Samacheer Kalvi Class 7 Maths Term 1 Chapter 1 Number System Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 1 chapter 1 Number System. Here students can easily find all the solutions for Number System Exercise 1.1, 1.2, 1.3, 1.4, 1.5 and 1.6. Also here our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 1 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Number System Exercise 1.1 Solutions :
(1) Fill in the blanks
Solution :
(i) ( -30) + 90 = 60
(ii) (-5) + (-95) = -100
(iii) (-52) + (-52) = -104
(iv) 22 + (-22) = 0
(v) 140 + (-70) = 70
(vi) 20 + 80 + (-100) = 0
(vii) 75 + (-25) = 50
(viii) 171 + (-171) = 0
(ix) [(-3) + (-12) ] + (-77) = (-3) + [ (-12) + (-77)]
(x) (-42) + [ 15 + (-23) ] = [ (-42) + 15 ] + (-23)
(2) Say true or false.
Solution :
(i) The additive inverse of (–32) is (–32)
= False
(ii) (−90)+( −30) = 60
= False
(iii) (−125)+25 = −100
= True
(3) Add the following
Solution :
(iii) (−100) + (−10)
= (−100) + (−10)
= -110
(iv) 20 + (-72)
= 20 + (-72)
= -52
(v) 82 + (-75)
= 82 + (-75)
= 7
(vi) −48 + (-15)
= −48 + (-15)
= -63
(vii) −225 + (−63)
= −225 + (−63)
= – 288
(4) Thenmalar appeared for competitive exam which has negative scoring of 1 mark for each incorrect answer. In paper I she answered 25 questions incorrectly and in paper II, 13 questions incorrectly. Find the total reduction of marks.
Solution :
Given, for 1 incorrect answer = -1 mark
In paper I she answered 25 questions incorrectly.
Thus , paper I = – 25 marks
Also, in paper II, 13 questions incorrectly
So, paper II = -13 marks
Therefore, total
= -25 + (-13)
= -38 marks.
Hence, the total reduction of marks is -38.
(5) In a quiz competition, Team A scored +30,–20, 0 and team B scored –20, 0, +30 in three successive rounds. Which team will win? Can we say that we can add integers in any order?
Solution :
Given, team A scored +30,–20, 0
Thus, total scored = +30 + (–20) + 0 = 10 …. (1)
Also, team B scored –20, 0, +30
So, total scored
= (–20) + 0 +30
= 10 ……. (2)
From 1 and 2, we have both teams are equal scored
Therefore, Team A = Team B
Yes, we say that we can add integers in any order.
(6) Are (11+7) + 10 and 11 + (7+10) equal. mention the property
Solution :
(11 + 7) + 10 = 18 + 10 = 28 … (1)
11 + ( 7 + 10 ) = 11 + 17 = 28 …. (2)
From 1 and 2, we have (11 + 7) + 10 and 11 + ( 7 + 10 ) be equal
It’s Associative Property under Addition.
Objective type questions :
(8) The temperature at 12 noon at a certain place was 18° above zero. if it decreases at the rate of 3° per hour at what time it would be 12° below zero
(i) 12 mid night
(ii) 12 noon
(iii) 10 am
(iv) 10 pm
Solution :
Alternative (iv) is correct.
The time will be 10 pm.
(9) Identify the problem with negative numbers as its answer
(i) −9 + (-5) + 6
(ii) 8 + (-12) – 6
(iii) -4 + 2 + 10
(iv) 10 + (–4) + 8
Solution :
Alternative (i) −9 + (-5) + 6 is correct.
(10) (-10) + (+7) = ____
(i) +3
(ii) –3
(iii) −17
(iv) +17
Solution :
The correct option – (ii)
= (-10) + (+7)
= -3
(11) (-8) + 10 + (-2) = ____
(i) 2
(ii) 8
(iii) 0
(iv) 20
Solution :
(-8) + 10 + (-2) = 0.
