Page 18 - Start Up Mathematics_6

Solution: Cloth required to stitch a shirt = 2 m 15 cm = 215 cm ( 1 m = 100 cm)

Total cloth available for stitching shirts = 40 m
= 4,000 cm 215 4,000 18
Number of shirts that can be stitched – 215

= 4,000 cm ÷ 215 cm 1,850
Since the quotient is 18 and remainder is 130 therefore, 18 shirts – 1,720
130
can be stitched and 130 cm or 1 m 30 cm cloth will remain.

Example 17: The distance between the school and the
house of a student is 1 km 875 m. Everyday
she walks both ways. Find the total distance
covered by her in six days.

Solution: Distance between the school and the house = 1 km 875 m

= 1,875 m ( 1 km = 1,000 m)
Distance covered by the student everyday = 1,875 × 2 3,750
= 3,750 m × 6

Total distance covered in 6 days = 3,750 × 6 = 22,500 m 22,500
Therefore, the total distance covered in 6 days = 22 km 500 m

Example 18: A large tanker of capacity 15 kL 500 L is full of petrol. How many petrol pumps
can be supplied each with a capacity of 720 L of petrol?
Solution: Total petrol in the tanker = 15 kL 500 L = 15,500 L ( 1 kL = 1,000 L)

Capacity of each petrol pump = 720 L
Number of petrol pumps that can be supplied with petrol 72 1,550 21
= 15,500 ÷ 720 – 144
= 1,550 ÷ 72 110

Since the quotient is 21 and remainder is 38 therefore, 21 petrol – 72
38
pumps can be supplied with petrol.

EXERCISE 1.2

1. Add 3,45,634; 2,37,854 and 1,23,00,870.
2. A businessman earned a profit of ` 2,34,561 in the month of January, ` 1,95,467 in the
month of February and ` 3,44,560 in the month of March. What is the total profit earned
in these months?

3. A man bought a piece of land for ` 18,70,900. He paid ` 13,45,670 for the construction
material and ` 2,25,000 to the labour who constructed the building. How much money did
he spend in all?

4. Fill in the blanks with suitable digits at each place.
(a) 100 million = ______________ crore (b) 1 billion = ______________ crore

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