The circumference of the front wheel of cart is 40 ft & tht of back wheel is 48 ft long.What is distance travelled by cart,when the front wheel has done 5 more revolutions than the rear wheel ?
nitya2903 SaysThe circumference of the front wheel of cart is 40 ft & tht of back wheel is 48 ft long.What is distance travelled by cart,when the front wheel has done 5 more revolutions than the rear wheel ?
i think we shld consider the displacement of center of the two wheels. it would be same.
in short, 40(x+5) = 48x
x = 25 revolutions
distance travelled = 48*25 = 1200 ft
:)
nitya2903 SaysThe circumference of the front wheel of cart is 40 ft & tht of back wheel is 48 ft long.What is distance travelled by cart,when the front wheel has done 5 more revolutions than the rear wheel ?
let n1 be rev for front wheel and n2 be the rev for back wheel.
Also, distance traveled by both wheels should be equal.
Then, n1*C1 = n2*C2
also n2 = n1-5
substituting we get,
n1C1 = n1C2 - 5*C2
=> 5 * C2 = n1 * (C2 - C1)
=> 5 * 48 = n1 * 8
=> n1 = 30
Distance travelled = n1 * C1 = 30*40 = 1200 ft.
please solve these and give solutions in detail......
from chpt-time and work
1.In a fort there was sufficient food for 200 students for 31 days.after 27 days 120 soldiers left the fort.for how many extra days will the rest of the food last for the remaining soldiers?
2.An engineer undertakes a project to build a road 15km long in 300 days and employs 45 men for the purpose.after 100 days,he finds only 2.5 km of the road has been completed.find the number of extra men he must employ to finish the work in time.
please solve these and give solutions in detail......
from chpt-time and work
1.In a fort there was sufficient food for 200 students for 31 days.after 27 days 120 soldiers left the fort.for how many extra days will the rest of the food last for the remaining soldiers?
800 units of food left
800/80 = 10 days ??
2.An engineer undertakes a project to build a road 15km long in 300 days and employs 45 men for the purpose.after 100 days,he finds only 2.5 km of the road has been completed.find the number of extra men he must employ to finish the work in time.
90000 units work
work completed -- 90000/6 = 15000 units
daily 150 units of work is being done by 45 workers
1 worker -- 150/45 = 10/3 units daily
75000 units in 200 days
375 units daily work
workers neeeded = 375*3/10 = 112.5
113 round off
68 more workers ??
the number of days required by a,b,c to work individually is 6,12,8 respectively.they started a work doing it alternatively .if a has started then followed by b and so on ,how many days are needed to complete the whole work ?
niksnayyar Saysthe number of days required by a,b,c to work individually is 6,12,8 respectively.they started a work doing it alternatively .if a has started then followed by b and so on ,how many days are needed to complete the whole work ?
let total amount of work be 24 units
then A will do-4 units of work in 1 day...
and similarly B-2units C-3units.......
now it is following a cycle
ABC ABC ABC....
1 CYCLE OF ABC does 9 units of work in 3 days...
so 2 cycles will be reqd...in that 18 units of work will be done....
then A will do 4 units of work....so total 22units completed...
then B will do 2 units hence total 24 will be completed...
hence answer is 8 days....
i hope it is clear.......
A number when divided by a divisor D leaves remainder of 13.When thrice the number is divided by D, the remainder obtained is 2. Find the number of possibilities for D.
1) 0
2) 1
3) 2
4) 3
My take is option 2.
what is the correct ans?
A number when divided by a divisor D leaves remainder of 13.When thrice the number is divided by D, the remainder obtained is 2. Find the number of possibilities for D.
1) 0
2) 1
3) 2
4) 3
N = Da + 13
3N = Db + 2
=> 3(Da + 13) = Db + 2
Or, D(b - 3a) = 37 = 37*1
=> D=37
Hence, option 2)
My take is option 2.
what is the correct ans?
correct..
Block 1 - Number system
Q1-For the no. 7200 find..
A-The sum and no. of factors divisible by 150
B-The sum and no. of factors divisible by 15
C-The sum and no. of factors not divisible by 75
D-The sum and no. of factors not divisible by 24
Q2-Find the No. of divisors 544 which are greater than 3?
Q3-Find the No. of divisors 544 excluding 1 and 544?
