Another Peek Beyond the Point Class 7 Solutions

(a) Here, 6 × 4 tenths = 24 tenths. (b) 7 × 0.3 = 7 × 3 tenths = 21 tenths (c) Here, 9 × 5 hundredths = 45 hundredths

(a) 27.34 × 6 = 164.04. (b) 4.23 × 3.7 = 15.651. (c) 0.432 × 0.23 = 0.09936.

Given: Thejus needs 1.65 m of cloth for a shirt. For 3 shirts, the total cloth needed = 1.65 × 3 = \frac {165}{100} × 3 = \frac {495}{100} = 4.95

Here cost of 1 notebook = ₹ 15.50 Therefore, Cost of 4 notebooks = 4 × 15.50 = \frac{4 \times 1550}{100} = \frac {6200}{100} = ₹ 62 and cost of 1 eraser = ₹ 2.75 Therefore, Cost of 3 erasers = 3 × ₹ 2.75 = \frac{3 \times 275}{100} = \frac {825}{100} = ₹ 8.25 Therefore, Total amount spent = 62 + 8.25 = ₹ 70.25

Thickness of 1 coin = 1.45 mm Total thickness of 36 coins = 36 × 1.45 = \frac{36 \times 145}{100} = \frac {5220}{100} = 52.2 mm Now 10 mm = 1 cm 1 mm = \frac {1}{10} cm Therefore, 52.2 mm = \frac {52.2}{10} = 5.22 cm.

Price of 1 kg of oranges = ₹ 56.50 Price of 2.250 kg of oranges = 56.50 × 2.250 = \frac{5650 \times 2250}{100 \times 1000} = \frac {12712500}{100000} = ₹ 127.125 Now 56.5 × 2.25 = 127.125 Hence, we will get the same product. The zeroes at the end of a decimal do not change its value. As we saw, 56.50 is the same as 56.5, and 2.250 is the same as 2.25.

Hence, the product of the two numbers will be the same.

Profit per notebook = Selling price – wholesale price = 30 – 23.6 = ₹ 6.4 Total profit = Profit per notebook × No. of notebooks = 6.4 × 50 = ₹ 320

(a) Here 18 × 12 = 216 …..(i) Now 18 × 1.2 = \frac{18 \times 12}{10} [Using (i)] = \frac {216}{10} = 21.6 (1 decimal place) (b) 18 × 0.12 = \frac{18 \times 12}{100} (Using (i)) = \frac {216}{100} = 2.16 (2 decimal places) (c) 1.8 × 1.2 = \frac{18}{10} \times \frac{12}{10}=\frac{216}{100} [Using (i)] = 2.16 (2 decimal places) (d) 0.18 × 0.12 = \frac{18}{100} \times \frac{12}{100}=\frac{216}{100 \times 100} [Using (i)] = 0.0216 (4 decimal places) (e) 1.8 × 12 = \frac{18 \times 12}{10}=\frac{216}{10} [Using (i)] = 21.6 (1 decimal place) When multiplying two numbers positive if both numbers are less than 1, their product will also be less than 1.

In (d) and (e) product is less than 1.

Yes, we can find the answer without actual multiplication, only by using decimal place values. Multiplying by a number greater than 1: The product is greater than the original number. Multiplying by a number between 0 and 1: The product is less than the original number. Multiplying two numbers between 0 and 1: The product will be less than both factors, and therefore definitely less than 1.

(a) Greater than 1. (b) Less than 1. (c) Greater than 1. (d) Less than 1.

Completed table: 5.7 gives 57, 570, 5700; 23.02 gives 230.2, 2302, 23020; 0.92 gives 9.2, 92, 920; 0.306 gives 3.06, 30.6, 306; and 24.67 gives 246.7, 2467, 24670 when multiplied by 10, 100, and 1000 respectively.

