Arithmetic Operations Questions and Answers

Answer: C

The frog needs 75 jumps.

Each full jump-and-slide gains \(70 - 30 = 40\) cm. But on the final jump the frog leaves the well and does not slide back, so stop counting slides before the last jump.

\[ \begin{aligned} 40(k - 1) + 70 &\ge 3000 \\ k - 1 &\ge 73.25 \Rightarrow k = 75 \end{aligned} \]

After 74 jumps (and slides) the frog is at \(74 \times 40 = 2960\) cm, and the 75th jump of 70 cm takes it to 3030 cm, out of the well. Dividing 3000 by 40 gives exactly 75 here, but only by coincidence; 76 and 78 come from ignoring the last-jump rule in other ways, and 73 stops short.

Answer: B

H stands for 7.

H3, H2 and H9 are two-digit numbers with tens digit H, so they are \(10H + 3\), \(10H + 2\) and \(10H + 9\).

\[ \begin{aligned} (10H + 3) - (10H + 2) + 81 - 62 \\ - 44 - (10H + 9) &= -103 \\ -33 - 10H &= -103 \Rightarrow H = 7 \end{aligned} \]

Check: \(73 - 72 + 81 - 62 - 44 - 79 = -103\). With H = 4, 5 or 9 the left side is −73, −83 or −123.

Answer: A

He can make 300 pieces.

Each metre of cloth gives 8 pieces, so the number of pieces is

\[ 37.5 \times 8 = 300 \]

The other options would need different amounts of cloth: 360 pieces need 45 m, 400 pieces need 50 m and 450 pieces need 56.25 m.

Answer: A

The ascending order is 2/3, 3/4, 5/6, 7/8.

Convert to a common denominator, 24:

\[ \begin{aligned} \frac{2}{3} &= \frac{16}{24},\ \frac{3}{4} = \frac{18}{24} \\ \frac{5}{6} &= \frac{20}{24},\ \frac{7}{8} = \frac{21}{24} \end{aligned} \]

Shortcut: each fraction is 1 minus a unit fraction (\(1 - \frac{1}{3}\), \(1 - \frac{1}{4}\), and so on), and the bigger the denominator, the smaller the gap from 1. The other options misplace 3/4 or 5/6, putting a larger fraction before a smaller one.

Answer: B

The correct order is Litre, decilitre, millilitre.

A decilitre is one tenth of a litre, and a millilitre is one thousandth of a litre, so from largest to smallest the units run litre (1 L), decilitre (0.1 L), millilitre (0.001 L).

The other orders put the millilitre above the decilitre or the litre below one of them, which reverses the prefixes deci (one tenth) and milli (one thousandth).

The prefix deci- means one tenth and milli- one thousandth, so 1 litre = 10 decilitres = 1000 millilitres.

Answer: A

Correct answer: 23

Square both sides of the given equation: \[ \left(x + \frac{1}{x}\right)^{2} = x^{2} + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^{2}} = x^{2} + \frac{1}{x^{2}} + 2. \] So \[ 25 = x^{2} + \frac{1}{x^{2}} + 2 \;\Rightarrow\; x^{2} + \frac{1}{x^{2}} = 23. \] The middle term is always exactly 2 because \( x \) and \( 1/x \) multiply to 1, and that is what makes the family of identities so quick to use. The related results are worth having ready. \( \left(x - \frac{1}{x}\right)^{2} = x^{2} + \frac{1}{x^{2}} - 2 \), so here \( x - \frac{1}{x} = \sqrt{21} \). And \( x^{3} + \frac{1}{x^{3}} = \left(x + \frac{1}{x}\right)^{3} - 3\left(x + \frac{1}{x}\right) = 125 - 15 = 110 \). The distractor 25 is the square of the given value, offered for a reader who forgets to subtract the 2.

Answer: A

Correct answer: 1

Simplify the numerator first. Four copies of 325 added together is \[ 325 + 325 + 325 + 325 = 4 \times 325 = 1300. \] The denominator is also \( 4 \times 325 = 1300 \), so the fraction is \[ \frac{1300}{1300} = 1. \] The point of the question is to be recognised rather than computed: repeated addition of the same quantity is multiplication, so the numerator and denominator are the same expression written two different ways, and the value must be 1 without any arithmetic at all. The distractor 4 is what comes of cancelling only the 325 and forgetting the 4, and 325 of cancelling only the 4. Reading the extent of the division bar carefully is what such questions test - \( \frac{325 \times 4}{4} \times \frac{4}{325} \) is a different expression with the value 4, and the placement of the brackets is all that separates them.

Answer: C

Correct answer: 2 : 3

To combine two ratios sharing a middle term, make that term equal in both. Here \( b \) appears as 4 in the first ratio and 8 in the second, so multiply the first throughout by 2: \[ a : b = 6 : 8, \qquad b : c = 8 : 9. \] The chain now reads \( a : b : c = 6 : 8 : 9 \), so \[ a : c = 6 : 9 = 2 : 3. \] The shortcut is to multiply along the chain: \( a : c = (3 \times 8) : (4 \times 9) = 24 : 36 = 2 : 3 \). The step people omit is reducing the final ratio to its lowest terms, which is why 24 : 36 must become 2 : 3. The same method extends to any number of linked ratios - equalise each shared term in turn, or multiply the corresponding entries along the whole chain and reduce once at the end.

Answer: B

Let the two numbers be x and y.

Given, x + y = 40 and x - y = 4.

Adding the two equations: 2x = 44, so x = 22.

Substituting x = 22 in the first equation: 22 + y = 40, so y = 18.

Their product is x × y = 22 × 18 = 396.

Answer: D

Correct answer: the sum of its digits is divisible by 9

A number is divisible by 9 exactly when the sum of its digits is divisible by 9 - and the test may be applied repeatedly until a single digit remains. For 4,617 the digits sum to 18, which is divisible by 9, so 4,617 is too. The divisibility rules worth having ready form a set. By 2: the last digit is even. By 3: the digit sum is divisible by 3. By 4: the last two digits form a multiple of 4. By 5: the last digit is 0 or 5. By 6: divisible by both 2 and 3. By 8: the last three digits form a multiple of 8. By 10: the last digit is 0. By 11: the alternating sum of the digits is 0 or a multiple of 11. The third option fails because divisibility by 3 and 6 together only guarantees divisibility by 6 - the two share a factor, so their product is not implied.