Arithmetic Operations Questions and Answers

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.

Answer: B

He has to pay ₹ 150.

From 7.00 a.m. to 4.30 p.m. is 9 hours 30 minutes. That falls in slab (d), more than 8 hours and up to 12 hours, which is charged a flat ₹ 150.

The options ₹ 130 and ₹ 135 come from wrongly extending slab (c), adding ₹ 10 an hour to ₹ 75 beyond its 8-hour limit, and ₹ 100 matches no slab. Slab (c) stops at 8 hours, after which the flat rate of slab (d) applies.

Answer: C

Sunita has ₹ 68 left.

Find the price of one of each item first.

\[ \begin{aligned} 1 \text{ samosa} &= 156 \div 8 = 19.50 \\ 1 \text{ idli} &= 98 \div 4 = 24.50 \\ 3 \times (19.50 + 24.50) &= 3 \times 44 = 132 \\ 200 - 132 &= 68 \end{aligned} \]

So she spends ₹ 132 and keeps ₹ 68. The half-rupee unit prices are the trap: rounding them before multiplying, or multiplying the price of 8 samosas by 3, gives an amount other than ₹ 68.

Answer: B

15 bottles hold 12 L 750 mL.

Convert to millilitres, find one bottle's share, then scale up.

\[ \begin{aligned} 24 \text{ L } 650 \text{ mL} &= 24650 \text{ mL} \\ 1 \text{ bottle} &= 24650 \div 29 = 850 \text{ mL} \\ 15 \text{ bottles} &= 15 \times 850 = 12750 \text{ mL} \end{aligned} \]

12750 mL is 12 L 750 mL. The other options are not multiples of 850 mL by 15, so they cannot be the contents of 15 equal bottles.

Answer: C

There are 408 lemons in the bag.

Work in grams and subtract everything that is not lemons from the total.

\[ \begin{aligned} \text{total} &= 46160 \text{ g} \\ \text{others} &= 17115 + 15325 + 120 \\ &= 32560 \text{ g} \\ \text{lemons} &= 46160 - 32560 = 13600 \text{ g} \end{aligned} \]

Since 30 lemons weigh 1000 g, 13.6 kg holds \(13.6 \times 30 = 408\) lemons.

Forgetting to subtract the 120 g bag gives 13720 g, about 411.6 lemons, which is not even a whole number; the empty bag's weight must be removed along with the potatoes and onions.

Answer: C

The day is Friday.

Compare each day with the day before it:

• Tuesday: 114 − 86 = 28
• Wednesday: 141 − 114 = 27
• Thursday: 95 − 141, a fall
• Friday: 124 − 95 = 29
• Saturday: 151 − 124 = 27

Only Friday is exactly 29 more than the previous day. Tuesday (28) is the near miss, and Wednesday and Saturday are each 27 more.

Answer: C

The descending order is \(\frac{4}{5}, \frac{8}{11}, \frac{5}{9}, \frac{3}{7}\).

Compare the fractions as decimals:

\[ \begin{aligned} \tfrac{4}{5} &= 0.8 \\ \tfrac{8}{11} &\approx 0.727 \\ \tfrac{5}{9} &\approx 0.556 \\ \tfrac{3}{7} &\approx 0.429 \end{aligned} \]

Each value is smaller than the one before, so this order is descending. The first option is the same list in ascending order. The second puts \(\frac{4}{5}\) after smaller fractions, and the fourth puts \(\frac{4}{5}\) second and \(\frac{8}{11}\) last, so neither is in order.

Answer: B

She started at 6.22 am.

Total time for the round trip:

\[ \begin{aligned} 4\text{ h }44\text{ min} + 3\text{ h }24\text{ min} &= 7\text{ h }68\text{ min} \\ &= 8\text{ h }8\text{ min} \end{aligned} \]

Counting back 8 h 8 min from 2.30 pm: 8 hours earlier is 6.30 am, and 8 minutes earlier still is 6.22 am.

6.24 am and 6.44 am come from mishandling the minutes (68 minutes is 1 h 8 min), and 6.00 am ignores the minutes altogether.

Answer: C

The total weight is 89.955 kg.

Convert each bag to kilograms: 9 kg 995 g = 9.995 kg. Then multiply by 9:

\[ \begin{aligned} 9 \times 9.995 &= 9 \times (10 - 0.005) \\ &= 90 - 0.045 = 89.955\ \text{kg} \end{aligned} \]

A quick check: each bag is just 5 g short of 10 kg, so nine bags are 45 g short of 90 kg. 89.595 kg and 89.795 kg come from misplacing digits, and 89.985 kg would mean only 15 g short in total.

Answer: D

The digits are A = 5, B = 1.

The two-digit number 2A times A ends in A, so \(A \times A\) must end in the digit A. Only 0, 1, 5 and 6 have that property.

\[ \begin{aligned} A = 5 &: 25 \times 5 = 125 \\ A = 6 &: 26 \times 6 = 156 \\ A = 1 &: 21 \times 1 = 21 \end{aligned} \]

Only 125 has the form B2A, with middle digit 2 and last digit 5, giving \(B = 1\). The product 156 has middle digit 5, so A = 6 fails, and A = 1 gives a two-digit product, so no option with A = 1 can work.