Solution :
Here in these types of questions you have to only find the GcD/HCF of the powers only !!
100= 2^2 X 5^5
120=2^3 X 5 X 3
hcf =20
Answer :2^20-1
and how about the term -1..we dont have to consider that????????
and how about the term -1..we dont have to consider that????????
divyaakanksha Sayswhat r the three nos
49-4=45
81-36=45
529-484=45...
divyaakanksha Saysand how about the term -1..we dont have to consider that????????
No !!
We ignore that
the sides of a triangle are 21,20 and 13 .find the area of the larger triangleinto which the given triangle id divided by the perpendicular upon the longest side from the opposite vertex?
a 72
b 96
c 168
d 144
e 150
plz post the approach
the sides of a triangle are 21,20 and 13 .find the area of the larger triangleinto which the given triangle id divided by the perpendicular upon the longest side from the opposite vertex?
a 72
b 96
c 168
d 144
e 150
plz post the approach
answer is 96
let BC is the largest side and AB is the smallest side
if AD is perpendicular from A to BC.then ADC is the bigger trinagle so formed..
now, find angle ABC by cosine rule. cos(ABC) comes 5/13.
from here we get BD =5 and DC = 16
now from triangle ABD ,which is a right angled triangle at B, find AD as AB and BD are known
so AD = 12
so area of bigger triangle ADC = 1/2*16*12=96 square units
the sides of a triangle are 21,20 and 13 .find the area of the larger triangleinto which the given triangle id divided by the perpendicular upon the longest side from the opposite vertex?
a 72
b 96
c 168
d 144
e 150
plz post the approach
its given that larger side is divided by a perpendicular., So let sides be x and 21-x
so apply pythogoroes theorem now.
h^2+x^2=20^2
h^2+(21-x)^2=13^2
20^2-x^2=13^2-(21-x)^2
20^2-13^2=x^2-(21-x)^2
x=16
so h^2=20^2-16^2
h=12
so area=1/2x12x16=96
PQRS is the diameter of a circle of radius 6 cm .The length PQ ,QR and RS are equal .Semi cirsles are drawn with PQ and QS as diameters as shown in the fig along side .Find the ration of the area of the shaded region to that of the unshaded region ?
a 1:2
b 25:121
c 5:18
d 5:13
e 5:12
m getting the and as 5:9 whr am i going wrong in this
now pqrs=6cm
so pq=qr=rs=2cm
area of circle=22/7 *3^2
semi circle=99/7
since rs=2cm area=22/7 * 2^2
half of its rea = 44/7
so shaded part = 99/7-44/7
=55/7
unshaded = 99/7+44/7
=143/7
=>55/7/143/7
=>5:13
plz helpme solve
2. how many pair of natural numbers are there the difference of whose square is 45?
a. 1
b. 2
c. 3
d. 4
e. 5
let A^2-B^2 = 45= 5*3^2
=>(A+B)*(A-B)= 5*3^2=(9*5)=(15*3)=(45*1)
=> A+B= 9 or 15 or 45
and
A-B= 5 or 3 or 1
by solving A+B with corresponding A-B we get
(A,B)={(7,2), (9,6), (23,22)}
three pairs..
This one's a ripper
The sum of the squares of the fifth and the eleventh term of an AP is 3 and the product of the second and the fourteenth term is equal to P.Find the product of the first and the 15th term of the AP.
a) (59P - 39)/45
b) (98P + 39)/72
c) (116P - 39)/90
d) (98P+39)/90
e) None of these
Find the sum of all possible whole number divisors of 720
a) 2012
b) 2624
c) 2210
d) 2418
e) 2520
Some more
1) If a1, a2 ...aN are in AP where aI>0, then find the value of the expression
1/{sqrt(a1) + sqrt(a2)} + 1/{sqrt(a2) + sqrt(a3) +.... till n terms
(Note: Does anyone think that it should be n-1 terms..)
a) (1-n)/{sqrt(a1) + sqrt(aN)}
b) (n-1)/{sqrt(a1) + sqrt(aN)}
2) If the first two terms of an HP are 2/5 and 12/13 respectively then which one of the following be the largest term.
a) 4th
b) 5th
c)7th
d) 8th
3) If p,q,r are three consecutive distinct numbers then the expression (q+r-p)(p+r-q)(p+q-r) is
a) positive
b non negative
c both
d ) negative
e non positive
This one's a ripper
The sum of the squares of the fifth and the eleventh term of an AP is 3 and the product of the second and the fourteenth term is equal to P.Find the product of the first and the 15th term of the AP.
a) (59P - 39)/45
b) (98P + 39)/72
c) (116P - 39)/90
d) (98P+39)/90
e) None of these
Find the sum of all possible whole number divisors of 720
a) 2012
b) 2624
c) 2210
d) 2418
e) 2520
1)a+4d+a+6d=3
=>2a+10d=3
also (a+d)x(a+13d)=p
required ax(a+14d)=>a^2+14ad
(a+d)x(a+13d)=p
=>a^2+14ad+13d^2=P
=>a^2+14ad=p-13d^2
from 1 d=3-2a/10
sub back to the eq
2)2^4x3^2x5
apply the formula for sum of divisors
=>(2^5-1)(3^3-1)(5^2-1)/2x4
=>2418
1)a+4d+a+6d=3
=>2a+10d=3
also (a+d)x(a+13d)=p
required ax(a+14d)=>a^2+14ad
(a+d)x(a+13d)=p
=>a^2+14ad+13d^2=P
=>a^2+14ad=p-13d^2
from 1 d=3-2a/10
sub back to the eq
I think you made the wrong equation with the first one...it is sum of squares and not just sum....please check bro !!
