Chapter 4: Another Peek Beyond the Point (Decimals)

Spices (1 kg = 1000 g): cinnamon $\frac{50}{1000}$ kg = 0.05 kg; cumin $\frac{100}{1000}$ kg = 0.1 kg; cardamom $\frac{25}{1000}$ kg = 0.025 kg; pepper $\frac{250}{1000}$ kg = 0.25 kg.

0.3, 0.04, 0.067, 4.57, 0.71, 0.43, 0.09; spices 0.05 kg, 0.1 kg, 0.025 kg, 0.25 kg.

0.0847, 1.73, 0.023

Put a decimal point at the end of the number, then move it left by as many places as there are zeroes in the divisor, adding zeroes in front if needed: $123÷10=12.3$, $24÷100=0.24$, $678÷1000=0.678$.

Move the decimal point left by the number of zeroes (adding zeroes in front if needed).

Yes to both: $2.5\times 0.4=1$, $1.25\times 0.8=1$; and $0.5\times 4=2$, $2.25\times 8=18$.

$5.96\times 24.8$: 2 + 1 = 3 decimal places, so 147.808.

Yes (e.g. 2.5 × 0.4 = 1, 0.5 × 4 = 2); 5.96 × 24.8 = 147.808.

(a) 24 tenths = 2.4 (b) 21 tenths = 2.1 (c) 45 hundredths = 0.45

(a) 2.4 (b) 2.1 (c) 0.45

(a) $2734\times 6=16404$ → 164.04

(b) $423\times 37=15651$ → 3 decimal places → 15.651

(c) $432\times 23=9936$ → 5 decimal places → 0.09936

(a) 164.04 (b) 15.651 (c) 0.09936

$1.65\times 3=$ 4.95 m

4.95 m

$4\times 15.50+3\times 2.75=62+8.25=$ ₹70.25

₹70.25

$36\times 1.45=52.2$ mm $=$ 5.22 cm

5.22 cm

$56.5\times 2.25=127.125$, so the cost is ₹127.125 (about ₹127.13).

Yes: zeroes at the end of the decimal part do not change the value (56.50 = 56.5, since 50 hundredths = 5 tenths), so the product is the same.

₹127.125 (≈ ₹127.13); yes, trailing zeroes do not change the numbers.

Profit per book $=30-23.6=6.4$; on 50 books $=6.4\times 50=$ ₹320.

₹320

(a) 21.6 (b) 2.16 (c) 2.16 (d) 0.0216 (e) 0.000216 (f) 21.6

Less than 1: (d) and (e).

21.6, 2.16, 2.16, 0.0216, 0.000216, 21.6; (d) and (e) are less than 1.

When both numbers are between 0 and 1, the product is less than both, hence less than 1: (b) (0.42) and (d) (0.0042). (a) and (c) are 4.2.

(b) and (d)

Move the decimal point right by 1, 2, 3 places (see table).

0.039 m $=3.9$ cm $=39$ mm.

3.9 cm = 39 mm; move the point left 1, 2, 3, 4 places (table).

(a) $\frac{18\times 2}{5\times 2}=\frac{36}{10}=$ 3.6 (b) $\frac{415\times 25}{100}=\frac{10375}{100}=$ 103.75 (c) $\frac{6085}{10}=$ 608.5 (d) $\frac{4827\times 125}{1000}=\frac{603375}{1000}=$ 603.375

(Long division gives the same: e.g. $4827÷8$: 603 remainder 3; 30 tenths ÷ 8 = 3 rem 6; 60 hundredths ÷ 8 = 7 rem 4; 40 thousandths ÷ 8 = 5: 603.375.)

