Numerical Computation Quiz-1A

Solution:

1397\times 1397

(1397)^{2}

  (1400-3)^{2}

(1400)^{2}+(1400)^{2}-(2\times 1400\times 3)

=1960000 + 9 - 8400

=1960009 - 8400

=1951609

Solution: Required sum = (2 + 3 + 5 + 7 + 11) = 28.

Note: 1 is not a prime number.

Solution:

Solution:

Let the two numbers are 15x and 11x
HCF = x
x = 13
Therefore, the numbers are (15 x 13 and 11 x 13), i.e. 195 and 143.

Solution:

Difference between numbers = 90-34 = 56, and 104- 90=14

HCF of 56 and 14 = 14

Hence, 14 is the largest number which divides the given number and leaves the same remainder in each case.

Solution:

Let the numbers be 29a and 29b. Then, 29a + 29b= 174 or 29(a + b) = 174 or, a + b =\frac {174}{29} = 6 . Values of co-primes (with sum 6) is(1, 5).
So, the possible pairs of numbers is (29 \times 1, 29 \times 5) i.e. 29 and 145

Solution:

L.C.M. of 2, 3, 4, 6 = 12 ; 12 = 2^{2}\times 3. So, the required number = 2^{2}\times 3 \times 2 \times 3^{2} = 216 .

Solution: Let the required number be x.

\frac{(–251 \times 21 \times (– 12))} {?} = 158.13

x=400

Solution:

\sqrt{{\sqrt{4900}}+\sqrt{5476}}

= \sqrt{70+74}

= \sqrt{144}

=12

Solution:

124% of 624.809 + 11.75% of 1279.43 = ?

? ≈ 125% of 625 + 12% of 1280

? ≈ (100% of 625 + 20% of 625 + 4% of 625) + (10% of 1280 + 2% of 1280)

? ≈ (625 + 125 + 25) + (128 + 25.6)

? ≈ 775 + 153.6 = 928.6 ≈ 925.