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Simplification Questions and Answers

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1.

Find the value of question mark (?) in the following :

\[ \begin{aligned} &1\tfrac{3}{4} + 5\tfrac{1}{8} \\ &- (?) \times \tfrac{4}{3} \div 2\tfrac{2}{3} \\ &= 5\tfrac{3}{4} \end{aligned} \]

Exam Year(s)

CBSE Assistant Secretary - 2024,

Answer: C

The missing value is \(2\frac{1}{4}\).

Follow BODMAS: the multiplication and division on (?) come before the addition and subtraction. First the known terms:

\[ \begin{aligned} 1\tfrac{3}{4} + 5\tfrac{1}{8} &= \tfrac{14}{8} + \tfrac{41}{8} = \tfrac{55}{8} \\ \tfrac{4}{3} \div 2\tfrac{2}{3} &= \tfrac{4}{3} \times \tfrac{3}{8} = \tfrac{1}{2} \end{aligned} \]

So \(\tfrac{55}{8} - \tfrac{1}{2}(?) = 5\tfrac{3}{4} = \tfrac{46}{8}\), giving \(\tfrac{1}{2}(?) = \tfrac{9}{8}\) and \((?) = \tfrac{9}{4} = 2\tfrac{1}{4}\).

The other options do not satisfy the equation; for example \(3\frac{1}{2}\) would make the left side \(5\frac{1}{8}\).

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2.

The value of \(\frac{4}{7} + \frac{5}{14} - \frac{7}{8}\) is :

Exam Year(s)

CTET - 2024,

Answer: A

The value is \(\frac{3}{56}\).

The denominators are 7, 14 and 8. Their LCM is 56, so

\[ \begin{aligned} \frac{4}{7} + \frac{5}{14} - \frac{7}{8} &= \frac{32}{56} + \frac{20}{56} - \frac{49}{56} \\ &= \frac{32 + 20 - 49}{56} = \frac{3}{56} \end{aligned} \]

Each fraction is converted by multiplying top and bottom by the same number: \(\frac{4}{7}\) by 8, \(\frac{5}{14}\) by 4 and \(\frac{7}{8}\) by 7.

\(\frac{5}{56}\), \(\frac{3}{14}\) and \(\frac{5}{28}\) come from conversion slips; \(\frac{3}{14} = \frac{12}{56}\), for instance, is four times the correct value.

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3.

The value of \(31096 \div (2704 \div 2) + 0.6738 + 0.63 + 0.059 - 0.79\) is equal to :

Exam Year(s)

CTET - 2024,

Answer: C

The value is 23.5728.

Work the brackets first, then the division, then add and subtract:

\[ \begin{aligned} 2704 \div 2 &= 1352 \\ 31096 \div 1352 &= 23 \\ 23 + 0.6738 + 0.63 + 0.059 &= 24.3628 \\ 24.3628 - 0.79 &= 23.5728 \end{aligned} \]

The other options differ from 23.5728 by 0.1 or 1, the usual result of misaligning decimal places in the additions.

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4.

Which one of the following is the best approximation of \( \frac{7.1 \times 3.85}{9.09} \)?

Exam Year(s)

CTET - 2018,

Answer: B

The best approximation is \( \frac{7 \times 4}{9} \).

Round each number to the nearest whole number: 7.1 is about 7, 3.85 is about 4, and 9.09 is about 9.

\[ \begin{aligned} \frac{7.1 \times 3.85}{9.09} &\approx \frac{7 \times 4}{9} \\ &\approx 3.1 \end{aligned} \]

The exact value is about 3.007. Rounding 3.85 down to 3 makes the numerator far too small, and rounding 9.09 up to 10 moves the denominator further than necessary, so the options with 3 or 10 are poorer approximations.

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5.

Evaluate :

17.5 × 3 - 21 ÷ 7 - 3 × 12.5

Exam Year(s)

CTET - 2019,

Answer: B

The value is 12.

By the order of operations, multiplication and division are done before subtraction:

\[ \begin{aligned} &17.5 \times 3 - 21 \div 7 - 3 \times 12.5 \\ &= 52.5 - 3 - 37.5 = 12 \end{aligned} \]

The option 52.5 is just the first product. Working strictly from left to right, ignoring the order of operations, gives a different, wrong value.

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6.

The price list of vegetables in a super market is given as follows :

ItemQuantityPrice (₹)
Tomato1 kg40
Potato2 kg25
Carrot250 g20
Bottlegourd1 kg10
Chillies100 g10
Lemon4 pieces10

Sanjay buys \(\frac{1}{2}\) kg tomatoes, 1 kg potatoes, \(\frac{1}{2}\) kg carrot, 250 g chillies and 6 lemons. He gives a note of ₹ 200 to the bill clerk at the counter. How much money will he get back ?

