EXERCISE 1.2
1). A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Ans: Number of tickets sold on first day = 1094
Number of tickets sold on second day = 1812
Number of tickets sold on third day = 2050
Number of tickets sold on fourth day = 2751
Therefore total number of tickets sold on all the four days
= 1094 + 1812 + 2050 + 2751
= 7707 tickets
2).Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Ans. Number of runs scored = 6980 runs
He wishes to complete = 10,000 runs
Runs needed to score 10,000 will be = 10,000 – 6980
= 3020 runs.
Shekhar needs 3020 more runs to score 10,000 runs.
3).In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Ans: Number of votes of the successful candidate = 5,77,500
Number of votes of the rival candidate = 3,48,700
Margin by which election was won = 5,77,500 – 3,48,700
= 2,28,800 votes
∴ The successful candidate won by a margin of 2,28,800 votes.
4. Kirti bookstore sold books worth Rs 2,85,891 in the first week of June. The bookstore sold books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Ans: Price of books sold in the first week = Rs 2,85,891
Price of books sold in the second week = Rs 4,00,768
Total sale of books = Rs 2,85,891 + Rs 4,00,768
= Rs 6,86,659
The sale was highest in the second week.
Difference in sale of both weeks = Rs 4,00,768 – Rs 2,85,891
= Rs 1,14,877
∴ The sale was greater in the 2nd week by Rs 1,14,877.
5). Find the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Ans. Greatest number = 76432
Least number = 23467
Difference between the two numbers = 76,432 – 23,467
= 52,965
∴The difference between the two numbers is 52,965.
6). A machine on an average, manufactures 2,625 screws a day. How many screws did it produce in the month of January 2006?
Ans. No of screws manufactured in a day = 2,825
No of days in January = 31 days
∴ No of screws produced in January = 31 x 2,825
= 87, 575 screws
∴ 87, 575 screws were manufactured in the month of January 2006.
7). A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?
Ans. Money with the merchant = Rs 78,592
Cost of one radio set = Rs 1,200
No of radio sets to be purchased = 40
∴ Cost of 40 radio sets = Rs 1,200 x 40
= Rs 48,000
Money left with the merchant = (Money had before) – (Money spent)
= Rs 78, 592 – Rs 48,000
= Rs 30, 592
∴ The amount remaining is Rs 30, 592.
8). A student multiplied 7236 by 65 instead of multiplying by 56. How much was his answer greater than the correct answer?
(Hint : You do not have to do both the multiplication).
Ans. Difference between 65 and 56 = 65 – 56 = 9
The difference between the correct and incorrect answer will be
= 7236 x 9 = 65124
∴ The answer was greater by 65,124 than the correct answer.
9). To stitch a shirt 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
Ans:
i) Total length of cloth = 40 m
Total length in cm = 40 x 100 cm
= 4000 cm
Cloth required for 1 shirt = 2 m 15 cm
= 2 x 100 cm + 15 cm
= 200 + 15
= 215 cm
No of shirts that can be stitched will be = 4000 cm ÷ 215 cm
= 18 shirts
ii) Length of remaining cloth will be
total length of cloth required for shirts = 215 cm x 18
= 3, 870 cm.
Length of cloth remaining = cloth available – cloth used
= 4000 cm – 3870 cm
= 130 cm = 1 m 30 cm
∴ 18 shirts can be stitched out of 40 m and the cloth that will remain will be 1 m 30 cm.
10). Medicine is packed in boxes, each weighting 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Ans. Weight of one box = 4 kg 500 g
= 4 x 1000 g + 500 g
=4, 500 g.
Maximum weight carried by the van = 800 kg
= 800 x 100 g
= 8,00,000 g
No of boxes that can be loaded in the van = 8,00,000 g ÷ 4500 g
= 177 boxes
∴ 177 boxes can be loaded in the van.
11). The distance between the school and the house of a student is 1 km 875 m. Everyday she walks both ways between her school and home. Find the total distance covered by her in six days.
Ans. Distance between the school = 1 km 875 m
And the house = 1000 m + 875 m
= 1875 m
Since she walks both ways in a day
Distance travelled in one day = 1875 x 2
= 3, 750 m
Distance travelled by the student is = 3750 x 6
= 22,500 m
= 22 km 500 m
∴The total distance covered by the student is 22 km 500 m.
12). A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity can it be filled?
Ans. Quantity of curd in the vessel = 4 litres 500 ml
= 4 x 1000 ml + 500 ml
= 4000 ml + 500 ml
= 4500 ml
Capacity of one glass = 25 ml
No of glasses that can be filled with curd = 4500 ml ÷ 25 ml
= 180 glasses
∴ 180 glasses can be filled with curd.