Infosys Number, Decimal and Fractions Questions and Answers Quiz – 1

Length of each piece = (1/8) m = 0.125 m

Required no of pieces = (37.5/0.125) = (375 * 100) / 125 = 300.

4\times A + \frac{7}{B} + 2\times C + \frac{5}{D} + 6\times E =47.2506
=> 4\times A + \frac{7}{B} + 2\times C + \frac{5}{D} + 6\times E = 40 + 7 + 0.2 + 0.05 + 0.0006
Comparing the terms on both sides, we get :
4*A =40, \frac{7}{B} = 7, 2*C = 0.2, \frac{5}{D} = 0.05, 6*E = 0.0006

Therefore:- A = 10, B = 1, C = 0.1, D = 100, E = 0.0001.
5*A + 3*B + 6*C + D + 3*E =(5*10) + (3*1) + (6*0.1) +100 +(3*0.0001)
= 50 + 3 + 0.6 + 100 + 0.003 = 153.6003.

Explanation:

Given expression is, 83.04 x 2.01 x 2.1
The multiplication value without decimal point = 8304 x 201 x 2 = 3338208
Sum of the number of decimal places in the given numbers = 2 + 2 + 1 = 5
Now, in the product, the decimal point is marked off to obtain as many places of decimal.
Then the answer is 33.38208.

Here, we have to find the value of 7/(8/9) - (7/8)/9.
(7 x 9/8) - [(7/(8 x 9)]
63/8 - 7/72
[(63 x 9) - 7] / 72
567 - 7/72 = 560/72 = 70/9 = 7 7/9.
Hence, 7/(8/9) is greater than (7/8)/9 by 7 (7/9).

Let the required fraction be p/q.

Given that, sum of the numerator and denominator of a fraction is 13;
i.e., p + q = 13 ...(1)

Adding 3 to numerator and 9 to the denominator, we have,

(P+3)/(q+9) = 2/3
3(P+3) = 2(q+9)
3p+9 = 2q+18
3p-2q = 9 ...(2)

Multiply eqn (1)by 2, 2p + 2q = 26...(3)

Adding (2) and (3),

5p = 35
p = 7.

Putting p value in (1), we get q = 13 - p = 13 - 7 = 6.

Hence the required fraction is p/q = 7/6

3\frac{2}{3} + 6\frac{7}{12} + 4\frac{9}{36} + 5 + 7\frac{1}{12}
Add only the fraction parts: \frac{2}{3} + \frac{7}{12} + \frac{9}{36} + \frac{1}{12}
=\frac{24 + 21 + 9 + 3 }{36} = \frac{57}{36} = \frac{19}{12} = 1\frac{7}{12}
Thus to make the whole number we should add 1 - \frac{7}{12} = \frac{5}{12}

Let a and b are the numbers.Then a+b is 70 and ab is 1221.
a = 70 - b…... (i)
ab = 1221
(70-b)b = 1221
70b-b²-1221 = 0
Multiplying the above equation by (-1)
Then, the equation becomes: b² - 70b + 1221 = 0
Solving the above equation we get through the middle term splitting method
b = 37
Substituting the value of b in equation (i) a = 33

The sum five consecutive integers can be written as n + (n + 1) + (n + 2)+(n+3)+(n+4) = 5n + 10
If the sum is 24, we need to solve the equation 5n+10 = 215;
=> 5n = 205;
=> n = 41
The greatest of 5 consecutive integers is n+4 = 41+4 = 45

Let the nos be “a” and “b”.
According to question,
a+b=15
and, a-b=3
a^{2} - b^{2} = (a+b)(a-b)
a^{2} - b^{2} = (15+3)(15-3)
a^{2} - b^{2} = (18)(12)
a^{2} - b^{2} = 216