@sparklingaubade said:N = 7^7*19^4*13^5How many factor of N are 6m+1 type?
8*5*6=240
@sparklingaubade said:N = 7^7*19^4*13^5How many factor of N are 6m+1 type?
Every factor of this number is 6m+1 type
So total factors =(7+1)(4+1)(5+1)=8*5*6=240 ANS
So total factors =(7+1)(4+1)(5+1)=8*5*6=240 ANS
In the given figure Ac is parallel to FD and FB is parallel to DC . Area of quadrilaterial bcde is 238 . Ab = 10 . BE = 5 . EF = 7. What's the area of triangle Abe ?
In the above figure pqrs is a parallelogram with pq = 10 , qr = 5 and PO = 4 . Find ST and area of pqts ?
@meenu05 said:In the given figure Ac is parallel to FD and FB is parallel to DC . Area of quadrilaterial bcde is 238 . Ab = 10 . BE = 5 . EF = 7. What's the area of triangle Abe ?
DC = 12 (BCDF is a parallelogram)
Area of ABE/ (Area of ABE + Area of BCDE) = (5/12)^2
x/(x + 238) = 25/144 => 144x = 25x + 25*238 => x = 238*25/(119) = (25)*(14)/(7) = 50
So, 50 ?
@meenu05 said:In the above figure pqrs is a parallelogram with pq = 10 , qr = 5 and PO = 4 . Find ST and area of pqts ?
Area of parallelogram = base * height = PQ*PO = PS * ST => ST = 10*4/5 = 8 ?
Area of PQTS = Area of parallelogram + Area of STR = 40 + 1/2*ST*TR = 40 + 1/2*6*8 = 64 ?
The marked prices of two articles are in the ratio of 1:2, their discount percentages are also in the ratio of 1:2 and the profit they get is also in the ratio of 1:2. What is the ratio of their cost price
when a three digit num divides 67588 and 63424 ,it leaves the same remainder in both cases..find the least possible value of the three digit number??
1.101
2.231
3.347
4.301
explain
@meenu05 said:In the given figure Ac is parallel to FD and FB is parallel to DC . Area of quadrilaterial bcde is 238 . Ab = 10 . BE = 5 . EF = 7. What's the area of triangle Abe ?
Area = 50 ?
|x| + 2|y| = 100
no. of integral solutions
@pavimai said:when a three digit num divides 67588 and 63424 ,it leaves the same remainder in both cases..find the least possible value of the three digit number??1.1012.2313.3474.301explain
Option 3 : 347
Since the nos. leaves the same remainder theri diff should be divisible by the required 3 digit no.
so 67588-63424/ ( required no ) = 0
from the options 347 satisfies the condition.
PS: sum1 plz post how to solve in the absence of options...
@bullseyes said:|x| + 2|y| = 100 no. of integral solutions
y = 0 => |x| = 100 -> 2 solutions
|y| = 1 => |x| = 98-> 4 solutions
and so on.. till
|y| = 49 => |x| = 1 -> 4 solutions
|y| = 50 => |x| = 0 -> 2 solutions
So, total solutions = 49*4 + 2 + 2 = 200 ?
find the remainder when the 100-digit numbers formed by writing the consecutive natural numbers starting from 1 next to each other is divided by 16
1.6
2.8
3.9
4.10
??
@pavimai said:when a three digit num divides 67588 and 63424 ,it leaves the same remainder in both cases..find the least possible value of the three digit number??1.1012.2313.3474.301 explain
67588 = N*k1 + r
63423 = N*k2 + r
Subtract => 4164 = N*(k1 - k2) => N is a factor of 4164 and is the least three digit factor
=> Using options, 347 ?
In a circle AB is a diameter and C lies on circumference of the circle such that ABC forms a triangle with
(area of circle)/(area of triangle ABC) = 2*pi
find tan(ABC) and tan(CAB)
@pavimai said:find the remainder when the 100-digit numbers formed by writing the consecutive natural numbers starting from 1 next to each other is divided by 161.62.83.94.10??
We need to find the last four digits for this
1- 9 -> 9 digits
10 - 54 -> 90 digits
=> Last four digits = 3545
=> N mod 16 = 3545 mod 16 = 9 ?
@pavimai said:find the remainder when the 100-digit numbers formed by writing the consecutive natural numbers starting from 1 next to each other is divided by 161.62.83.94.10??
9 ??
@bullseyes said:In a circle AB is a diameter and C lies on circumference of the circle such that ABC forms a triangle with (area of circle)/(area of triangle ABC) = 2*pi find tan(ABC) and tan(CAB)
C lies of the circumference => ABC is a right angled triangle => Area = 1/2*AC*BC
=> (pi*(AB)^2)/4) / (1/2*AC*BC) = 2pi
=> AB^2 = 4*AC*BC
4*sin(ABC)*cos(ABC) = 1 => sin(2*ABC) = 1/2 => 2*ABC = 30 => ABC = 15 degree
tan(ABC) = root( [1 - cos(2ABC)]/[1 + cos(2ABC)] ) = 2 - _/3 ?
tan(CAB) = 2 + _/3 ?