Hence, the correct answer to this question is option (iii)
Number System Exercise 1.2 Solutions :
(1) Fill in the blanks :
(i) −44 + __ =−88
(ii) __ − 75 = −45
(iii) __ − (+50) = −80
Solution :
(i) -44 + (-44) = -88
(ii) 30 – 75 = -45
(iii) (-30) – (+50) = -80
(2) Say true or false
(i) (-675) – (-400) = -1075
(ii) 15 – (-18) is the same as 15 + 18
(iii) (-45) – (-8) = (-8) – (-45)
Solution :
(i) → False
(ii) → True
(iii) → False
(3) Find the value of the following
(i) – 3 – ( – 4) using number line
(ii) 7 – (– 10) using number line
(iii) 35 – (-64)
(iv) −200 – (+100)
Solution :
(iii) 35 – (-64)
= 35 + 64
= 99
(iv) −200 – (+100)
= −200 – 100
= – 300
(4) Kabilan was having 10 pencils with him. He gave 2 pencils to Senthil and 3 to Karthik. Next day his father gave him 6 more pencils, from that he gave 8 to his sister. How many pencils are left with him?
Solution :
Initially Kabilan have = 10 pencils
He gave 2 pencils to Senthil and 3 to Karthik
= -2 + (-3)
= -5 pencils
Next day father give him = +6 pencils
Also, he gave sister = -8 pencils
Now, left pencils
= 10 + (-5) + 6 + (-8)
= 5 -2
= 3
Hence, there are 3 pencils are left with him
(5) A lift is on the ground floor. If it goes 5 floors down and then moves up to 10 floors from there. Then in which floor will the lift be?
Solution :
Initially lift = ground floor
Number of floors down = -5
Number of floors move up = +10
Now, -5 + 10
= +5
Hence, lift be at 5 floors from up side of ground floor
(6) When Kala woke up, her body temperature was 102°F. She took medicine for fever. After 2 hours it was 2°F lower. What was her temperature then?
Solution :
At wake up temperature = 102° F
After medicine down temperature = – 2°F
Thus, 102° F – 2°F
= 100°F
Therefore, after medicine Kala temperature is 100°F
(7) What number should be added to (-17) to get (-19)?
Solution :
Let, x be this number.
Now, x + (–17) = (–19)
x = (–19) – (–17)
= (–19) + 17
= – 2
Therefore, a number is (-2)
(8) A student was asked to subtract (–12) from (–47). He got −30.Is he correct? Justify.
Solution :
No, the above statement is not correct.
Justification :
(–47) – (–12)
= (–47) + 12
= -35
Hence, he got (-35).
Objective type question Solutions :
(9) (-5) – ( – 18) = _____
(i) 23
(ii) –13
(iii) 13
(iv) –23
Solution :
(-5) – ( – 18) = 13
So, the correct answer of this question is option (iii)
(10) (-100) – 0 + 100 = _____
(i) 200
(ii) 0
(iii) 100
(iv) −200
Solution :
(-100) – 0 + 100 = 0
So, the correct answer of this question is option (ii)
Number System Exercise 1.3 Solutions :
(1) Fill in the blanks :
(i) −80 × __ = − 80
(ii) (-10) × __ = 20
(iii) (100 ) × __ = −500
(iv) __ × (−9) = -45
(v) __ × 75 = 0
Solution :
(i) −80 × 1 =− 80
(ii) (-10) × -2 = 20
(iii) (100 ) × (-5) = −500
(iv) 5 × (−9) = -45
(v) 0 × 75 = 0
(2) Say true or false
(i) (-15) × 5 = 75
(ii) (-100) × 0 × 20 = 0
(iii) 8 × (−4) = 32
Solution :
(i) → False
(ii) → True
(iii) → False
(3) What will be the sign of the product of the following.
(i) 16 times of negative integer.
Solution :
Since, 16 is even number.
Thus, ‘+’ sign of the product.
(ii) 29 times of negative integer.
Solution :
Since, 29 is odd number.
Thus, ‘-‘ sign of the product.