Q4-Find the No. of divisors 544 which are perfect squares?
Q5-Find the number of zeroes in
100^1* 99^2* 98^3 * 97^4.....................* 1^100
Q1-For the no. 7200 find..
A-The sum and no. of factors divisible by 150
B-The sum and no. of factors divisible by 15
C-The sum and no. of factors not divisible by 75
D-The sum and no. of factors not divisible by 24
Q2-Find the No. of divisors 544 which are greater than 3?
Q3-Find the No. of divisors 544 excluding 1 and 544?
Q4-Find the No. of divisors 544 which are perfect squares?
Q5-Find the number of zeroes in
100^1* 99^2* 98^3 * 97^4.....................* 1^100
1080 means 2^3 * 3^3 * 5
therefore the number of divisors of 1080 is (3+1)*(3+1)*(1+1)= 32 ;
Now , from 2^3 * 3^3 * 5 , we get the perfect squares as 2^2 , 3^2 , and (2*3)^2 , these are possible perfect squares . so excluding them all , we will get 32-3 = 29 divisors . So 29 is the correct answer .. :clap:
Answer to this is 28 as given in arun sharma....pg23 Block 1
because if you exclude perfect squares the we are left with 2^1*3^1*5^1=30 minus 2( for 2^3 and 3^3)
=28...
Correct me if i am wrong..
Two people A and B start from P and Q(distance= D) at the same time towards each other. They meet at a point R, which is at a distance of 0.4D from P. They continue to move two and fro between the two points. Find the distance from point P at which the fourth meeting takes place.
a) 0.8D
b) 0.6D
c) 0.3D
d) 0.4D
7200 = 2^5 * 3^2 * 5^2
1.
no. of factors = (5+1)(2+1)(2+1) = 54 -------------------------(f1)
sum of factors = (2^6 - 1)(3^3 - 1)(5^3 - 1)/(2-1)(3-1)(5-1)-----(s1)
3.
no. of odd factors = factors of 3^2*5^2 = (2+1)(2+1) = 9 -----------(f3)
sum of factors ... use the same process -----------------------(s3)
2.
no. of even factors = (f1) - (f3)
sum of even factors = (s1)-(s3)
4.
7200 = 25 * 2^5*3^2 ... 25 should be in each factor ...
no. of factors div by 25 = no of factors of 2^5 * 3^2
work on similar lines for all the rest ...
For the number 7200 find.
1 The sum and number of all factors
2. The sum and number of even factors
3. The sum and number of odd factors
4. The sum and number of factors divisible by 25
5. The sum and number of factors divisible by 40
6. The sum and number of factors divisible by 150
7. The sum and number of factors not divisible by 75
8. The sum and number of factors not divisible by 24
So 6 answer should be
150= 15*10= 5*3*5*2 = 5^2*3*2
7200= 2^5 * 3^2 * 5^2
Sum= (2^1+2^2+2^3+2^4+2^5)(3^1+3^2)
No.= (2^4 * 3^1)
Kindly let me know if this is right or not?
Can someone plzz send me link of Arun Sharma's book for quant,verbal and di....If some better book is there plzz suggest me.
Thanx in advance
sorry for double post
1- Find the number of divisors 544 which are greater than 3.
2- Find the number of divisors 544 excluding 1 and 544.
3- Find the number of divisors 544 which are perfect squares.
4- Find the number of zeros in 100^1*99^2*98^3*97^4*...................*1^100.
5- Find the max. value of n such that 477x42x37x57x30x90x70x2400x2402x243x343 is perfectly divisible by 21^n.
6- 100! + 200! =
7- 57*60*30*15625*4096*625*875*975=
8- 1!*2!*3!*4!*..................50! =
9- Find the remainder when 51^203 is divided by 7?
10- Find the remainder when 67^99 is divided by 7?
11- Find the remainder when 75^80 is divided by 7?
12- Find the remainder when 54^124 is divided by 17?
13- Find the remainder when 25^102 is divided by 17?
14- Find the units digit in each of the following case-
1^1*2^2*3^3........................*100^100
173^45x152^77x777^999
82^43*83^44*84^97*86^98*87^105*88^94
15- Find number of numbers between 300-400 both included that are not divisible by 2,3,4 and 5.
2- Find the number of divisors 544 excluding 1 and 544.
3- Find the number of divisors 544 which are perfect squares.
