Mental Maths<\/strong><\/p>\n Question 1. Question 2. Multiple Choice Questions<\/strong><\/p>\n Choose the correct answer from the given four options (3 to 14): Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Value Based Questions<\/strong><\/p>\n Question 1. Higher Order Thinking Skills (HOTS)<\/strong><\/p>\n Question 1. Question 2.
\nFill in the blanks:
\n(i) The simplest form of the ratio \\(\\frac { 1 }{ 6 }\\) : \\(\\frac { 1 }{ 4 }\\) is ………
\n(ii) 75 cm : 1.25 m = ……..
\n(iii) If two ratios are equivalent, then the four quantities are said to be in ………
\n(iv) If 8, x, 48 and 18 are in proportion then the value of x is ………
\n(v) If the cost of 10 pencils is \u20b9 15, then the cost of 6 pencils is ……..
\n(vi) If a cyclist is travelling at a speed of 15 km\/h, then the distance covered by him in 20 minutes is ……..
\nSolution:
\n(i) The simplest form of the ratio \\(\\frac { 1 }{ 6 }\\) : \\(\\frac { 1 }{ 4 }\\) is 2 : 3.
\n(ii) 75 cm : 1.25 m = 3 : 5
\n(iii) If two ratios are equivalent,
\nthen the four quantities are said to be in proportion.
\n(iv) If 8, x, 48 and 18 are in proportion then the value of x is
\n8 : x :: 48 : 18
\nx = 3
\n(v) If the cost of 10 pencils is \u20b9 15, then the cost of 6 pencils is
\nCost of 10 pencils = \u20b9 15
\nLet cot of 6 pencils = \u20b9 x
\n10 : 6 15 : x
\n10x = 6 \u00d7 5
\nx = 3
\nCost of 6 pencils = \u20b9 3
\n(vi) If a cyclist is travelling at a speed of 15 km\/ h,
\nthen the distance covered by him in 20 minutes is ……
\nSpeed of a cyclist = 15 km\/h
\nDistance travelled in 20 minutes = \\(\\frac { 20 }{ 60 }\\) \u00d7 15 km = 5 km<\/p>\n
\nState whether the following statements are true (T) or false (F):
\n(i) A ratio is always greater than 1.
\n(ii) Ratio of half an hour to 20 seconds is 30 : 20.
\n(iii) The ratio 5 : 7 is greater than the ratio 5 : 6.
\n(iv) If the numbers 3, 5, 12 and x are in proportion then the value of x is 20.
\n(v) The ratio 3 : 2 and 4 : 5 are equivalent.
\nSolution:
\n(i) A ratio is always greater than 1. (False)
\nCorrect:
\nRatio can be less than 1 or equal to 1.
\nIt is not necessary that it is greater than 1.
\n(ii) Ratio of half an hour to 20 seconds is 30 : 20. (False)
\nCorrect:
\n= 30 \u00d7 60 : 20 = 1800 : 20 = 90 : 1
\n(iii) The ratio 5 : 7 is greater than the ratio 5 : 6. (False)
\nCorrect:
\n\\(\\frac { 5 }{ 7 }\\), \\(\\frac { 5 }{ 6 }\\)
\n\\(\\frac { 30:35 }{ 42 }\\)
\n30 < 35
\n5 : 7 is not greater than the ratio 5 : 6
\n(iv) If the numbers 3, 5, 12 and x are in proportion
\nthen the value of x is 20. (True)
\n3, 5, 12 and x are in proportion then
\nx \u00d7 3 = 5 \u00d7 12
\nx = 20
\n(v) The ratio 3 : 2 and 4 : 5 are equivalent. (False)
\nCorrect:
\n\\(\\frac { 3 }{ 2 }\\) \u2260 \\(\\frac { 4 }{ 5 }\\) as 15 \u2260 8<\/p>\n
\nQuestion 3.
\nA ratio equivalent to 6 : 10 is
\n(a) 3 : 4
\n(b) 18 : 30
\n(c) 12 : 40
\n(d) 5 : 3
\nSolution:
\n6 : 10 = 3 : 5
\n18 : 30 = 3 : 5
\n6 : 40 is equivalent to 18 : 30 (b)<\/p>\n
\nA ratio equivalent to the ratio \\(\\frac { 2 }{ 3 }\\) : \\(\\frac { 3 }{ 4 }\\) is
\n(a) 4 : 6
\n(b) 8 : 9
\n(c) 6 : 8
\n(d) 9 : 8
\nSolution:
\n\\(\\frac { 2 }{ 3 }\\) : \\(\\frac { 3 }{ 4 }\\)
\n\\(\\frac { 8:9 }{ 12 }\\)
\n= 8 : 9 (b)<\/p>\n
\nThe ratio of 75 mL to 3 litres is
\n(a) 25 : 1
\n(b) 40 : 1
\n(c) 1 : 40
\n(d) 3 : 200
\nSolution:
\n75 mL to 3 litres
\n= 75 : 3000
\n= 1 : 40 (c)<\/p>\n
\nThe ratio of the number of sides of a rectangle to the number of edges of a cuboid is
\n(a) 1 : 2
\n(b) 1 : 3
\n(c) 2 : 3
\n(d) none of these
\nSolution:
\nRatio of number of sides of a rectangle
\nto the number of edges of a cuboid = 4 : 12 = 1 : 3 (b)<\/p>\n
\nIn a class of 35 students, the number of girls is 20. The ratio of number of boys to the number of girls in the class is
\n(a) 3 : 4