(4) Find the product of
(i) (-35) × 22
(ii) (-10) × 12 × (-9)
(iii) (-9) × (-8) × (-7) × (-6)
(iv) (-25) × 0 × 45 × 90
(v) (-2) × (+50) × (-25) × 4
Solution :
(i) = (-35) × 22
= – 770
(ii) (-10) × 12 × (-9)
= (-10) × (-108)
= 1080
(iii) = (-9) × (-8) × (-7) × (-6)
= 72 × 42
= 3024
(iv) = (-25) × 0 × 45 × 90
= 0 × 45 × 90
= 0 × 90
= 0
(v) = (-2) × (+50) × (-25) × 4
= (-100) × (-100)
= 10000
(5) Check the following for equality and if they are equal, mention the property
(i) (8 – 13 ) × 7 and 8 – (13 × 7)
(ii) [ (-6) – (+8) ] × (-4) and (-6) – [8 × (-4)]
(iii) 3 × [ (-4) × (-10) ] and [ 3 × (-4) + 3 × (-10)]
Solution :
(i) (8 – 13 ) × 7 = (-5) × 7 = -35 … (1)
8 – (13 × 7) = 8 – 91 = -83 …. (2)
From, 1 and 2 we have
(8 – 13 ) × 7 and 8 – (13 × 7) are not equal.
(ii) [ (-6) – (+8) ] × (-4) = (-14) × (-4) = 56 … (1)
(-6) – [8 × (-4) ] = (-6) – (-32) = (-6) + 32 = 26 … (2)
From 1 and 2, we have
[ (-6) – (+8) ] × (-4) and (-6) – [8 × (-4) ] are not equal
(iii) 3 × [ (-4) + (-10) ] = 3 × (-12) = -42 ….. (1)
[ 3 × (-4) + 3 × (-10) ] = (-12) + (-30) = -42 … (2)
From 1 and 2, we have
3 × [ (-4) × (-10) ] and [ 3 × (-4) + 3 × (-10) ] are equal with property distributive property of multiplication over addition.
(6) During summer, the level of the water in a pond decreases by 2 inches every week due to evaporation. What is the change in the level of the water over a period of 6 weeks?
Solution :
Given, during summer, the level of the water in a pond decreases by 2 inches every week due to evaporation.
Thus, rate of water decreases = -2 inches
Total weeks = 6
Therefore,
= 6 × (-2)
= -12
Hence, the level of the water in a pond decreases by 12 inches over a period of 6 weeks.
Objective type question Solutions :
(8) Which of the following expressions is equal to –30
(i) –20 – (–5 x 2)
(ii) (6 x 10) – (6 x 5)
(iii) (2 x 5) + (4 x 5)
(iv) (–6) x (+5)
Solution :
= (–6) x (+5) = – 30
Hence, the correct answer to this question is option → (iv)
(9) Which property is illustrated by the equation
(i) Commutative
(ii) Closure
(iii) Distributive
(iv) Associative
Solution :
The correct option is – (iii)
Distributive property is illustrated by the equation.
(10) 11 × (–1)= _____
(i) –1
(ii) 0
(iii) +1
(iv) –11
Solution :
11×(–1) = –11
Hence, alternative (iv) –11 is the correct answer to this question.
(11) (-12) × (-9) = ____
(i) 108
(ii) – 108
(iii) +1
(iv) –1
Solution :
(-12) × (-9) = 108
Hence, option (i) 108 is the correct answer of this question.
Number System Exercise 1.4 Solutions :
(1) Fill in the blanks
(i) (-40) ÷ __ = 40
(ii) 25 ÷ __ = -5
(iii) __ ÷ (-4) = 9
(iv) (-62) ÷ (-62) = __
Solution :
(i) (-40) ÷ (-1) = 40
(ii) 25 ÷ (-5) = -5
(iii) (-36) ÷ (-4) = 9
(iv) (-62) ÷ (-62) = 1
(2) Say true or false.
(i) (-30) ÷ (-6) = -6
(ii) (-64) ÷ (-64) is 0
Solution :
(i) False
(ii) False
(3) Find the values of the following
(i) (-75) ÷ 5
(ii) (-100) ÷ (-20)
(iii) 45 ÷ (-9)
(iv) (-82) ÷ 82
Solution :
(i) (-75) ÷ 5 = -15
(ii) (-100) ÷ (-20) = 5
(iii) 45 ÷ (-9) = -5
(iv) (-82) ÷ 82 = -1
(4) The product of two integers is −135. If one number is −15, Find the other integer.
