hi this is Ani,,,,,can anyone help me out with the topics related to clocks......
@vivaciousgal said:Ques 1: Find the number of terms of the4 series 1/81, -1/27, 1/9............-729?
(a) 11 (b) 12 (c) 10 (d) 13 (e) 14Ques 2: A and B are two numbers whose AM is 25 and GM is 7. Which of the following may be a value of A?(a) 10 (b) 20 (c) 49 (d) 25 (e) 24Ques 3: Two numbers A and B are such that their GM is 20% lower than their AM. Find the ratio between the numbers.??(a) 3:2 (b) 4:1 (c) 2:1 (d) 3:1 (e) 5:1Ques 4: Find the sum of all numbers in between 10-50 excluding all those numbers which are divisible by 8.(a) 1070 (b) 1220 (c) 1320 (d) 1160 (e) 1060Ques 5: The 7th and 21st terms of an AP are 6 and -22 respectively. Find the 26th term.(a) -34 (b) -32 (c) -12 (d) -10 (e) -16
1. The ratio is -3. Now 1/81 has to be multiplied by (-3) 4 times to get to 1 and another 6 times to get to 729. Hence ten terms after the first so total 11 terms. So (a)
2. AM [i.e. (a+b)/2] is 25 so sum is 50. GM [i.e.sqrt(ab)] is 7 so product is 49. Looking at this, the numbers should be 1 and 49 (we can solve it mathematically but it is longer and quite painful). Hence (c)
3. Use answer choices. 3 and 2: AM = 2.5, GM = sqrt(6). 4 and 1: AM = 2.5, GM = 2 which satisfies the condition as 2 is 25% less than 2.5. (Can check others and show that they don't satisfy) Hence (b)
4. um of numbers from 10 - 50 = 41 numbers, average 30 which sum to 41 x 30 = 1230. Numbers divisible by 8 = 16, 24, 32, 40, 48 i.e. 5 numbers with average 32 which sum to 32 x 5 = 160. Hence the required sum is 1230 - 160 = 1070. Hence (a)
5. 7th to 21st = 14 times common difference = -28. Hence diff = -2. Now 26th term is 5 steps of (-2) further which is to say -22 +5 (-2) = -32. Hence (b)
~scrabbler
1 Like
@erm said:1 what is unit of 7^11^22^33?? 2 what is the sum of digits of the least multiple of 13 , which when diveded by 6, 8 & 12 leaves a remainder of 5,7,11 as the remainder???? PLZ elaborate your method of solving
Answers are 7 and 8? {First one is 7^(4k+1) and second number is 143 - has to be a multiple of 13 as well as of (24k - 1) form, using trial and error it is 143}
~scrabbler
~scrabbler
A shopkeeper always weighs 20% less than the correct weight. One day, he weighed 20% more than the quantity usually weighed by him. If the profit on the correct weight is 20%, what is the effective percentage of profit in this transaction?
@[510729:erm] 1). unit digit of 7^11^22^33 = 7
for finding last digit when 7 is divided by 10 we get remainder 7/10=7 : 7^2/10= 9 ; 7^3 /10 =3 ; 7^4/10 =1 ; 7^5/10= 7...so on.........this follows a cyclic pattern of 4 (i.e.after every 4th power of 7 ,digits repeat) there we should whether 11 ^22^23 is divisible by 4 or gives a remainder ..
....when we divide 11 by 4 we find cyclicity of 2 i.e. 11/4= 3 ; 11^2/4= 1....then again repetition 11^3 /4=3 so on.
.........now we have to check for 22^23 whether it is divided by 2 or not.....as 22 is even no. therefore any power of 22 is divisible by 2 therefore it gives remainder digit of 11^ 22^23 = 1
therefore 7^ 11^22^23 equates to 7^1 therefore last digit is 7^1 = 7........
1 Like
@[510729:erm] 2). lowest multiple of 13 divided by 6,8,12 to give remainder 5,7,11 respectively is 143.
@[510729:erm] 2). for this question take the desired factor of 13 as 13m where m is a natural no.
if u notice when we divide 13m by 6 it gives remainder 5 .......where 6 -5 =1
similarly 13m/8 gives remainder 7 where 8 -7 =1
similarly 13m/12 => remainder 11 where 12-11 =1
all differences between divisor and remainder =1 =q
the general form of such numbers equal to LCM( All divisors) - difference....... therefore according to relation the general form should be 24k- 1
this general form 24k-1 is equal to 13m ........simplifying k = (13m +1)/24 ....here the lowest value of 13m comes out getting the lowest value of 13m +1 divisible by 24 by........substituting different values of m =0,1,2,3,4,5,6.................. we get 13m+1 =144 .which is divisible by 24 ..therefore 143 is answer.
is x is odd???
I 3x is odd.
II 15x is odd.
a) statement I is sufficent to answer & not II
b)statement II is sufficent to answer & not I
c) both
d) none of these
@j.rajat2000 said:A shopkeeper always weighs 20% less than the correct weight. One day, he weighed 20% more than the quantity usually weighed by him. If the profit on the correct weight is 20%, what is the effective percentage of profit in this transaction?
let the correct weight be 100 units..he always weighs 80 units..but one day he weighed 96 units (20% more)..
profit - on correct weight 20..
new profit = 24/96*100 = 25%
@[395917:abhibansal123]
from where u r getting 24?
@j.rajat2000 said: @abhibansal123from where u r getting 24?
let the cost of 100 units be Rs.100
So, he sells it for 120 (20%profit)
But he sells 96 units instead of 100 units..
so cost of 94 units is Rs.94
Profit = SP (120) - CP(94) = 24
Constantly getting percentiles in the 50s in the quant section in the proc mocks n AIMCATs... Very tensed with just two months to go...! Help needed...:-(
How many 4-digits numbers can be formed using the digits 5,7,3,2 exactle once, such that the number formed is divisible by 11?
@[571137:jashholmes] is it 0???
@[103935:chandrakant.k] no
Let abcd be a four digit number. For it to be divisible by 11 either (b+d) - (a+c) = 0 or the difference between them must be a multiple of 11. In this case i don't think so any of them will hold true. So we cannot have a number using these digits which is a multiple of 11.
IMO
@[571137:jashholmes] See @[557270:HydRocker] explaination.. It has to be 0..
is ten digit number pqrstu8672 divisible by 32?????????
how to check ??? the number is divisible by 8 and 4 which are factors of 32
last 5 digits should be divisible by 32 .
but the the number is divisible by 4 and 8