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Arithmetic Operations Questions and Answers

Important Instructions
111.

The parking rates of car in a railway station parking are depicted as follows :

(a) Upto 2 hrs - ₹ 50
(b) More than 2 hrs and upto 5 hrs - ₹ 75
(c) After 5 hrs - ₹ 10 per extra hour upto 8 hrs.
(d) More than 8 hrs and upto 12 hrs - ₹ 150
(e) More than 12 hrs and upto 24 hrs - ₹ 250

Rajeev parks his car at 7.00 a.m. and comes back to pick it up at 4.30 p.m. on the same day. How much money does he have to pay ?

Exam Year(s)

CTET - 2019,

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.

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112.

The cost of 8 Samosas is ₹ 156 and that of 4 Idlis is ₹ 98. Sunita has a ₹ 200 note only and buys 3 Samosas and 3 Idlis. How much money is left with her now?

Exam Year(s)

CTET - 2026,

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.

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113.

24 L 650 mL of milk is filled in 29 bottles each of same size. The quantity of milk in 15 such bottles is

Exam Year(s)

CTET - 2026,

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.

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114.

A bag containing potatoes, onions and lemons weighs 46 kg 160 g. The weight of potatoes is 17 kg 115 g and that of onions is 15 kg 325 g. The weight of empty bag is 120 g. If 30 lemons weigh 1 kg, then what is the number of lemons in the bag?

Exam Year(s)

CTET - 2026,

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.

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115.

The following table shows the number of visitors to a library in the first six days of a week :

DaysNumber of visitors
Monday86
Tuesday114
Wednesday141
Thursday95
Friday124
Saturday151

On which day, the number of visitors to the library was 29 more than its previous day?

Exam Year(s)

CTET - 2026,

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.

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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.

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117.

Amita starts from the place A and reaches place B in 4 hours 44 minutes. From there, she returns to the starting place A in 3 hours 24 minutes. If she reaches back at 2.30 pm, then her starting time from place A was :

Exam Year(s)

CTET - 2026,

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.

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118.

Radha bought 9 bags of rice each weighing 9 kg 995 g. Total weight of rice in the bags is :

Exam Year(s)

CTET - 2026,

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.

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119.

If

\[ \begin{array}{r} 2\,A \\ \times\, A \\ \hline B\,2\,A \end{array} \]

where A and B are digits, then

Exam Year(s)

CTET - 2026,

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.

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120.

Asmita reaches school for a meeting 15 minutes before 8.30 am. She reached half an hour earlier than her colleague who is 40 minutes late for meeting. What is the scheduled time of the meeting ?

Exam Year(s)

CTET - 2021,

Answer: D

The meeting was scheduled for 8.05 am.

Work through the times step by step:

\[ \begin{aligned} \text{Asmita arrives} &= 8.30 - 15\ \text{min} = 8.15 \\ \text{Colleague arrives} &= 8.15 + 30\ \text{min} = 8.45 \\ \text{Meeting} &= 8.45 - 40\ \text{min} = 8.05 \end{aligned} \]

The colleague is 40 minutes late, so the meeting began 40 minutes before 8.45 am. Note that Asmita, arriving at 8.15, was herself 10 minutes late. The option 8.15 am is only her arrival time, 8.45 am is the colleague's arrival, and 9.10 am adds the 40 minutes instead of subtracting them.

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