Hector1 SaysI think you made the wrong equation with the first one...it is sum of squares and not just sum....please check bro !!
oh yeah thx
This one's a ripper
The sum of the squares of the fifth and the eleventh term of an AP is 3 and the product of the second and the fourteenth term is equal to P.Find the product of the first and the 15th term of the AP.
a) (59P - 39)/45
b) (98P + 39)/72
c) (116P - 39)/90
d) (98P+39)/90
e) None of these
Find the sum of all possible whole number divisors of 720
a) 2012
b) 2624
c) 2210
d) 2418
e) 2520
answers are: 1.c 2.d
solution:
1.let a is the 1st term and d is the common diff. of AP
given, (a+d)(a+13d) = p..............(1)
and (a+4d)^2+(a+10d)^2=3........(2)
find d^2 in terms of a*(a+14d) { assume a*(a+14d) as k for simplicity} from equation (1) and substitute this value in equation (2)
in equation 2 also put a*(a+14d) =k
find the value of k this is the desired result
2..
720=(2^4)*(3^2)*5
sum of all whole no. devisors = (2^0+2^1+2^2+2^3+2^4)*(3^0+3^1+3^2)*(5^0+5^1) = 2418
1)a+4d+a+6d=3
=>2a+10d=3
also (a+d)x(a+13d)=p
required ax(a+14d)=>a^2+14ad
(a+d)x(a+13d)=p
=>a^2+14ad+13d^2=P
=>a^2+14ad=p-13d^2
from 1 d=3-2a/10
sub back to the eq
2)2^4x3^2x5
apply the formula for sum of divisors
=>(2^5-1)(3^3-1)(5^2-1)/2x4
=>2418
answers are: 1.c 2.d
solution:
1.let a is the 1st term and d is the common diff. of AP
given, (a+d)(a+13d) = p..............(1)
and (a+4d)^2+(a+10d)^2=3........(2)
find d^2 in terms of a*(a+14d) { assume a*(a+14d) as k for simplicity} from equation (1) and substitute this value in equation (2)
in equation 2 also put a*(a+14d) =k
find the value of k this is the desired result
2..
720=(2^4)*(3^2)*5
sum of all whole no. devisors = (2^0+2^1+2^2+2^3+2^4)*(3^0+3^1+3^2)*(5^0+5^1) = 2418
Hi,
Thnx !!
Arun Sharma did not have that sum formula but I found that now.....there is one more request.
I hope you would be knowing that there is formula for questions like
a^n + b^n % (a+b) where n is even/odd and the sign between a and b is positive/negative
Can you please post those formulas if you have those. They are not in Arun Sharma and I can't seem to find them elsewhere.
Hi,
Thnx !!
Arun Sharma did not have that sum formula but I found that now.....there is one more request.
I hope you would be knowing that there is formula for questions like
a^n + b^n % (a+b) where n is even/odd and the sign between a and b is positive/negative
Can you please post those formulas if you have those. They are not in Arun Sharma and I can't seem to find them elsewhere.
a^n - b^n is always divisible by a-b
a^n - b^n is divisible by a+b when n is even.
a^n + b ^n is divisible by a+b when n is odd.
a^n + b^n + c^n +... is divisible by a+b+c+.... when n is odd.
when last n digits of a no. are divided by 2^n, the remainder is same as the remainder when the entire no. is divided by 2^n.
Edit: At hector do use pm facility at PG :D
Ganu02 Saysyep dude..Missed it out..
no problem....that question is such a beauty..anyone can miss
Hi,
Thnx !!
Arun Sharma did not have that sum formula but I found that now.....there is one more request.
I.
page no. 22 and 23rd or the 2nd edition deals with these formulas..read these pages carefully because it covers atleast 5 question profiles
a^n - b^n is always divisible by a-b
a^n - b^n is divisible by a+b when n is even.
a^n + b ^n is divisible by a+b when n is odd.
a^n + b^n + c^n +... is divisible by a+b+c+.... when n is odd.
when last n digits of a no. are divided by 2^n, the remainder is same as the remainder when the entire no. is divided by 2^n.
Thnx bro....can you tell me from which book you picked these up !!
P.S. Would highly appreciate if you can shed some light on some good books for CAT preparation (LR/DI and VA). It seems like you are a PRO.... 😃