### 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)

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
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
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
• okunola nifemi: nyc guy Waldon weldone.
6 years ago
• hibramed: U are a guru
5 years ago
• suntoozee: amsu
1 year ago
1 year 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
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
OMG am loving this site so much
Follow
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
is this really jamb pst questions?
Follow
Dont be confused,just apply the law of indices,you will get d answer
Follow
i dnt understand any of the solution pls help
Follow
what a judicious question
Follow
But this is not the actual questions we are 2 answered in jamb.no matrix,no integration,no co odinate geometry?
Follow
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
the answer is c ; be it -25/4 or -13/2
Follow
P=5/2 , q=-25/4 and r=-17/4 not -9/2
Thank u for ur compliance 😃
Follow