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Q1. To make a miniature truck, we need tires with a diameter between 1.465 cm and 1.472 cm. Will a tire that is 1.4691 cm in diameter work? Explain why or why not. 

Solution

We must compare and order these decimals to help us solve this problem. Specifically, we need to determine if the third decimal is between the first two. The given diameters are 1.465 cm and 1.472 cm The required measurement of diameter of tire = 1.4691 cm  Let us order these decimals from least to greatest. i.e., 1.465 < 1.4691 < 1.472 Thus, a tire that is 1.4691 cm in diameter will work, since 1.4691 is between 1.465 cm and 1.472 cm.

Q2. In a factory, 9.2 kilograms of pumpkin pie filling is made per minute. How many kilograms of pie filling will be made in 6 minutes?

Solution

Pie filling made in 1 minute = 9.2 kg Pie filling made in 6 minutes = 6 x 9.2 kg = 55.2 kg

Q3. Meena and Seema both got fever. Meena's temperature is 103.1 degrees and Seema's temperature is 0.99 degrees less than Meena. Find Seema's temperature.

Solution

Temperature of Meena = 103.1 degrees Temperature of Seema = 0.99 degrees less than Seema's temperature Thus, to find Seema's temperature, we need to subtract 0.99 from 103.1, i.e., Temperature of Seema = (103.1 - 0.99) = 102.11 degrees

Q4. Five swimmers entered into a competition. Four of the swimmers have had their turns. Their timings are 9.8 sec, 9.75 sec, 9.79 sec, and 9.81 sec. What timing must the last swimmer set in order to win the competition?  

Solution

In order to win the competition, the last swimmer has to set a time less than the least timing set by other contestants.   For that, we have to compare the timings obtained by arranging them in the increasing order of magnitude.   i.e., 9.75 < 9.79 < 9.80 < 9.81   Since the least decimal value is 9.75, the last swimmer must set a time less than 9.75 sec, in order to win.    

Q5. Find the missing factor : (a) 2 x ___ = 0.08 (b) ___ x 0.09 = 0.63 (c) 12 x ___ = 0.6

Solution

Here we have to find the missing factor as a multiple of the other number. (a) 2 x ___ = 0.08 Here if we multiply 2 with 4, we get 8. But since there are 2 decimal places in 0.08, we will have to multiply 2 with a number that has equal no. of decimal places to get the desired product. i.e., 2 x 0.04 = 0.08 (b) ___ x 0.09 = 0.63 Here we already have 2 decimal places in one of the factors. Thus, we need to multiply it with a whole no. only. i.e., 7 x 0.09 = 0.63 (c) 12 x ___ = 0.6 Here the desired factor must have 2 decimal places as 6 is not a multiple of 12 but 60 is. So, 0.6 can be written as 0.60. i.e., 12 x 0.05 = 0.60

Q6. Rani poured 0.75 l of curd in a glass and added 1.25 l of water in it to make a buttermilk. She drank 0.5 l of buttermilk from it. How much litres of buttermilk is left in glass?  

Solution

Here, we need to add the quantity of curd and water in glass and then subtract the quantity drank by Rani. Curd added in glass = 0.75 l Water added in glass = 1.25 l Thus, Quantity in glass = 0.75 + 1.25 = 2.00 l Quantity of Buttermilk drank by Rani = 0.5 l Thus, Quantity of buttermilk left in glass = Quantity in glass - Quantity drank by Rani = 2.00 - 0.50 = 1.50 l  

Q7. Write as a decimal number: (a) Sixty-two and eighteen hundredths  (b) Seventy-four and four tenths (c) Fifty-two and thirty-three hundredths (d) Twenty-seven and six hundredths

Solution

Here, we will use a place value chart. First, we write the whole number part and then the decimal or the fractional part. (a) Tens Ones Tenths Hundredths 6 2 . 1 8 Sixty-two and eighteen hundredths is 62.18 (b) Tens Ones Tenths 7 4 . 4 Seventy-four and four tenths is 74.4 (c) Tens Ones Tenths Hundredths 5 2 . 3 3 Fifty-two and thirty-three hundredths is 52.33 (d) Tens Ones Tenths Hundredths 2 7 . 0 6 Twenty-seven and six hundredths is 27.06

Q8. The highest recorded temperature in Manipur was 107 degrees Fahrenheit on August 27, 1975. The average monthly high temperature is 80.1 degrees Fahrenheit. How many degrees hotter than average was the temperature on August 27, 1975?  