\n(b) 4 : 3
\n(c) 7 : 4
\n(d) 7 : 3
\nSolution:
\nTotal number of students = 35
\nNo. of girls = 20
\nNo. of boys = 35 – 20 = 15
\nRatio in boys and girls = 15 : 20 = 3 : 4 (a)<\/p>\n
\nThe ratio of number of girls to the number of boys in a class is 6 : 7. If there are 21 boys in the class, then the number of girls in the class is
\n(a) 39
\n(b) 24
\n(c) 18
\n(d) 13
\nSolution:
\nGirls : boys = 6 : 7
\nNo. of boys = 21
\nLet no. of girls = x
\n6 : 7 :: x : 21
\n7 \u00d7 x = 6 \u00d7 21
\nx = 18
\nNo. of boys = 18 (c)<\/p>\n
\nTwo numbers are in the ratio 3 : 5. If the sum of the numbers is 144, then the largest number is
\n(a) 48
\n(b) 54
\n(c) 72
\n(d) 90
\nSolution:
\nRatio between two numbers = 3 : 5
\nSum of numbers 144
\nThen larger number = \\(\\frac { 144 }{ 3+5 }\\) \u00d7 5
\n= \\(\\frac { 144 }{ 8 }\\) \u00d7 5 = 90 (d)<\/p>\n
\nIf x, 12, 8 and 32 are in proportion, then x is
\n(a) 6
\n(b) 4
\n(c) 3
\n(d) 2
\nSolution:
\nx, 12, 8 and 32 are in proportion, then
\nx \u00d7 32 = 12 \u00d7 8
\nx = 3 (c)<\/p>\n
\nIf 3, 12 and x are in continued proportion, then x is
\n(a) 4
\n(b) 6
\n(c) 16
\n(d) 48
\nSolution:
\n3, 12 and x are in continued proportion, then
\n3 : 12 :: 12 : x
\n3x = 12 \u00d7 12
\nx = 48 (d)<\/p>\n
\nIf the weight of 5 bags of sugar is 27 kg, then the weight of one bag of sugar is
\n(a) 5.4 kg
\n(b) 5.2 kg
\n(c) 5.4 kg
\n(d) 5.6 kg
\nSolution:
\nWeight of 5 bags of sugar = 27 kg
\nWeight of one bag of sugar
\n= 5 : 1 :: 27 : x
\n5x = 1 \u00d7 27
\nx = \\(\\frac { 27 }{ 5 }\\) = 5.4 kg (c)<\/p>\n
\nSonali bought one dozen notebooks for \u20b9 66. What did she pay for one notebook?
\n(a) \u20b9 6.50
\n(b) \u20b9 6.60
\n(c) \u20b9 5.60
\n(d) \u20b9 5.50
\nSolution:
\nCost of 12 books = \u20b9 66
\nLet cost of one book = x
\n12 : 1 = 66 : x
\n12 \u00d7 x = 1 \u00d7 66
\nx = 5.5
\nCost of 1 book = \u20b9 5.50 (d)<\/p>\n
\nThe speed of 90 km\/h is equal to
\n(a) 10 m\/sec
\n(b) 18 m\/sec
\n(c) 25 m\/sec
\n(d) none of these
\nSolution:
\nSpeed of 90 km\/h = \\(\\frac { 90\\times 5 }{ 18 }\\) m\/sec
\n= 25 m\/sec (c)<\/p>\n
\nSudhanshu divided his property into two parts in the ratio 8 : 5. If the first part is \u20b9 1,60,000 and second part is donated to an orphanage, find the amount donated to the orphanage. What values are being promoted?
\nSolution:
\nRatio in two parts of a property = 8 : 5
\nFirst part = \u20b9 1,60,000
\nSecond part = \\(\\frac { 160000\\times 5 }{ 8 }\\) = \u20b9 10,00,000
\nIt is a good to donate the needy people and support them.<\/p>\n
\nPresent ages of Rohit and Mayank are in the ratio 11 : 8. 8 years hence the ratio of their ages will be 5 : 4. Find their present ages.
\nSolution:
\nRatio in the present ages of Rohit and Mayank = 11 : 8
\nLet age of Rohit = 11x, then that of Mayank = 8x
\n8 years hence, their ages will be
\nAge of Rohit = 11x + 8 years
\nand age of Mayank = 8x + 8 years
\nand the ratio of their ages after 8 years = 5 : 4
\n\\(\\frac { 11x+8 }{ 8x+8 }\\) = \\(\\frac { 5 }{ 4 }\\)
\n\u21d2 44x + 32 = 40x + 40
\n\u21d2 44x – 40x = 40 – 32
\n\u21d2 4x = 8
\n\u21d2 x = 2
\nPresent age of Rohit = 11x = 11 \u00d7 2 = 22 years
\nand age of Mayank = 8x = 8 \u00d7 2 = 16 years<\/p>\n
\nRatio of length and breadth of a rectangle is 3 : 2. If the length of the rectangle is 5 m more than the breadth, find the perimeter of the rectangle.
\nSolution
\nRatio in length and breadth of a rectangle = 3 : 2
\nLet length = 3x m and Breadth = 2x
\nAlso, l of rectangle is 5 m more than the breadth
\ni.e. 3x = 2x + 5
\n\u21d2 3x – 2x = 5
\n\u21d2 x = 5
\nLength = 3x = 3 \u00d7 5 = 15m
\nBreadth = 2x = 2 \u00d7 5 = 10m
\nPerimeter = 2(Length + Breadth) = 2 (15 + 10) m = 2 \u00d7 25 = 50 m<\/p>\nML Aggarwal Class 7 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"