#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Q.1 If xyz=1, x + 1/z =5, y + 1/x =29, then find z + 1/y Q.2 If a+b+c=2, a^2 + b^2 + c^2 =6, a^3 + b^3 + c^3 =8, then find a^4 + b^4 + c^4.

10 years ago

Dear Abhinav,

we can write x+(1/z) as (1/yz) + (1/z) solving this we get(1+y)/yz = 5 ................(1)

now from the second equation we get y + yz = 29

or, yz = 29-y putting it in (1) we get (1+y) = 5(29-y)

or, 1+y = 145 - 5y6y = 144or, y = 24now again putting it in (1) we get 25 = 120z

or, z = 5/24so x = 1/yz or, x = 1/5 notice that this fulfills the requirement xyz = 1so now z + 1/x =  z + zy = 5/24 + (5/24)(1/5) = 5/24 + 1/24 = 1/4 answer.

BEST OF LUCK..!!!!

Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

AMAN BANSAL

10 years ago

xyz=1, thus yz=1/x. x+1/z=5 so z=1/[5-x ] y+1/x=29 so y=[ 29x-1 ]/x Now, yz=1/x [ [29x-1 ]/x]*[1/[5-x ]] =1/x On solving, x=1/5 yz=5, y=24 z+1/y=[yz+1]/y. Substitute the values of yz and y. The answer u wud get as 1/4.