Arithmetic Operations Questions and Answers

Answer: D

The properties used are commutative and associative both.

Going from (x + 5) + 7y to 5 + (x + 7y) takes two steps:

\[ \begin{aligned} (x + 5) + 7y &= (5 + x) + 7y \quad \text{commutative} \\ &= 5 + (x + 7y) \quad \text{associative} \end{aligned} \]

The order of x and 5 changes, which is the commutative property, and the grouping changes, which is the associative property. No multiplication is distributed over addition, so the distributive property (B, C) plays no part, and commutativity alone (A) would not regroup the terms.

Answer: C

The sum is \( 1\frac{7}{15} \).

Use the common denominator 15:

\[ \begin{aligned} \frac{2}{3} + \frac{4}{5} &= \frac{10}{15} + \frac{12}{15} \\ &= \frac{22}{15} = 1\frac{7}{15} \end{aligned} \]

\( \frac{6}{8} \) is the common error of adding numerators and denominators separately. \( \frac{8}{15} \) multiplies the numerators over the common denominator, and \( 1\frac{6}{15} \) is one fifteenth short of the true sum.

Answer: A

52 packets can be made.

\[ \begin{aligned} 3\frac{1}{4} \div \frac{1}{16} &= \frac{13}{4} \times 16 \\ &= 13 \times 4 = 52 \end{aligned} \]

Each kilogram makes 16 packets, so 3 kg make 48 and the extra quarter kilogram makes 4 more. 48 forgets that quarter kilogram, 64 counts 4 whole kilograms, and 12 multiplies 3 by 4.

Answer: A

The journey took 3 hours 40 minutes.

\[ \begin{aligned} 18{:}40 \to 21{:}40 &= 3\ \text{hours} \\ 21{:}40 \to 22{:}20 &= 40\ \text{minutes} \end{aligned} \]

Subtracting column by column, 20 minutes is less than 40, so an hour is borrowed as 60 minutes: 80 - 40 = 40 minutes and 21 - 18 = 3 hours. '3 hours 80 minutes' is not a proper time, since 80 minutes is more than an hour. Checking the others: starting at 18 : 40, 4 hours 40 minutes would end at 23 : 20 and 3 hours 20 minutes at 22 : 00, not 22 : 20.

Answer: D

The combination that does not work is 6, 3.

\[ 6 \times 200 + 3 \times 300 = 1200 + 900 = 2100\ \text{mL} \]

That overflows the 2000 mL bucket. The other combinations fill it exactly: 1 × 200 + 6 × 300 = 2000; 4 × 200 + 4 × 300 = 2000; and 7 × 200 + 2 × 300 = 2000.

Answer: B

II costs more.

\[ \begin{aligned} \text{I} &= 200 \times 250 = 50000 \\ \text{II} &= 20 \times 12 \times 250 = 240 \times 250 = 60000 \end{aligned} \]

Twenty dozen is 240 items, more than 200 packets at the same price, so II costs ₹10,000 more. The costs are not equal, and they can be calculated from the data given.

Answer: A

The sum is zero.

Pair the terms: \( (1 - 1) + (1 - 1) + \cdots \). With an even number of terms every 1 is matched by a \( -1 \), so each pair is 0 and the total is 0.

With an odd number of terms one extra +1 is left over, giving 1; that is the trap in the +1 option. −1 and 2 cannot be reached by any number of terms.

Answer: C

The value of y is 4.

The two-digit number 7y multiplied by 6 must give a three-digit number with all three digits equal to y. Trying y = 4:

\[ 74 \times 6 = 444 \]

which fits. A short way to see it: 6 × y must end in y, which holds only for y = 0, 2, 4, 6 or 8, and 70 × 6 = 420 means the product is between 420 and 474, so it must be 444.

Check the others: 78 × 6 = 468, 76 × 6 = 456 and 72 × 6 = 432, none of which repeats one digit.

Answer: D

The travel time is 40 hours 15 minutes.

Count in steps from 16:30 on 29 August:

\[ \begin{aligned} \text{to 16:30 on 31 Aug} &= 48\ \text{h} \\ \text{08:45 is } 7\ \text{h}\ 45\ \text{min earlier} &\Rightarrow 40\ \text{h}\ 15\ \text{min} \end{aligned} \]

Equivalently, 7 h 30 min to midnight on 29 August, 24 h on 30 August, and 8 h 45 min on 31 August make 40 h 15 min. The other options come from miscounting a day or the minutes.

Answer: D

The decreasing order is 8500 mm, 80 dm 8 cm, 800 cm 8 mm, 8 m.

Convert all four lengths to centimetres:

• 8500 mm = 850 cm
• 80 dm 8 cm = 800 cm + 8 cm = 808 cm
• 800 cm 8 mm = 800.8 cm
• 8 m = 800 cm

From largest to smallest: 850, 808, 800.8, 800, which is the fourth option. The first option puts 800 cm 8 mm before 80 dm 8 cm, and the others start with a smaller length.