Solution

The highest temperature in Manipur = 107 degrees Average monthly high temperature = 80.1 degrees We have to subtract 80.1 from 107 to get the increase in temperature on August 27, 1975 Thus,  107 - 80.1 =              107.0          -    80.1                          26.9           So, 26.9 degrees Fahrenheit hotter than average was the temperature on August 27, 1975.

Q9. Find which is greater? (a) 0.2 or 0.02 (b) 5.06 or 5.60 (c) 0.9 or 0.99

Solution

To find which is greater, first we have to check the whole number part. If that is equal then check the numbers one by one to the right of decimal point. The greater the number, the greater will be the value of the decimal number. (a) 0.2 or 0.02 In 0.2, the number just after the decimal point (tenths place) is 2, whereas in 0.02, it is 0. Since, 2 > 0, we get 0.2 > 0.02 (b) 5.06 or 5.60 In 5.06, the number just after the decimal point is 0, whereas in 5.60, it is 6. Since, 6 > 0, we get 5.60 > 5.06 (c) 0.9 or 0.99 In 0.9, the number just after the decimal point (tenths place) is 9, whereas in 0.99, it is also 9. Since both are equal, we check the number at the hundredths place of 0.9, i.e., 0 and in 0.99, it is 9. Since, 9 > 0 , we get 0.99 > 0.9

Q10. Saee has saved Rs. 5165.80. She gave away Rs. 1500.50 to charity and bought a new bicycle of Rs. 3125.75. Calculate the money left with Saee. 

Solution

Total saved money = Rs. 5165.80 Money given to charity = Rs. 1500.50 Money spent for bicycle = Rs. 3125.75 Thus, total money spent = Money given to charity + Money spent for bicycle = 1500.50 + 3125.75 = Rs. 4626.25 Money left with Saee = Total saved money - Total spent money = 5165.80 - 4625.25 = Rs. 539.55 Hence, Saee has Rs. 539.55 left with her.

Q11. The sum of three numbers is 14.593. If two of the numbers are 2.831 and 8.04 respectively. Find the third number.  

Solution

Let the unknown no. be p It is given that :-   2.831 + 8.04 + p = 14.593 Thus, in order to find p, we have to add the first two given nos. and subtract the result from the total sum. i.e.,     2.831        + 8.040                 10.871        Now, subtract 10.871 from 14.593   14.593 - 10.871 = 3.722 Thus, the third number is 3.722      

Q12. Add the following: 3.25, 0.075 and 5  

Solution

First, we need to line up the decimals as follows: 3.25 = 3.250 0.075 = 0.075 5 = 5.000 Now, adding them gives    3.250 + 0.075 + 5.000    8.325    

Q13. A rectangle has length 5.6 cm and breadth 3.9 cm. Find the area of the rectangle.  

Solution

Given that :-   Length of the rectangle = 5.6 cm   Breadth of the rectangle = 3.9 cm   We know that :-   Area of the rectangle = length x breadth   = 5.6 x 3.9   Multiply as you would multiply whole numbers   = 56 x 39   = 2184   Count the number of decimal places. There is 1 decimal place in 5.6 and 1 decimal place in 3.9. So, move the decimal point 2 places to the left in the answer.   2184. → 21.84   Thus,area of the rectangle is 21.84 cm²    

Q14. Multiply : (a) 1.55 x 2.6 (b) 0.17 x 4.92

Solution

Multiply as you would multiply whole numbers (b) 1.55 x 2.6 155 x 26 = 4030 Count the number of decimal places. There are 2 decimal places in 1.55 and 1 decimal place in 2.6. So, move the decimal point 3 places to the left in the answer. 4030. → 4.030 Thus, 1.55 x 2.6 = 4.030 (b) 0.17 x 4.92 17 x 492 = 8364 Count the number of decimal places. There are 2 decimal places in 0.17 and 2 decimal places in 4.92. So, move the decimal point 4 places to the left in the answer. 8364. → 0.8364 Thus, 0.17 x 4.92 = 0.8364

Q15. Find the area of a square park with side 45.50 m.