(a) Given \frac {18}{5} To convert the denominator 5 into 10, multiply both the numerator and Dr by 2. \frac{18 \times 2}{5 \times 2}=\frac{36}{10} = 3.6 Verification 18 ÷ 5 Dividing 1 ten and 8 ones into 5 equal parts. 1 < 5 It means we need to regroup 1 ten as 10 ones, i.e., 10 + 8 = 18 ones 18 ones ÷ 5 3 ones remain.

To divide 3 ones into 5 equal parts. Regroup the 3 ones as 30 Tenths. (Place a decimal while regrouping ones into tenths). 30 Tenths ÷ 5 = 6 Then, 18 ÷ 5 = 3.6 Hence verified. (b) Given \frac {415}{4} To convert the denominator 4 into 100, multiply both the Nr and Dr by 25. \frac{415 \times 25}{4 \times 25}=\frac{10375}{100} = 103.75 Verification By following the steps Therefore, 414 ÷ 4 = 103.75 Hence verified.

(c) Given \frac {1217}{2} To convert the denominator 2 into 10, multiply both the Nr and Dr by 5. \frac{1217 \times 5}{2 \times 5}=\frac{6085}{10} = 608.5 Verification By following the steps: Therefore, 1217 ÷ 2 = 608.5 Hence verified. (d) Given \frac {4827}{8} To convert the denominator 8 into 1000, multiply both the Nr and Dr by 125. \frac{4827 \times 125}{8 \times 125}=\frac{603375}{1000} = 603.375 Verification By following the steps: We get 4827 ÷ 8 = 0603.375 Hence verified.

(a) \frac {1526}{4} By using the Long Division Method: Hence, option (iii) is correct. (b) Given \frac {3567}{8} By using the Long Division Method: Hence, option (iii) is correct.

(a) \frac {132}{4} Therefore, Quotient = 33 (b) \frac {13.2}{4} = \frac {132}{40} Therefore, Quotient = 3.3 (c) Here \frac{1.32}{4}=\frac{132}{400} Therefore, Quotient = 0.33 (d) Here \frac{0.132}{4}=\frac{132}{4000} Therefore, Quotient = 0.033

(a) \frac {126}{8} Hence quotient = 15.75 (b) \frac {12.6}{8} Hence quotient = 1.575 (c) Here 1.26 ÷ 8 Hence quotient = 0.1575 (d) Here 0.126 ÷ 8 Hence quotient = 0.01575 (e) Here 0.0126 ÷ 8 Hence quotient = 0.001575

(a) \frac {2}{5} Multiply both the Nr and Dr by 2. \frac{2}{5} \times \frac{2}{2}=\frac{4}{10} Now, put decimal \frac {4}{10} = 0.4 Hence \frac {2}{5} in decimal form is 0.4. (b) \frac {13}{4} Multiply both the Nr and Dr by 25. \frac{13}{4} \times \frac{25}{25}=\frac{325}{100} Now, put a decimal \frac {325}{100} = 3.25 Hence \frac {13}{4} in decimal form is 3.25.

(c) \frac {4}{50} Multiply both the Nr and Dr by 2. \frac{4}{50} \times \frac{2}{2}=\frac{8}{100} Now, put the decimal \frac {8}{100} = 0.08 Hence \frac {4}{50} in decimal form is 0.08. (d) \frac {5}{8} Multiply both the Nr and Dr by 125. \frac{5}{8} \times \frac{125}{125}=\frac{625}{1000} Now, put the decimal \frac {625}{1000} = 0.625 Hence \frac {5}{8} in decimal form is 0.625.