4- Find the number of zeros in 100^1*99^2*98^3*97^4*...................*1^100.
5- Find the max. value of n such that 477x42x37x57x30x90x70x2400x2402x243x343 is perfectly divisible by 21^n.
6- 100! + 200! =
7- 57*60*30*15625*4096*625*875*975=
8- 1!*2!*3!*4!*..................50! =
9- Find the remainder when 51^203 is divided by 7?
10- Find the remainder when 67^99 is divided by 7?
11- Find the remainder when 75^80 is divided by 7?
12- Find the remainder when 54^124 is divided by 17?
13- Find the remainder when 25^102 is divided by 17?
14- Find the units digit in each of the following case-
1^1*2^2*3^3........................*100^100
173^45x152^77x777^999
82^43*83^44*84^97*86^98*87^105*88^94
15- Find number of numbers between 300-400 both included that are not divisible by 2,3,4 and 5.
Please help me with the solutions for these questions on Progressions.
Thank you :)
1)The sum of 5 nos in AP is 30 and the sum of the squares is 220. Which is the 3rd term?
Ans:9
2) 4 Geometric means are inserted between 1/8 and 128 . Find the third Geometric mean.
Ans: 8
3) Two nos A and B are such that their GM is 20% lower than their AM.
Find the ratio betn nos.
Ans: 4:1
4) A no is divided into 4 parts that are in AP such that the ratio of product of 1st and 4th to the product of 2nd and 3rd is 2:3.
Find the largest part.
Ans: 8
5)The sum of an infinite GP whose common ratio is less than 1 is 32 and the sum of the first two terms is 24. What will be the 3rd term?
Ans: 4
6) What will be the value of x^(1/2) * x^(1/4) * x^(1/... to infinity.
Ans:x
7) Determine the first term of GP, the sum of whose 1st term and 3rd term is 40 and the sum of 2nd and 4th term is 80
ans: 8
Ans: -1
9) The sum of the first three terms of AP is 30 and the sum of the squares of the 1st and 2nd term is 116. Find the 7th term of the progression if its 5th term is known to be exactly divisible by 14.
ANs: 40
10) A is the sum of n terms of series 1+ 1/4 + 1/16 + ...
B is the sum of 2n terms of series 1+ 1/2 + 1/4 + ....
Find A/B.
Ans: 2/3
11) Find the sum of series: 1/(1*5) + 1/(5*9) + 1/(9*13) .... + 1/(221*225)
Ans: 56/225
12) If in a decreasing AP, sum of all its terms except for the first term is equal to -36. The sum of all its terms except for the last term is 0.
And the difference of the 10th and the 6th term is equal to -16.
Then what will be the first term of the series?
Ans: 16
13)The 1st and 3rd terms of an AP are equal, respectively, to the 1st and 3rd term of GP. And the 2nd term of AP exceeds the 2nd term of GP by 0.25
Calculate the sum of the first 5 terms of the AP is its 1st term is equal to 2.
Ans: 2.5 or 2.75
14) The sum of the squares of 5th and 11th term of an AP is 3 and the product of 2nd and 14th term is Q. Find the product of 1st and 15th term of AP.
ANs: (98Q+39)/90
15) The ration of harmonic mean of 2 nos to the Geometric mean is 12:13,
find the ratio of the nos.
Ans: 3/4 or 4/5
16)Find the sum of all whole number divisors of 720.
Ans:2624
17) If 3 positive real nos are in AP such that xyz=4, then what will be the minimum value of y?
Ans:2^(2/3)
1
Ans: 8
19) Find the sum of all 3 digit whole nos less than 500 that leave a remainder of 2 when they are divided by 3.
Ans: 39767
20) If the first two terms of a HP are 2/5 and 12/13 resp, which of the folln terms is the largest term?
a)4th term b)5th c)6th d)7th
Ans: 4th
For Q.18 - The only possible answer to this is 2*3 = 6. The answer given in Arun Sharma (
is definitely incorrect. 
Two people A and B start from P and Q(distance= D) at the same time towards each other. They meet at a point R, which is at a distance of 0.4D from P. They continue to move two and fro between the two points. Find the distance from point P at which the fourth meeting takes place.
a) 0.8D
b) 0.6D
c) 0.3D
d) 0.4D
is the answer 0.8D....option a...