Home
Class 14
MATHS
29.8% "of" 260+60.01% "of" 510-103.57=?...

`29.8% "of" 260+60.01% "of" 510-103.57=?`

A

450

B

320

C

210

D

280

Text Solution

Verified by Experts

The correct Answer is:
D

The difference between two nearest values is 70 (210 and 280) . So round off the number to the nearest integers. `29.8% of 260 ~~ 30% of 260, 60.01 % of 510 ~~60% of 510 and 103.57~~104`
Now the equation will become
30% of 260 + 60% of 510-104=?
30/100`xx`60/100xx510-104=?
`78+306-104=?`
`?=384-104=280`
Promotional Banner

Topper's Solved these Questions

  • AREA AND PERIMETER

    IBPS & SBI PREVIOUS YEAR PAPER|Exercise MCQ|75 Videos

Similar Questions

Explore conceptually related problems

30.21% "of" 260+59.84% "of" 510-103.57=?

0.01% of ₹1000

30.1% "of" 259.99+59.9% "of" 510-103.57=?

125%"of"260+? % "of" 700=500

23% "of" 6783+57% "of" 8431=?

41% "of" 801-150.17=?-57% "of" 910

39.8% of 400 + ?% of 350 = 230

12.8 % of 715=?

What is (0.08% "of " 0.008%" of " 8)^(1/9) ?

?% "of" 180.02=sqrt((24.01)^(2)+(18.21)^(2)+60.01% "of" 660.17)