(a) Here 24.86 ÷ 1.2 Converting division into a fraction, we get \frac{24.86}{1.2}=\frac{2486 \times 10}{12 \times 100}=\frac{2486}{120} Now Therefore, Quotient = 20.7166… (b) Converting division into a fraction, we get Therefore, Quotient = 3.76

Given 156 × 12 = 1872 ……(i) ⇒ 156 = \frac {1872}{12} …….(ii) ⇒ 12 = \frac {1872}{156} …….(iii) (a) Now converting division into a fraction 15.6 × 1.2 = \frac{156 \times 12}{10 \times 10}=\frac{1872}{100} = 18.72 [using (i)] (b) Converting division into a fraction 187.2 ÷ 1.2 187.2 ÷ 1.2 = \frac{187.2}{1.2}=\frac{1872}{12} = 156 [using (ii)] (c) Converting division into a fraction 18.72 ÷ 15.6, we get 18.72 ÷ 15.6 = \frac{18.72}{15.6}=\frac{1872 \times 10}{156 \times 100} = \frac {12}{10} = 1.2 [Using (iii)] (d) Here 0.156 × 0.12 = \frac{156 \times 12}{1000 \times 100} [Using (i)] = \frac{1872}{1000 \times 100} = 0.01872

(a) 1000, because 25 ÷ 1000 = 0.025. (b) 0.1. (c) 10. (d) 0.1. (e) 10. (f) 100.

Converting 2.46 ÷ 1.5 into fraction \frac{2.46}{1.5}=\frac{246 \times 10}{15 \times 100}=\frac{246}{150} Therefore, Quotient = 1.64 (b) Converting 2.46 ÷ 0.15 into a fraction, we get \frac{2.46}{0.15}=\frac{246 \times 100}{15 \times 100}=\frac{246}{15} Therefore, Quotient = 16.4 (c) Converting 2.46 ÷ 0.015 into a fraction, we get \frac{2.46}{0.015}=\frac{246 \times 1000}{15 \times 100}=\frac{2460}{15} Therefore, Quotient = 164 Now \frac{24.6}{1.5}=\frac{246 \times 10}{15 \times 10}=\frac{246}{15} and \frac{2.46}{0.15}=\frac{246 \times 100}{15 \times 100}=\frac{246}{15} Both are the same.

Hence quotient obtained in 24.6 ÷ 1.5 is the same as the quotient obtained in 2.46 ÷ 0.15.

Here total length = 4 m No. of pieces = 5 Length of each piece = \frac{\text { Total length }}{\text { No. of pieces }} = \frac {4}{5} = 0.8 m

Here Perimeter = 208.8 cm No. of sides = 12 Length 0f a side = \frac{\text { Perimeter }}{\text { No. of sides }} = \frac {208.8}{12} = 17.4 cm

Here total quantity of juice = 3 litres No. of friends = 8 Therefore, Juice per friend = \frac {3}{8} litre = \frac {3}{8} × 1000 ml = \frac {3000}{8} = 375 ml

Given total distance = 234.45 km Total petrol = 12.6 litres Hence, the total distance travelled per litre of petrol is 18.607 km.

Total quantity of flour = 13.5 kg No. of students = 15 Flour per student = \frac {13.5}{15} = 0.9 kg Now Each student receives = 0.9 kg.

Given cost of a peanut is ₹ 70.5 per 210 grams. Therefore, Cost per gram = \frac {70.5}{210} = 0.3357 and cost of potato chips is ₹ 33.25 for 110 grams. Therefore, Cost per gram = \frac {33.25}{110} = 0.3023 because 0.3023 < 0.3357. Hence, potato chips are cheaper.

(i) Here number line is divided into 10 equal parts. Difference between 3.2 and 3.1 = 3.2 – 3.1 = 0.1 Value of each mark = \frac {0.1}{10} = 0.01 Now the arrow is on the sixth mark after 3.1 Therefore, Decimal number at the arrow mark = 3.1 + 6 × 0.01 = 3.1 + 0.06 = 3.16 (ii) Here number line is divided into 10 equal parts between 2.15 and 2.17.