Solution

Length of the side of the park = 45.50 m We know that, a square has all its sides of equal length. Thus, area of the square = (side × side) = 45.50 x 45.50 (Here we can neglect the zeroes on the hundredth place because zeroes at the end of the decimal part has no value) = 45.5 x 45.5 = 2070.25 Thus, area of the square park = 2070.25 m²

Q16. Mary's car gets 44.8 miles per gallon on the highway. If her fuel tank holds 15.4 gallons, then how far can she travel on one full tank of gas?   

Solution

Distance covered by Mary's car with 1 gallon fuel = 44.8 miles   Capacity of her fuel tank = 15.4 gallons   Thus, distance travelled on one full tank of gas = 44.8 x 15.4   Multiply as you would multiply whole numbers    = 448 x 154   = 68992     Move 2 places from right to the left to place the decimal point because there is 1 decimal place in 44.8 and 1 decimal place in 15.4   i.e., 689.92     Thus, she can travel 689.92 miles on one full tank of gas.   

Q17. Set the numbers in the order from the least to the greatest: (a) 8.9, 8.1, 5.3, 5.6, 9.2 (b) 2.25, 2.50, 2.75, 2.20

Solution

First we have to check the whole number part. If that is equal, then check the numbers one by one to the right of decimal point. The greater the number, the greater will be the value of the decimal number. (a) 8.9, 8.1, 5.3, 5.6, 9.2 5.3 < 5.6 < 8.1 < 8.9 < 9.2 (b) 2.25, 2.50, 2.75, 2.20 2.20 < 2.25 < 2.50 < 2.75

Q18. Simplify : (a) 0.2852 x 100 (b) 1.919 x 10 (c) 10.006 x 1000 (d) 63.501 x 100

Solution

In the above cases, we have to move the decimal places to the right as many places as there are zeroes over one. (a) 0.2852 x 100 Since there are 2 zeroes in 100, we move 2 decimal places to the right. 0.2852 x 100 = 28.52 (b) 1.919 x 10 = 19.19 (Since there is 1 zero in 10) (c) 10.006 x 1000 = 10006 (Since there are 3 zeroes in 1000) (d) 63.501 x 100 = 6350.1 (Since there are 2 zeroes in 100)

Q19. Find : (a) 2 x 1.3 (b) 9 x 0.08

Solution

Multiply as you would multiply whole numbers (a) 2 x 1.3 Now, 13 x 2 = 26 Count the number of decimal places in the factors. There is 1 decimal place in 1.3. Move the decimal point 1 place to the left in the answer. 26. → 2.6 Thus, 1.3 x 2 = 2.6 (b) 9 x 0.08 9 x 8 = 72 Count the number of decimal places. There are 2 decimal places in 0.08. Move the decimal point 2 places to the left in the answer. 72. → 0.72 Thus, 9 x 0.08 = 0.72

Q20. A chemist has several beakers full of different liquids that he will use to make a solution. The chemist records the amount of liquid in each beaker: 640.6 milliliters, 908.44 milliliters, 1.5553 liters, and 0.6 liters. How many milliliters of solution will the chemist have after he mixes together the liquids in the beakers?  

Solution

In order to add the measure of solutions, the amount of liquid in each beaker must have same units. Since the measures of all the beakers do not have the same units, we convert liters into milliliters to make all the units similar.  We know that :-        1 liter = 1000 milliliters  So,  1.5553 liters =  1555.30 milliliters  And  0.6 liters = 600 milliliters  Now, adding all the measurements we get,               _{2 1 1 1}                640.60          +    908.44          +  1555.30          +    600.00                          3704.34              Thus, the chemist will get 3704.34 milliliters of solution after he mixes together the liquids in the beakers.