### If $$\frac{(a^2 b^{-3}c)^{\frac{3}{4}}}{a^{-1}b^{4}c^{5}}=a^{p} b^{q} c^{r}$$ What is the value of p+2q?

If $$\frac{(a^2 b^{-3}c)^{\frac{3}{4}}}{a^{-1}b^{4}c^{5}}=a^{p} b^{q} c^{r}$$ What is the value of p+2q?

• A. (5/2)
• B. -(5/4)
• C. -(25/4)
• D. -10
##### Explanation

Hint: apply basic mathematics rules beginning from BODMAS to algebra, and follow solution carefully to arrive at p =(5/2), q = -(25/4) and r = -(17/4).

Then p+2q will give you $$\frac{5}{2}+2\left(\frac{-25}{4}\right)= -10$$

#### Contributions (65)

Waldon95
7 years ago
Soln

[a^2(3/4)b^-3(3/4)c^(3/4)]divided by (a^-1b^4c^5)

===>(a^3/2b^-9/4c^3/4)divided by (a^-1b^4c^5).

Note a^p =3/2-(-1)==>p=5/2

q=-9/4-(4)===>q=-25/4

r=3/4-(5)===>r=-17/4

now p+2q

5/2+2(-25/4)

:- p+2q=-10.

Make use of indice...
Follow
• lamboy001: guy,i dey feel u,i undastand it nw
6 years ago
• TAIWO KHADIJAT: pls guyz explain to me
3 years ago
• psamson.sh8: Thanks for your explanation. God bless you
3 years ago
• Hannat001: Is it waec question?
9 months ago
• Ngene Benjamin: thanks for this wonderful explanation,i apperciate
7 months ago
sunday felicia
4 years ago
I SWEAR HEE NO EASY AT ALL.IS DIS D KIND OF QUESTION WE ARE GOIN TO FACE INSIDE HALL. BEFORE I FINISHE SOLVING DIS QUESTION INSIDE EXAM HALL,DERE WOULD BE ANYTIME LEFT.MAY GOD HELP US NI O
Follow
• suntoozee: abi na he no easy at all,, person suppose to agonize b4 solving it.
1 year ago
• Favour: u are right may God help us.
1 year ago
• Kenny: i dey tell u
1 year ago
Waldon95
7 years ago
Soln

[a^2(3/4)b^-3(3/4)c^(3/4)]divided by (a^-1b^4c^5)

===>(a^3/2b^-9/4c^3/4)divided by (a^-1b^4c^5).

Note a^p =3/2-(-1)==>p=5/2

q=-9/4-(4)===>q=-25/4

r=3/4-(5)===>r=-17/4

now p+2q

5/2+2(-25/4)

:- p+2q=-10.

Make use of indice...
Follow
Felix Powell
6 years ago
Soln

{(a^2b^-3c)^3\4 /a^-1b^4c^5} =a^pb^qc^r

from the power law of indices.

(a^2b^-3c)^3\4 =a^2*3\4.b^-3*3\4.c^1*3\4

=a^3\2.b^-9\4.c^3\4

from the divisional law of indices

(a^3\a.b^-9\4.c^3\4)/a^-1.b^4.c^5 =

a^3\2+1.b^-9\4-4.c^3\4-5

=a^5\2.b-25\4.c^-17\4

Now by comparing the powers in a^p.b^q.c^r

p=5\2,b=-25\4,c=-17\4

substituting the values of p and q in the equation p+2q,we will have

5\2 + 2(-25\4) = 5\2- 25\2

=(5-25)\2 = -20\2 = -10 Ans.
Follow
maxi957
7 years ago
This question is really great.what you need to first is to open the bracket then use indices to remove the division sign to subtraction sign then after which you compare to get p,q and r.then u can substitute it into p+2q.that's it.you just need to tink hard.
Follow
• Bukatti: i followed what you said but i approximated it to get -10 am i right
1 year ago
6 years ago
OMG am loving this site so much
Follow
Omotayo isaac
7 years ago
Is this what we want to solve in UTME......yea!I don suffer :(
Follow
• sammy6829: u re nt serious, u have to be a man.
5 years ago
• lily.u.: each time I see past UTME maths ?s my hrt always beat twice faster than normal.
5 years ago
prof kc
4 years ago
is this really jamb pst questions?
Follow
ahmedhalidu
6 years ago
Dont be confused,just apply the law of indices,you will get d answer
Follow
viole3t@valerie
5 years ago
i dnt understand any of the solution pls help
Follow
tenten
6 years ago
what a judicious question
Follow
ahmedhalidu
6 years ago
But this is not the actual questions we are 2 answered in jamb.no matrix,no integration,no co odinate geometry?
Follow
Remmy2
5 years ago
Trying very hard to understand the question n the answer,, still to no avail,,, i love learning pls make it easy for people like me thanks
Follow
ruthbaiby
4 years ago
the answer is c ; be it -25/4 or -13/2
Follow
7 months ago
P=5/2 , q=-25/4 and r=-17/4 not -9/2
Thank u for ur compliance 馃槂
Follow

#### Quick Questions

Please don't post or ask to join a "Group" or "Whatsapp Group" as a comment. It will be deleted. To join or start a group, please click here

Soln

[a^2(3/4)b^-3(3/4)c^(3/4)]divided by (a^-1b^4c^5)

===>(a^3/2b^-9/4c^3/4)divided by (a^-1b^4c^5).

Note a^p =3/2-(-1)==>p=5/2

q=-9/4-(4)===>q=-25/4

r=3/4-(5)===>r=-17/4

now p+2q

5/2+2(-25/4)

:- p+2q=-10.

Make use of indice...
Waldon95