Solution :
Let x be other integer.
Thus,
x = (-135) ÷ (-15)
x = 9
Therefore, other integer is 9.
(5) In 8 hours duration, with uniform decrease in temperature, the temperature dropped 24°. How many degrees did the temperature drop each hour?
Solution :
Given, the temperature dropped 24°
Total duration = 8 hours
Since, In 8 hours duration, with uniform decrease in temperature.
Therefore, 24 ÷ 8
= 3
Hence, 3 degrees did the temperature drop each hour.
(7) A person lost 4800 calories in 30 days. If the calory loss is uniform, calculate the loss of calory per day.
Solution :
Given, a person lost 4800 calories in 30 days.
Since, the calory loss is uniform.
Thus,
= 4800 ÷ 30
= 480 ÷ 3
= 160
Hence, the loss of calorie per day is 160 calories.
(8) Given, 168 × 32 = 5376, find (-5376) ÷ (-32)
Solution :
Given, 168 × 32 = 5376
Thus, 5376 ÷ 32 = 168
Hence, (-5376) ÷ (-32) = 168
(10) (-400) divided into 10 equal parts gives ____
Solution :
(-400) ÷ 10 = -40
Therefore, (-400) divided into 10 equal parts gives -40.
Objective Type Question Solutions :
(11) Which of the following does not represent an Integer
(i) 0 ÷ (-7)
(ii) 20 ÷ (-4)
(iii) (-9) ÷ 3
(iv) (12) ÷ 5
Solution :
(12) ÷ 5 does not represent an Integer.
Hence, the correct answer to this question is option (iv) (12) ÷ 5
Number System Exercise 1.5 Solutions :
(1) One night in Kashmir, the temperature is −5°C. next day the temperature is 9°C. what is the increase in temperature?
Solution :
Given, one night in Kashmir, the temperature is −5°C.
Thus, In night temperature = −5°C.
Next day temperature = 9°C
Therefore,
= 9°C – (−5°C)
= 14 °C
Hence, temperature is increase by 14°C
(2) An atom can contain protons which have a positive charge (+) and electrons which have a negative charge (-). when an electron and a proton pair up, they become neutral (0) and cancel the charge out. now, determine the net charge
(i) 5 electrons and 3 protons → -5 + 3 = −2 that is 2 electrons
(ii) 6 Protons and 6 electrons →
(iii) 9 protons and 12 electrons →
(iv) 4 protons and 8 electrons →
(v) 7 protons and 6 electrons →
Solution :
(i) 5 electrons and 3 protons → -5 + 3 = -2 that is 2 electrons.
(ii) 6 Protons and 6 electrons → +6 + (-6) =0 that is 0 electron.
(iii) 9 protons and 12 electrons → 9+(-12) = – 3 that is 3 electrons.
(iv) 4 protons and 8 electrons → 4 + (-8) = -4 that is 4 electrons.
(v) 7 protons and 6 electrons → 7 + (-6) = 1 that is 1 proton.
(3) Scientists use the Kelvin Scale (K) as an alternative temperature scale to degrees Celsius (°C) by the relation T°C = (T + 273)K. convert the following to kelvin
(i) -275°C
(ii) 45°C
(iii) -400°C
(iv) -273 °C
Solution :
(i) -275°C
Given, temperature -275°C.
We know, T°C = ( T + 273) K
Thus, -275°C = ( -275 + 273) K = -2 K
(ii) 45°C
Given, temperature 45°C.
We know, T°C = ( T + 273) K
Thus, 45°C = ( 45 + 273) K = 318 K
(iii) -400°C
Given, temperature -400°C.
We know, T°C = ( T + 273) K
Thus, -400°C = ( -400 + 273) K = -127 K
(iv) -273 °C
Given, temperature -273°C.
