Simplification Questions and Answers

Answer: C

The value is approximately 350.

Round each part to a convenient number:

\[ \begin{aligned} 13.08 \times 24.98 &\approx 13 \times 25 = 325 \\ \sqrt{624} &\approx \sqrt{625} = 25 \\ 325 + 25 &= 350 \end{aligned} \]

The exact value is about 351.7, so 350 is the nearest option. 375 and 380 overestimate the product, and 320 drops the square root term.

In approximation questions the options are usually far enough apart that rounding each term to the nearest convenient value is enough; here the gap between 350 and its neighbours is 25 or more.

Answer: B

The value is \(\frac{10}{11}\).

Evaluate each root (the index of each radical is printed beside it):

Numerator: \(\sqrt[3]{729} = 9\), \(\sqrt[3]{27} = 3\), \(\sqrt[2]{16} = 4\). Denominator: \(\sqrt[3]{512} = 8\), \(\sqrt[3]{343} = 7\), \(\sqrt[4]{256} = 4\).

\[ \frac{9 - 3 + 4}{8 + 7 - 4} = \frac{10}{11} \]

\(\frac{11}{10}\) is the same fraction turned upside down, and \(\frac{9}{10}\) and \(\frac{12}{11}\) come from misreading one of the roots, for example taking the fourth root of 16 (2) instead of its square root.

Answer: D

The value is 8.5.

Work inside each bracket first, doing division and multiplication before subtraction:

\[ \begin{aligned} 1350 \div 15 - 5 &= 90 - 5 = 85 \\ 47.5 - 15 \times 2.5 &= 47.5 - 37.5 = 10 \\ 85 \div 10 &= 8.5 \end{aligned} \]

85 is the first bracket alone, and 10 is the second bracket alone. 10.5 comes from an arithmetic slip in the second bracket.

Answer: C

The sum is 5.75 + 4.8 + 7.375 = 17.925. To reach the whole number 18 we need 0.075, which is 3/40, option C.

1/40, 2/40 and 4/40 would leave the total at a non-whole number.

To add the fractions, use the common denominator 40: 3/4 = 30/40, 4/5 = 32/40 and 3/8 = 15/40, so the fractional parts add up to 77/40, or 1 and 37/40. The whole numbers add to 16, so the total is 17 and 77/40, which needs 3/40 more to make 18.

Answer: C

The expression has the form \[\tfrac{2(a^{2}+b^{2})}{a^{2}+b^{2}}=2\] with a = 963 and b = 476, so the value is 2, option C.

1449, 497 and 4 do not come from this identity.

This is the identity \((a+b)^{2}+(a-b)^{2}=2(a^{2}+b^{2})\). With a = 963 and b = 476 the numerator is 2(963 squared + 476 squared) and the denominator is 963 x 963 + 476 x 476, which is the same sum without the 2, so the fraction is exactly 2 without any arithmetic.

Answer: C

Work the fractions step by step. First part:

Second part:

Division:

The brace is 2/3 + 1/3 = 1, so 2/5 of it is 2/5 = 32/80. The result is 147/80 - 32/80 = 115/80 = 23/16 = 1 7/16, option C.

The English line of the question is clipped at the top of the page, and the numerator 4 - 2/(4 - 1/2) is read from the Tamil printing of the same question.

Answer: B

The cube is \(-\frac{1}{64}\).

Cube the numerator and the denominator, keeping track of the sign. A negative number raised to an odd power stays negative:

\[ \begin{aligned} \left(-\tfrac{1}{4}\right)^{3} &= -\frac{1^{3}}{4^{3}} \\ &= -\frac{1}{64} \end{aligned} \]

\(\frac{1}{64}\) loses the sign, which would be right only for an even power. \(\pm\frac{1}{16}\) is the square, \(4^{2} = 16\), not the cube.

Answer: A

The answer is 37.

Division is carried out from left to right:

\[ \begin{aligned} 666 \div 6 &= 111 \\ 111 \div 3 &= 37 \end{aligned} \]

111 is the result after only the first division. 333 comes from dividing 666 by 2, which happens if you wrongly work out \(6 \div 3\) first and then divide 666 by that. 84 does not come from any order of these operations.

Answer: B

Mrs. Tripathi weighed 102 kg to begin with.

Let her starting weight be W. Losses are subtracted and the gain is added:

\[ \begin{aligned} W - 5.5 + 2.25 - 3.75 &= 95 \\ W - 7 &= 95 \\ W &= 102 \end{aligned} \]

Check: 102 − 5.5 = 96.5, then 96.5 + 2.25 = 98.75, then 98.75 − 3.75 = 95. 106.5 kg is what you get by adding all three changes to 95, treating the gain as a loss too.

Answer: B

The value is \(7\frac{5}{16}\).

Work the brackets from the inside out, then 'of', then division, then addition:

\[ \begin{aligned} 2 - \tfrac{1}{3} &= \tfrac{5}{3},\ \ 3 - \tfrac{5}{3} = \tfrac{4}{3} \\ 4 \times \tfrac{4}{3} &= \tfrac{16}{3},\ \ 6 \div \tfrac{16}{3} = \tfrac{9}{8} \\ \tfrac{33}{4} \times \tfrac{3}{4} &= \tfrac{99}{16} \end{aligned} \]

Adding, \(\frac{99}{16} + \frac{18}{16} = \frac{117}{16} = 7\frac{5}{16}\). All four options end in \(\frac{5}{16}\), so the question turns on the whole number; \(6\frac{5}{16}\) is \(\frac{99}{16}\) alone, forgetting to add the \(\frac{9}{8}\).