(a) 3.6 (b) 103.75 (c) 608.5 (d) 603.375

(a) $1526÷4=381.5$: (iii) (b) $3567÷8=445.875$: (iii)

(a) (iii) 381.5 (b) (iii) 445.875

(a) 33 (b) 3.3 (c) 0.33 (d) 0.033

33, 3.3, 0.33, 0.033

(a) 15.75 (b) 1.575 (c) 0.1575 (d) 0.01575 (e) 0.001575

15.75, 1.575, 0.1575, 0.01575, 0.001575

$10÷9=1.111\dots$ (remainder 1 every time); $100÷11=9.0909\dots$ (remainders 1 and 10 alternate).

In $1÷7$ the remainder is always one of 1, 2, 3, 4, 5, 6 (never 0), so within at most 6 steps a remainder must repeat, and from then on the same steps repeat forever: $\frac{1}{7}=0.142857142857\dots$

Multiplying by 1 to 6 gives the same digits in the same cyclic order, just starting at a different place; multiplying by 7 gives 999999.

$1÷17=0.05882352941176470588\dots$, so 0588235294117647 is another cyclic number (16 digits).

1.111…, 9.0909…; the remainders 1–6 must repeat; 142857 × 1…6 cycles its digits and × 7 = 999999; 1/17 gives the cyclic number 0588235294117647.

No, only when the divisor is less than 1. For example $128÷0.4=320$ (greater), but $128÷2.5=51.2$ (smaller).

Quotient > dividend only when the divisor is between 0 and 1; equal when it is 1; smaller when it is more than 1.

(a) 0.4 (b) 3.25 (c) 0.08 (d) 0.625

(a) 0.4 (b) 3.25 (c) 0.08 (d) 0.625

(a) $\frac{248.6}{12}=20.71666\dots$ (the 6 repeats), i.e. about 20.72.

(b) $\frac{572.8}{152}=3.7684\dots$ (it does not end), i.e. about 3.77.

(a) 20.7166… ≈ 20.72 (b) 3.7684… ≈ 3.77

(a) 18.72 (b) 156 (c) 1.2 (d) 0.01872

(a) 18.72 (b) 156 (c) 1.2 (d) 0.01872

(a) 1000 (b) 0.1 (c) 10 (d) 0.1 (e) 10 (f) 100

(a) 1000 (b) 0.1 (c) 10 (d) 0.1 (e) 10 (f) 100

(a) 1.64 (b) 16.4 (c) 164

Yes, $24.6÷1.5=16.4=2.46÷0.15$: multiplying both the dividend and the divisor by 10 does not change the quotient.

(a) 1.64 (b) 16.4 (c) 164; yes, both are 16.4.

$4÷5=$ 0.8 m (80 cm)

0.8 m

$208.8÷12=$ 17.4 cm

17.4 cm

$3÷8=0.375$ L $=$ 375 mL

375 mL

$234.45÷12.6=18.607\dots$, i.e. about 18.61 km per litre.

About 18.61 km per litre.

$13.5÷15=$ 0.9 kg (900 g)

0.9 kg

$\frac{1}{32}=0.03125$ and $\frac{1}{3125}=0.00032$.

Pattern: the digits of $\frac{1}{2\times \cdots \times 2}$ are the powers of 5 (5, 25, 125, 625, 3125), and the digits of $\frac{1}{5\times \cdots \times 5}$ are the powers of 2 (2, 4, 8, 16, 32), with as many decimal places as factors.

Why: $2\times 5=10$. So $\frac{1}{2^n}=\frac{5^n}{2^n\times 5^n}=\frac{5^n}{10^n}$ and $\frac{1}{5^n}=\frac{2^n}{10^n}$.

0.03125 and 0.00032; since 2 × 5 = 10, 1/2ⁿ = 5ⁿ/10ⁿ and 1/5ⁿ = 2ⁿ/10ⁿ.

100 years, one leap day every 4 years: $100\times 365+25=36,525$, or $75\times 365+25\times 366=36,525$, or $25\times (3\times 365+366)=36,525$ days.

10,000 years (Gregorian rule): leap years $=2500-100+25=2425$ (divisible by 4, minus those divisible by 100, plus those divisible by 400).

Calendar days $=10,000\times 365+2425=36,52,425$.

