from LOD2 Time n Work...
According to a plan, a drilling team had to drill to a depth of 270m.
For first 3 days they drilled according to the plan.After that they started
drilling 8m more than the plan every day.Therefore, a day before the planned date they had drilled a depth of 280m. What was the actual planned drilling per day?
a)38m b)30m c)27m d)none of these
let they drill for x m everyday normally for k days.
3x+(x+
(k-4)=2803x+kx-4x+8k=312
but,kx=270
8k-x=42
x^2+42x-2160=0
(x+72)(x-30)=0
x=30 m per day
option (B)
Hi Guys,
Help me in this problem...
Number System LOD 3 Ques 54) What is the remainder when 50^51^52 is divided by 11.
Thanks & Regards
Vidhyasagar
Hi Guys,
Help me in this problem...
Number System LOD 3 Ques 54) What is the remainder when 50^51^52 is divided by 11.
Thanks & Regards
Vidhyasagar
Euler's theorem can be used, as 50 and 11 are co prime.
Need to express 51 as 10x + k.
51 -> last digit always 1 -> always of the form 10k + 1
so we only need remainder of 50 from 11 which is 6
PS : Already posted solution to this question on another thread in reply to this user; posting here again for benefit of junta
Euler's theorem can be used, as 50 and 11 are co prime.
Need to express 51 as 10x + k.
51 -> last digit always 1 -> always of the form 10k + 1
Hi r11gupta,
A doubt may be a silly one...
what happened to the 50^51^52...
Do you mean irrespective of the value present in the highlighted part, the answer will remain the same....
In this case it may me true as the equation will reduce into
*
-> *
->6
can you explain this case 32^33^34
if possible can someone solve these problems by remainder theorem...
Cheers
Vidhyasagar
Hi Guys,
Help me in this problem...
Number System LOD 3 Ques 54) What is the remainder when 50^51^52 is divided by 11.
Thanks & Regards
Vidhyasagar
see,50 and 11 are coprime.also,Euler number of 11 is 10.
so,we have to express 51^52 as 10k+something
51^52 is (50+1)^52
(10k+1)
so,50^(10k+1) mod 11
50 mod 11
remainder is 6.
yes irrespective of the value,the answer remains the same.this is because of the 51 part.
now,
32^33^34
here,
it is same as 2^33^34
(30+3)^34
3^34
3^34 mod 10
comes out to be 9.
2^(10p+9) mod 11
2^9 mod 11
512 mod 11
6 is the remainder.
Hi r11gupta,
A doubt may be a silly one...
what happened to the 50^51^52...
Do you mean irrespective of the value present in the highlighted part, the answer will remain the same....
In this case it may me true as the equation will reduce into
*
-> *
->6
can you explain this case 32^33^34
if possible can someone solve these problems by remainder theorem...
Cheers
Vidhyasagar
Kindly refer to the perfect explanation by shashank in the post above :
Hi Guys,
Help me in this problem...
Number System LOD 3 Ques 54) What is the remainder when 50^51^52 is divided by 11.
Thanks & Regards
Vidhyasagar
digits ending with 1 have last 2 digits......1 at unit place always and multiplication of last digit of power &tens; digit of the base
51^52=2*5=10....so last 2 digits=01
50^1/11=6 remainder
Q)Find the number of divisors of 1080 excluding the throughout divisors,which are perfect squares..
Options...: 28/29/30/31..
I am getting answer as 29..While the book gives 28...
Help...
see,50 and 11 are coprime.also,Euler number of 11 is 10.
so,we have to express 51^52 as 10k+something
51^52 is (50+1)^52
(10k+1)
so,50^(10k+1) mod 11
50 mod 11
remainder is 6.
yes irrespective of the value,the answer remains the same.this is because of the 51 part.
now,
32^33^34
here,
it is same as 2^33^34
(30+3)^34
3^34
3^34 mod 10
comes out to be 9.
2^(10p+9) mod 11
2^9 mod 11
512 mod 11
6 is the remainder.
hey for 32^33^34 mod 11...correct me if im doing something wrong..
here we have to express 33^34 in power of 10(euler for 11 as 32 and 11 are coprime)
Thus,33^34 mod 10=9
so 32^33^34= 32^9 mod 11=10