We know, T°C = ( T + 273) K
Thus, -273°C = ( -273 + 273) K = 0 K
(4) Find the amount that is left in the student’s bank account, if he has made the following transaction in a month. his initial balance is 690
(i) Deposit (+) of ₹485
(ii) Withdrawal (−) of ₹500
(iii) Withdrawal ( −) of ₹350
(iv) Deposit (+) of ₹89
(v) If another ₹ 300 was withdrawn, what would the balance be?
Solution :
(i) Deposit (+) of ₹ 485
initial balance is ₹ 690
Since, deposit (+) of ₹ 485
Therefore, ₹690 + ₹485 = ₹1175
∴ ₹1175 amount that is left in the student’s bank account
(ii) Withdrawal (−) of ₹ 500
Now balance is ₹ 1175
Since, withdrawal (−) of ₹ 500
Therefore, ₹1175 – ₹ 500 = ₹675
∴ ₹675 amount that is left in the student’s bank account
(iii) Withdrawal ( −) of ₹ 350
Now balance is ₹ 675
Since, withdrawal (−) of ₹ 350
Therefore, ₹675 – ₹ 350 = ₹325
Hence, ₹325 amount that is left in the student’s bank account
(iv) Deposit (+) of ₹ 89
Now balance is ₹325
Since, deposit (+) of ₹ 89
Therefore, ₹325 + ₹89 = ₹414
∴ ₹414 amount that is left in the student’s bank account
(v) If another ₹ 300 was withdrawn, what would the balance be?
Now balance is ₹414
Since, withdrawal (−) of ₹ 300
Therefore, ₹414 – ₹ 300 = ₹114
∴ ₹114 amount that is left in the student’s bank account
(5) A poet Tamizh Nambi lost 35 pages of his ‘lyrics’ when his file had got wet in the rain. use integers, to determine the following
(i) If Tamizh Nambi wrote 5 page per day, how many day’s work did he lose?
Solution :
Given, a poet Tamizh Nambi lost 35 pages
Total pages lost = 35
Per day wrote = 5
Thus, 35 ÷ 5 = 7
∴ 7 day’s work did he lose
(ii) If four pages contained 1800 characters, (letters) how many characters were lost?
Solution :
Given, a poet Tamizh Nambi lost 35 pages
Total pages lost = 35
Since, four pages contained 1800 characters
Thus, for 4 pages = 1800 characters
So, for 1 page = ( 1800 ÷ 4 ) = 450 characters
Therefore, for total 35 pages = 450 × 35 = 15,750 characters
(iii) If Tamizh Nambi is paid ₹250 for each page produced, how much money did he lose?
Solution :
Given, a poet Tamizh Nambi lost 35 pages
Total pages lost = 35
Since, he is paid ₹250 for each page produced
For 1 page = ₹250 produced
Therefore, for 35 pages = ₹(250 × 35) = ₹8750
∴ ₹8750 money did he lose.
(iv) If Kavimaan helps Tamizh Nambi and they are able to produce 7 pages per day, how many days will it take to recreate the work lost?
Solution :
Given, Kavimaan helps Tamizh Nambi and they are able to produce 7 pages per day
Thus, for per day Kavimaan helps = 7 pages
So, let x days Kavimaan help to produce = 35 pages
Therefore, x = 35 ÷ 7 = 5
∴5 days will it take to recreate the work lost.
(v) Tamizh Nambi pays Kavimaan ₹100 per page for his help. How much money does kavimaan receive?
Solution :
Given, a poet Tamizh Nambi lost 35 pages
Total pages lost = 35
Since, Tamizh Nambi pays Kavimaan ₹100 per page for his help
Thus, for 1 page = ₹100
∴ for 35 pages = ₹(35 × 100) = ₹3500.
(6) Add 2 to me. then multiply by 5 and subtract 10 and divide now by 4 and I will give you 15! who am I
Solution :
Let I am is x.
According to question,
(x + 2) × 5 – 10 /4 = 15
(x + 2) × 5 – 10 = 15×4
(x + 2) × 5 – 10 = 60
(x + 2) × 5 = 60 + 10
(x + 2) × 5 = 70
(x + 2) = 70/5
(x + 2) = 14
x = 14 – 2
x = 12.
Hence, I am 12.