Exam Year(s)

CTET - 2019,

Answer: B

He gets back ₹ 87.50.

Work out each item from the unit prices:

  • \(\frac{1}{2}\) kg tomatoes: 40 ÷ 2 = 20
  • 1 kg potatoes (2 kg cost 25): 12.50
  • \(\frac{1}{2}\) kg carrot (250 g cost 20): 40
  • 250 g chillies (100 g cost 10): 25
  • 6 lemons (4 cost 10): 15

\[ \begin{aligned} \text{Bill} &= 20 + 12.50 + 40 + 25 + 15 = 112.50 \\ \text{Change} &= 200 - 112.50 = 87.50 \end{aligned} \]

The option ₹ 112.50 is the bill itself, not the change; the others come from slips in one item.

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7.

If −12 × (−3) + [20 ÷ (−4) − (−24) ÷ 8] − [16 ÷ (−2)] = (−28 ÷ 7) + x,

then the value of x is

Exam Year(s)

CTET - 2019,

Answer: C

The value of x is 46.

Work out the brackets first, then divisions and multiplications, then additions and subtractions:

\[ \begin{aligned} -12 \times (-3) &= 36 \\ 20 \div (-4) - (-24) \div 8 &= -5 + 3 = -2 \\ 16 \div (-2) &= -8 \\ 36 + (-2) - (-8) &= 42 \end{aligned} \]

The right-hand side is \(-28 \div 7 + x = -4 + x\), so

\[ -4 + x = 42,\quad x = 46 \]

Taking \(-(-8)\) as \(-8\) gives 26 + 4 = 30, and forgetting to move the −4 across gives 42; the options 29, 39 and 47 come from similar sign slips.

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8.

The value of

\[ \begin{aligned} &6\tfrac{2}{3} \div 2\tfrac{1}{2} \times 3\tfrac{3}{4} - 5\tfrac{1}{2} \times 4\tfrac{1}{4} \\ &+ 1\tfrac{2}{3}\left(\tfrac{7}{8} + \tfrac{3}{4} \times \tfrac{2}{3}\right) \end{aligned} \]

is

Exam Year(s)

CTET - 2019,

Answer: A

The value is \(-11\frac{1}{12}\).

Convert the mixed numbers to improper fractions and follow the order of operations, working left to right for ÷ and ×:

\[ \begin{aligned} \tfrac{20}{3} \div \tfrac{5}{2} \times \tfrac{15}{4} &= \tfrac{8}{3} \times \tfrac{15}{4} = 10 \\ \tfrac{11}{2} \times \tfrac{17}{4} &= \tfrac{187}{8} \\ \tfrac{5}{3}\left(\tfrac{7}{8} + \tfrac{1}{2}\right) &= \tfrac{5}{3} \times \tfrac{11}{8} = \tfrac{55}{24} \end{aligned} \]

\[ \begin{aligned} 10 - \tfrac{187}{8} + \tfrac{55}{24} &= \tfrac{240 - 561 + 55}{24} \\ &= -\tfrac{266}{24} = -11\tfrac{1}{12} \end{aligned} \]

The positive option \(11\frac{1}{12}\) loses the sign; the result is negative because the subtracted product, about 23.4, exceeds the other terms.

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9.

If 1001 × 121 = 11 × 12111 − 275 × ___________, then what is the number at the blank space ?

Exam Year(s)

CTET - 2026,

Answer: C

The number at the blank is 44.

Work out both products and find the difference.

\[ \begin{aligned} 1001 \times 121 &= 121121 \\ 11 \times 12111 &= 133221 \\ 275 \times n &= 133221 - 121121 = 12100 \\ n &= 12100 \div 275 = 44 \end{aligned} \]

Check: \(275 \times 44 = 12100\), and \(133221 - 12100 = 121121\). Option 22 is half of the answer and 88 double it, which is what a slip in the division by 275 would give.

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10.

If 112 ones + 12 thousands − 11012 = □ × 25, then the number in the place of □ is :

Exam Year(s)

CTET - 2026,

Answer: D

The number in the box is 44.

Write each part as a number: 112 ones is 112 and 12 thousands is 12,000.

\[ \begin{aligned} 112 + 12000 - 11012 &= 1100 \\ \square \times 25 &= 1100 \\ \square &= 1100 \div 25 = 44 \end{aligned} \]

Check: \(44 \times 25 = 1100\). The trap is reading '12 thousands' as 1,200 or '112 ones' as 1,120; either slip leads to none of the options.

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