Therefore, Difference between 2.17 and 2.15 = 2.17 – 2.15 = 0.02 Value of each mark = \frac {0.02}{10} = 0.002 Arrow is on the sixth mark after 2.15 Therefore, Decimal number at the arrow mark = 2.15 + 6 × 0.002 = 2.150 + 0.012 = 2.162

Given that Shyamala bought 3 kg of bananas at ₹ 30 per kg. Total cost = 3 × 30 = ₹ 90 She sold 35 bananas for ₹ 5 each. Total revenue = 35 × 5 = ₹ 175 Profit = Total revenue – Total cost = 175 – 90 = ₹ 85

Given that, the teacher wanted to place 80 textbooks, each is 2.5 cm thick. Total thickness required = 80 × 2.5 = 200 cm The bookshelf is 160 cm long. The thickness of one book is 2.5 cm. Number of books that can fit = \frac{160 \mathrm{~cm}}{2.5 \mathrm{~cm}} = 64 64 books could be placed on the shelf.

The total thickness of these books = 64 × 2.5 = 160 cm, which is the full length of the shelf. The teacher wanted to place 80 textbooks, but only 64 textbooks can fit. Therefore, there is no space left after placing the maximum number of books.

Here, (i) 1 km = 1000 m Therefore, 5.5 km = 5.5 × 1000 = 5500 m (ii) 1 m = 100 cm Therefore, 35 cm = \frac {35}{100} = 0.35 m (iii) 1 cm = 10 mm Therefore, 14.5 cm = 14.5 × 10 = 145 mm (iv) 1 kg = 1000 g Therefore, 68 g = \frac {68}{1000} = 0068 kg (v) 1 m = 1000 mm Therefore, 9.02 m = 9.02 × 1000 = 9020 mm (vi) 1 l = 1000 ml Therefore, 125.5 ml = \frac {125.5}{1000} = 0.1255 l

Therefore, Quotient = 17.21

(a) Here 1.2 × 2 = 2.4 and 0.4 × 6 = 2.4 (b) Here 2.9 × 5 = 14.5 and 14.5 × 1 = 14.5

(a) 21. (b) 21. (c) 2100. (d) 0.21. (e) 2100. (f) 21. These follow by shifting both dividend and divisor by the same power of 10 before using 756 ÷ 36 = 21.

Here table given here represents division, where the value in each cell is the result of dividing the number in the corresponding ‘a’ column b the number in the corresponding row ‘b’. Given 1517 ÷ 37 = 41 \frac {151.7}{3.7} = 41 \frac {15.17}{0.37} = 41 \frac {1.517}{37} = 0.041

One valid set is: (a) maximum product, 82.0 × 5.4 = 442.8; (b) minimum product, 45.8 × 0.2 = 9.16; (c) a product greater than 150, 85.4 × 2.0 = 170.8; (d) product nearest to 100, 20.5 × 4.8 = 98.4; (e) product nearest to 5, 45.8 × 0.2 = 9.16.

Let A = 245.05, B = 0.942368, C = 7.9682 We note that B < 1 and C > 1 (a) Now A × B = 245.05 × 0.942368 < 245.05 (Therefore, Multiplying a number by a value less than 1 results in a smaller number) (b) A × C = 245.05 × 7.9682 > 245.05 (Multiplying a number by a value greater than 1 results in a larger number) (c) A ÷ C = 2.4505 ÷ 7.9682 < 245.05 (Dividing a number by a value greater than 1 results in a smaller number) (d) A ÷ B = 245.05 ÷ 0.942368 > 245.05 (Dividing a number by a value greater than 1 results in a larger number) (e) Now, expression less than 245.05 Therefore, 0.942368 is closer to 1 than 7.9682 is to 1.

0.942368 will result in a value closer to 245.05 than dividing by 7.9682. Therefore, (c) < (a) Again, 0.942368 is closer to 1 than 7.9682 is to 1. Therefore, Dividing by 0.942368 will result in a value closer to 245.05 than multiplying by 7.9682. Therefore, (d) < (b) Also, (f) 7.9682 is significantly smaller than 245.05; it will be the smallest value.

(f) Therefore, 7.9682 < (e) 245.05 ÷ 7.9682 Combining all, we get (f), (c), (a), (e),(d), (b).