Simplification Questions and Answers

Answer: B

Surbhi is left with ₹ 77.

Find the unit prices first.

\[ \begin{aligned} \text{1 pen} &= 116 \div 8 = 14.50 \\ \text{1 pencil} &= 67.50 \div 15 = 4.50 \end{aligned} \]

\[ \begin{aligned} \text{cost} &= 6 \times 14.50 + 8 \times 4.50 \\ &= 87 + 36 = 123 \\ \text{left} &= 200 - 123 = 77 \end{aligned} \]

So she spends ₹ 123 and keeps ₹ 77. The option ₹ 80 is what rounding the pen price down to ₹ 14 gives (84 + 36 = 120), so the half-rupee in each price matters.

Answer: B

The difference is 31.

Add the two pairs of days from the table and subtract.

\[ \begin{aligned} \text{Tue} + \text{Thu} &= 390 + 411 = 801 \\ \text{Mon} + \text{Fri} &= 382 + 388 = 770 \\ 801 - 770 &= 31 \end{aligned} \]

Wednesday's figure, 405, is not needed. The near options 32 and 33 come from small slips in the addition, so it is worth checking each sum: 390 + 411 = 801 and 382 + 388 = 770.

Answer: D

125 boxes are required.

Since 20 laddoos weigh 1 kg, 50 kg contains

\[ \begin{aligned} 50 \times 20 &= 1000\ \text{laddoos} \\ 1000 \div 8 &= 125\ \text{boxes} \end{aligned} \]

The division is exact, so no part-filled box is left over. 100 boxes would hold only 800 laddoos and 120 boxes 960; 150 boxes would hold 1,200, more than there are.

Answer: B

Statements (i), (iii) and (iv) are true, which is option 2. Work each one out in the same unit.

(i) \(6\) days in minutes: \(6 \times 24 \times 60 = 8640\). \(2\frac{2}{5}\) hours in seconds: \(2.4 \times 3600 = 8640\). Equal, so true.

(ii) \(10\) days in minutes: \(10 \times 1440 = 14400\). \(120\) hours in minutes: \(120 \times 60 = 7200\). Not equal, so false.

(iii) \(2\) hours in seconds: \(2 \times 3600 = 7200\). \(5\) days in minutes: \(5 \times 1440 = 7200\). Equal, so true.

(iv) \(5\) days in hours: \(5 \times 24 = 120\). \(2\) minutes in seconds: \(2 \times 60 = 120\). Equal, so true.

Option 1 and option 3 leave out (iv), and option 4 includes the false (ii) while leaving out (i) and (iv).

Answer: A

The number in the box is 1326.

Evaluate the left side first:

\[ \begin{aligned} 101 \times 210 &= 21210 \\ 120 \times 102 &= 12240 \\ 21210 - 12240 &= 8970 \end{aligned} \]

The right side is \(99 \times 104 = 10296\), so

\[ \begin{aligned} 10296 - \square &= 8970 \\ \square &= 10296 - 8970 = 1326 \end{aligned} \]

Check: \(10296 - 1326 = 8970\). Options 1327, 1329 and 1336 would make the two sides differ by 1, 3 and 10.

Answer: D

The expression equals \(3\dfrac{61}{112}\).

By the order of operations (BODMAS), 'of' is done first, then division and multiplication from left to right, then addition and subtraction.

\[ \begin{aligned} \tfrac{5}{7} \text{ of } \tfrac{35}{12} &= \tfrac{25}{12} \\ \tfrac{25}{12} \div \tfrac{28}{9} &= \tfrac{25}{12} \times \tfrac{9}{28} = \tfrac{75}{112} \end{aligned} \]

\[ \begin{aligned} \tfrac{96}{15} \div \tfrac{8}{5} &= \tfrac{96}{15} \times \tfrac{5}{8} = 4 \\ 4 \times \tfrac{3}{4} &= 3 \end{aligned} \]

\[ \begin{aligned} \tfrac{75}{112} + 3 - \tfrac{7}{56} &= 3 + \tfrac{75 - 14}{112} \\ &= 3\tfrac{61}{112} \end{aligned} \]

The order is what the question tests: 'of' binds the two fractions next to it before the division sign is applied, and the subtraction of \(\frac{7}{56}\) comes last. Working strictly left to right without these rules gives a different value, which is how the other options arise.

Answer: B

The difference is \(\frac{81}{112}\).

The mixed number \(1\frac{2}{7} = \frac{9}{7}\) is the only value above 1, so it is the largest. Among the others, \(\frac{4}{7} \approx 0.571\), \(\frac{9}{16} = 0.5625\) and \(\frac{3}{5} = 0.6\), so \(\frac{9}{16}\) is the smallest.

\[ \begin{aligned} \frac{9}{7} - \frac{9}{16} &= \frac{144 - 63}{112} \\ &= \frac{81}{112} \end{aligned} \]

Taking \(\frac{4}{7}\) as the smallest gives \(\frac{9}{7} - \frac{4}{7} = \frac{5}{7}\), which is not offered. \(\frac{25}{28}\) and \(\frac{31}{56}\) come from pairing the wrong fractions.

Answer: B

The value is \(3\frac{3}{4}\).

The other number is the product divided by the known number.

\[ \begin{aligned} -\frac{40}{3} \div \left(-\frac{5}{2}\right) &= \frac{40}{3} \times \frac{2}{5} \\ &= \frac{16}{3} \end{aligned} \]

Its reciprocal is \(x = \frac{3}{16}\).

\[ \begin{aligned} 8x + 2\tfrac{1}{4} &= \frac{3}{2} + \frac{9}{4} \\ &= \frac{15}{4} = 3\tfrac{3}{4} \end{aligned} \]

Using the other number itself instead of its reciprocal would give \(8 \times \frac{16}{3}\), far larger than any option.

Answer: C

The value of k is 0.1071.

Multiplication comes before addition and subtraction, so first work out \(1.001 \times 1.1\), then simplify the left side.

\[ \begin{aligned} 1.001 \times 1.1 &= 1.1011 \\ 0.139 + 0.75 + 2.105 &= 2.994 \\ 2.994 - 1.1011 &= 1.8929 \end{aligned} \]

\[ \begin{aligned} 1.8929 &= 2 - k \\ k &= 2 - 1.8929 = 0.1071 \end{aligned} \]

Doing the operations strictly left to right, subtracting 1.001 before multiplying, gives a different and wrong value; 0.1075 and the others do not satisfy the equation once the product is taken first.

Answer: B

The value is 64.0.

By the order of operations, carry out multiplication and division first, then subtraction from left to right:

\[ \begin{aligned} 25.3 \times 5 &= 126.5 \\ 35 \div 5 &= 7 \\ 3 \times 18.5 &= 55.5 \\ 126.5 - 7 - 55.5 &= 64.0 \end{aligned} \]

Working strictly from left to right, ignoring the order of operations, gives 283.05 (126.5, then 91.5, 18.3, 15.3 and finally 15.3 × 18.5), which is the trap in option 1.