Sunil has ₹ 800 . He spends 1 4 of the money on books and 1 5 of the remaining on a geometry box. Find the fraction of money left with him.

Let's break down the problem step by step: Sunil spends 1 4 of his money on books. This means he spends 1 4 × ₹ 800 = ₹ 200 on books. Now, he has ₹ 800 - ₹ 200 = ₹ 600 left. Sunil spends 1 5 of the remaining money on a geometry box. This means he spends 1 5 × ₹ 600 = ₹ 120 on the geometry box. Now, he has ₹ 600 - ₹ 120 = ₹ 480 left. Find the fraction of money left with Sunil. To express this as a fraction, we need to compare the amount left with the original amount of ₹ 800 . The fraction of money left is ₹ 480 ₹ 800 . To simplify this fraction, we can divide both the numerator and denominator by their greatest common divisor, which is 80 . ₹ 480 ₹ 800 = 480 ÷ 80 800 ÷ 80 = 6 10 = 3 5 Therefore, the fraction of money left with Sunil is 3 5 .

Sunil has ₹ 800 . He spends 1 4 of the money on books and 1 5 of the remaining on a geometry box. Find the fraction of money left with him.

Let's break down the problem step by step: Sunil spends 1 4 of his money on books. This means he spends 1 4 × ₹ 800 = ₹ 200 on books. Now, he has ₹ 800 - ₹ 200 = ₹ 600 left. Sunil spends 1 5 of the remaining money on a geometry box. This means he spends 1 5 × ₹ 600 = ₹ 120 on the geometry box. Now, he has ₹ 600 - ₹ 120 = ₹ 480 left. Find the fraction of money left with Sunil. To express this as a fraction, we need to compare the amount left with the original amount of ₹ 800 . The fraction of money left is ₹ 480 ₹ 800 . To simplify this fraction, we can divide both the numerator and denominator by their greatest common divisor, which is 80 . ₹ 480 ₹ 800 = 480 ÷ 80 800 ÷ 80 = 6 10 = 3 5 Therefore, the fraction of money left with Sunil is 3 5 .

A car runs at a speed of 66 2 3 km per hour. What is the distance covered by the car in 12 hours ?

To find the distance covered by the car in 12 hours, we need to multiply the speed of the car by the time it travels. Given that the car runs at a speed of 66 2 3 km per hour, we can convert this mixed fraction into an improper fraction: 66 2 3 = 66 × 3 + 2 3 = 200 3 km per hour. Now, to find the distance covered in 12 hours, we multiply the speed by the time: Distance = Speed × Time = 200 3 × 12 = 2400 3 = 800 km . Therefore, the car covers a distance of 800 km in 12 hours.

Fifteen boards are stacked on top of each other. The thickness of each board is 4 2 3 cm . Find the height of the stack.

To find the height of the stack, we need to multiply the thickness of each board by the number of boards. Given that the thickness of each board is 4 2 3 cm , we can convert this mixed fraction into an improper fraction: 4 2 3 = 4 × 3 + 2 3 = 14 3 cm . Now, to find the height of the stack, we multiply the thickness by the number of boards: Height = Thickness × Number of Boards = 14 3 × 15 = 70 cm Therefore, the height of the stack is 70 cm .

What should be added to 5 1 2 to get 10 ?

To find the value that should be added to 5 1 2 to get 10 , we can subtract 5 1 2 from 10 : 10 - 5 1 2 = 10 - 5 × 2 + 1 2 = 10 - 11 2 = 20 - 11 2 = 9 2 = 4 1 2 Therefore, 4 1 2 should be added to 5 1 2 to get 10 .

In a division problem the divisor is 4 5 , the quotient is 5 8 of the divisor and the remainder is zero. Find the dividend.

To find the dividend in a division problem, we can use the relationship between the divisor, quotient, and dividend. Given that the divisor is 4 5 and the quotient is 5 8 of the divisor, we can calculate the quotient as follows: Quotient = 5 8 × 4 5 = 20 40 = 1 2 . Since the remainder is zero, we know that the dividend is divisible by the divisor without any remainder. In other words, the dividend is a multiple of the divisor. To find the dividend, we can multiply the quotient by the divisor: Dividend = Quotient × Divisor = 1 2 × 4 5 = 4 10 = 2 5 . Therefore, the dividend in this division problem is 2 5 .