Earth's 10,000 revolutions $=10,000\times 365.2422=36,52,422$ days.

The calendar runs about 3 days ahead in 10,000 years. A fix: also make years divisible by 4000 ordinary years (removes 2 more days), leaving only about 1 day of difference in 10,000 years.

(The book's "364.2422 days" in one question should read 365.2422.)

36,525 days in 100 years; in 10,000 years the calendar has 36,52,425 days against 36,52,422, about 3 days too many. Skipping the leap day in years divisible by 4000 would fix most of it.

Cost per gram: chikki $70.5÷210\approx 0.336$ rupee; chips $33.25÷110\approx 0.302$ rupee. The potato chips are cheaper per gram.

The chips (≈ ₹0.30 per g against ≈ ₹0.34 per g).

First line: each part $=0.01$; the arrow is at the 6th mark: 3.16.

Second line: each part $=0.02÷10=0.002$; the arrow is at the 6th mark: $2.15+0.012=$ 2.162.

3.16 and 2.162

Cost = $3\times 30$ = ₹90. Selling price = $35\times 5$ = ₹175. Profit = ₹175 − ₹90 = ₹85.

₹85

$160÷2.5=64$ books fill the shelf exactly. So 64 books can be placed, no space is left, and 16 books will not fit.

64 books; no space left (16 books do not fit).

5500 m; 0.35 m; 145 mm; 0.068 kg; 9020 mm; 0.1255 l

5500, 0.35, 145, 0.068, 9020, 0.1255

$6.25÷2.5=62.5÷25=$ 2.5

$60.25÷3.5=602.5÷35=17.2142857142857\dots$, i.e. $17\frac{3}{14}$. The decimal never ends (look: the block 142857 repeats!), about 17.21.

2.5; and 17.2142857… = 17 3/14.

(a) $1.2\times 2$; $0.6\times 4$; $0.8\times 3$; $4.8\times 0.5$

(b) $2.9\times 5$; $1.45\times 10$; $0.5\times 29$; $2.5\times 5.8$

E.g. (a) 1.2 × 2, 0.6 × 4 (b) 2.9 × 5, 1.45 × 10

(a) 21 (b) 21 (c) 2100 (d) 0.21 (e) 2100 (f) 21

21, 21, 2100, 0.21, 2100, 21

See the table (every cell is 41 with the decimal point shifted).

(A computer check of all arrangements was used.)

(a) 82.0 × 5.4 = 442.8

(b) 45.8 × 0.2 = 9.16

(c) e.g. 84.5 × 2.0 = 169, 54.0 × 2.8 = 151.2

(d) 20.5 × 4.8 = 98.4

(e) The smallest possible product is 9.16 (from (b)), so 45.8 × 0.2 = 9.16 is nearest to 5.

(a) 82.0 × 5.4 = 442.8 (b) 45.8 × 0.2 = 9.16 (c) e.g. 84.5 × 2.0 = 169 (d) 20.5 × 4.8 = 98.4 (e) 45.8 × 0.2 = 9.16

Multiplying by a number below 1 makes 245.05 smaller; dividing by a number below 1 makes it larger. Approximately: (a) 230.9 (b) 1952.6 (c) 30.75 (d) 260.0.

Increasing order: (f) < (c) < (a) < (e) < (d) < (b)

(f), (c), (a), (e), (d), (b)

Puzzle 1 (5 × 5, numbers 1 to 25):

Puzzle 3 (the "tree", numbers 1 to 56):

Puzzle 2 (the one with 1 and 39): as printed in our copy, this grid has only 37 squares but must hold the numbers 1 to 39, so it cannot be completed exactly. (A computer search also shows that 29 cannot be placed next to 30 in that layout.) Check the grid in your book against the printed numbers, or ask your teacher for a corrected version.

Puzzles 1 and 3 are solved above (both have a single solution); puzzle 2, as printed, has 37 squares for 39 numbers and cannot be completed.