Solution: (d)
The difference between two nearest values is 70 (210 and 280).
So round off the numbers to the nearest integers.
29.8% of 260 \(\displaystyle \approx \)30% of 260; 60.01% of 510 \(\displaystyle \approx \)60% of 510 and 103.57 \(\displaystyle \approx \) 104
Now the equation will become
30% of 260 + 60% of 510 – 104 = ?
\(\displaystyle \frac{{30}}{{100}}\times 260+\frac{{60}}{{100}}\times 510-104=?\)
78 + 306 – 104 = ?
? = 384 – 104 = 280
Solution: (d)
Taking approximate values,
\(\displaystyle 5500\div 2300\text{ }\times 200=?\)
\(\displaystyle \approx 480=?\)
So, by approximation we can see that the answer is (d)
Solution: (a)
By approximation, take nearest approximate values
\(\displaystyle 66\div 11\text{ }\times 5=?\)
\(\displaystyle 6\times 5=?\)
\(\displaystyle \approx 30=?\)
39) \(\displaystyle 1002\div 49\times 99-1299=?\)
(a) 700
(b) 600
(c) 900
(d) 250
(e) 400
Solution: (a)
It can be rounded off to the nearest ten’s places.
1002 \(\displaystyle \approx \) 1000; 49 \(\displaystyle \approx \) 50; 99 \(\displaystyle \approx \) 100 and 1299 \(\displaystyle \approx \)1300
Now the equation will become
\(\displaystyle 1000\div 50\times 100-1300=?\)
\(\displaystyle 20\times 100-1300=?\)
\(\displaystyle 2000-1300=?\)
\(\displaystyle ?=700\)
Solution: (a)
\(\displaystyle {{21.98}^{2}}\approx {{22}^{2}}\)
\(\displaystyle {{25.02}^{2}}\approx {{25}^{2}}\)
And \(\displaystyle {{13.03}^{2}}\approx {{13}^{2}}\)
The equation will becomes
\(\displaystyle {{22}^{2}}+{{25}^{2}}+{{13}^{2}}=?\)
\(\displaystyle 484-625+169=?\)
\(\displaystyle 653-625=?\)
\(\displaystyle 28=?\)
so the nearest value is 25
