☰ अधिक माहितीसाठी येथे क्लीक करा

#### Question 1:

In the table below, write the place value of each of the digits in the number 378.025.
 Place Hundreds Tens Units Tenths Hundredths Thousandths 100 10 1 110$\frac{1}{10}$ 1100$\frac{1}{100}$ 11000$\frac{1}{1000}$ Digit 3 7 8 0 2 5 Place value 300 010=0$\frac{0}{10}=0$ 51000=0.005$\frac{5}{1000}=0.005$

 Place Hundreds Tens Units Tenths Hundredths Thousandths 100 10 1 110$\frac{1}{10}$ 1100$\frac{1}{100}$ 11000$\frac{1}{1000}$ Digit 3 7 8 0 2 5 Place value 300 7 × 10 = 70 8 × 1 = 8 010=0$\frac{0}{10}=0$ 2100=0.02$\frac{2}{100}=\mathbf{0}\mathbf{.}\mathbf{02}$ 51000=0.005$\frac{5}{1000}=0.005$

#### Question 2:

Solve.
(1) 905.5 + 27.197
(2) 39 + 700.65
(3) 40 + 27.7 + 2.451

(i)
$\begin{array}{cccccccc}& & \overline{)1}& & & & & \\ & 9& 0& 5& .& 5& 0& 0\\ +& & 2& 7& .& 1& 9& 7\\ & 9& 3& 2& .& 6& 9& 7\\ & & & & & & & \end{array}$
∴ 905.5 + 27.197 = 932.697

(ii)
$\begin{array}{ccccccc}& & 3& 9& .& 0& 0\\ +& 7& 0& 0& .& 6& 5\\ & 7& 3& 9& .& 6& 5\\ & & & & & & \end{array}$
∴ 39 + 700.65 = 739.65

(iii)
$\begin{array}{cccccccc}& & \overline{)1}& \overline{)1}& & & & \\ & & 4& 0& .& 0& 0& 0\\ & & 2& 7& .& 7& 0& 0\\ +& & & 2& .& 4& 5& 1\\ & & 7& 0& .& 1& 5& 1\\ & & & & & & & \end{array}$
∴ 40 + 27.7 + 2.451 = 70.151

#### Question 3:

Subtract.
(1) 85.96 – 2.345
(2) 632.24 – 97.45
(3) 200.005 – 17.186

(i)
$\begin{array}{ccccccc}& 8& 5& .& 9& {}^{5}\overline{)6}& {}^{10}\overline{)0}\\ -& & 2& .& 3& 4& 5\\ & 8& 2& .& 6& 1& 5\\ & & & & & & \end{array}$
∴ 85.96 − 2.345 = 82.615

(ii)
$\begin{array}{ccccccc}& {}^{5}\overline{)6}& {}^{12}\overline{)3}& {}^{11}\overline{)2}& .& {}^{11}\overline{)2}& {}^{14}\overline{)4}\\ -& & 9& 7& .& 4& 5\\ & 5& 3& 4& .& 7& 9\\ & & & & & & \end{array}$
∴ 632.24 − 97.45 = 534.79

(iii)
$\begin{array}{cccccccc}& {}^{1}\overline{)2}& {}^{9}\overline{)0}& {}^{9}\overline{)0}& .& {}^{9}\overline{)0}& {}^{9}\overline{)0}& {}^{15}\overline{)5}\\ -& & 1& 7& .& 1& 8& 6\\ & 1& 8& 2& .& 8& 1& 9\\ & & & & & & & \end{array}$
∴ 200.005 − 17.186 = 182.819

#### Question 4:

Avinash travelled 42 km 365 m by bus, 12 km 460 m by car and walked 640 m. How many kilometres did he travel altogether? (Write your answer in decimal fractions.)

Distance travelled by bus = 42 km 365 m = 42 km + 365 m = 42 km + $\frac{365}{1000}$ km = 42 km + 0.365 km = 42.365 km         (1 km = 1000 m)

Distance travelled by car = 12 km 460 m = 12 km + 460 m = 12 km + $\frac{460}{1000}$ km = 12 km + 0.460 km = 12.460 km

Distance travelled by walking = 640 m = $\frac{640}{1000}$ km = 0.640 km

∴ Total distance travelled altogether

= Distance travelled by bus + Distance travelled by car + Distance travelled by walking

= 42.365 km + 12.460 km + 0.640 km

= 55.465 km

$\begin{array}{cccccccc}& & & \overline{)1}& & \overline{)1}& & \\ & & 4& 2& .& 3& 6& 5\\ & & 1& 2& .& 4& 6& 0\\ +& & & 0& .& 6& 4& 0\\ & & 5& 5& .& 4& 6& 5\\ & & & & & & & \end{array}$
Thus, the total distance travelled by Avinash altogether is 55.465 km.

#### Question 5:

Ayesha bought 1.80 m of cloth for her salwaar and 2.25 m for her kurta. If the cloth costs 120 rupees per metre, how much must she pay the shopkeeper?

Length of cloth bought for salwaar = 1.80 m

Length of cloth bought for kurta = 2.25 m

∴ Total length of cloth bought

= Length of cloth bought for salwaar + Length of cloth bought for kurta

= 1.80 m + 2.25 m

= 4.05 m

$\begin{array}{ccccc}& \overline{)1}& & & \\ & 1& .& 8& 0\\ +& 2& .& 2& 5\\ & 4& .& 0& 5\\ & & & & \end{array}$
Rate of cloth = Rs 120/m

∴ Amount paid to the shopkeeper

= Total length of cloth bought × Rate of cloth

=  4.05 m × ₹ 120/m

= ₹ 486

$\begin{array}{cccccc}& & & 4& 0& 5\\ & & ×& 1& 2& 0\\ & & & 0& 0& 0\\ & & 8& 1& 0& ×\\ & 4& 0& 5& ×& ×\\ & 4& 8& 6& 0& 0\\ & & & & & \end{array}$
Thus, Ayesha paid ₹ 486 to the shopkeeper.

#### Question 6:

Sujata bought a watermelon weighing 4.25 kg and gave 1 kg 750g to the children in her neighbourhood. How much of it does she have left?

Total weight of the watermelon = 4.25 kg

Weight of the watermelon given to the children = 1 kg 750 g = 1 kg + 750 g = 1 kg + $\frac{750}{1000}$ kg = 1 kg + 0.75 kg = 1.75 kg

∴ Weight of the watermelon left with her

= Total weight of the watermelon − Weight of the watermelon given to the children

= 4.25 kg − 1.75 kg

= 2.5 kg

$\begin{array}{ccccc}& {}^{3}\overline{)4}& .& {}^{12}\overline{)2}& 5\\ -& 1& .& 7& 5\\ & 2& .& 5& 0\\ & & & & \end{array}$
Thus, the weight of watermelon left with Sujata is 2.5 kg.

#### Question 7:

Anita was driving at a speed of 85.6 km per hour. The road had a speed limit of 55 km per hour. By how much should she reduce her speed to be within the speed limit?

Original driving speed of Anita = 85.6 km/h

Speed limit for driving = 55 km/h

∴ Speed reduced by her to be within the speed limit

= Original driving speed of Anita − Speed limit for driving

= 85.6 km/h − 55 km/h

= 30.6 km/h

$\begin{array}{ccccc}& 8& 5& .& 6\\ -& 5& 5& .& 0\\ & 3& 0& .& 6\\ & & & & \end{array}$
Thus, Anita needs to reduce her speed by 30.6 km/h to be within the speed limit.
Std 6 th Mathematics- 5. Decimal Fractions page1 Reviewed by Amol Uge on January 08, 2019 Rating: 5