(7) Kamatchi, a fruit vendor sells 30 apples and 50 pomegranates. if she makes a profit of 8 per apple and loss 5 per pomegranate, what will be her overall profit (or) loss?
Solution :
Given, Kamatchi, a fruit vendor sells 30 apples and 50 pomegranates.
Since, she makes a profit of ₹ 8 per apple and loss ₹ 5 per pomegranate
Thus, for 1 apple = ₹8 profit.
So, for 30 apples ₹ (30 × 8) = ₹240 profit
Now, for 1 pomegranate = ₹5 loss
So, for 50 pomegranate = ₹(5 × 50) = ₹250 loss
Therefore, 250 – 240 = 10
Hence, She has ₹ 10 overall lose.
(7) During a drought, the water level in a dam fell 3 inches per week for 6 consecutive weeks. what was the change in the water level in the dam at the end of this period?
Solution :
Given, during a drought, the water level in a dam fell 3 inches per week for 6 consecutive weeks.
Thus, for per week water level = 3 inches fell.
For 6 week water level = 6 × 3 = 18 inches fell.
Hence, the change in the water level in the dam at the end of this period is 18 inches fell.
Number System Exercise 1.6 all Questions Solution
(1) What Should be added to -1 to get 10?
Solution :
Let, x be add
Thus, x + (-1) = 10
x = 10 – (-1)
= 10 + 1
= 11
He hence, 11 be add to -1 to get 10
(2) (Subtract 94860 from (− 86945)?
Solution :
= (−86945) – 94860
= –1,81,805
(4) Find the value of (-25) + 60 + (-95) + (-385)
Solution :
(-25) + 60 + (-95) + (-385)
= 35 + (-95) + (-385)
= (-60) + (-385)
= -445
(5) Find the sum of (-9999) (-2001) and (-5999)
Solution :
(-9999) + (-2001) + (-5999)
= (-12000) + (-5999)
= -17999
(6) Find the product of (-30) × (-70) × 15
Solution :
(-30) × (-70) × 15
= 2100 × 15
= 31500
(7) Divide (−72) by 8
Solution :
= (-72) ÷ 8
= – 9
Check the following for equality
(i) (11+7) + 10 and 11 + (7+10)
Solution :
(11+7) + 10 = 18 + 10 = 28 … (1)
11 + (7 + 10) = 11 + 17 = 28 …. (2)
From 1 and 2, we have
(11+7) + 10 and 11 + (7+10) are equal.
(ii) (8-13) × 7 and 8 – (13 × 7)
Solution :
(8-13) × 7 = (-5) × 7 = -35 …. (1)
8 – (13 × 7) = 8 – 91 = -84 …. (2)
From 1 and 2, we have
(8-13) × 7 and 8 – (13 × 7) are not equal.
(iii) [ (-6) – (+8) ] × (-4) and (-6) – [ 8 × (-4) ]
Solution :
[ (-6) – (+8) ] × (-4) = -14 × (-4) = 56 … (1)
(-6) – [ 8 × (-4) ] = (-6) – (-32) = 26 …(2)
From 1 and 2, we have
[ (-6) – (+8) ] × (-4) and (-6) – [ 8 × (-4) ] are not equal.
(iv) 3 × [ (-4) + (-10) ] and [ 3 × (-4) + 3 × (-10) ]
Solution :
3 × [ (-4) + (-10) ] = 3 × (-14) = -42 … (1)
[ 3 × (-4) + 3 × (-10) ] = -12 + (-30 ) = -42 …. (2)
From 1 and 2, we have
3 × [ (-4) + (-10) ] and [ 3 × (-4) + 3 × (-10) ] are equal.
(10) Kalaivani had ₹ 5000 in her bank account on 01.01.2018. She deposited ₹ 2000 in January and withdrew ₹ 700 in February. What was Kalaivani’s bank balance on 01.04.2018, if she deposited ₹ 1000 and withdrew ₹ 500 in March?
Solution :
Kalaivani had ₹5000 in her bank account on 01.01.2018.
Initially balance in 01.01.2018 = ₹5000
Deposited in January = +₹2000
Withdrew in February = – ₹700
Deposited in March = +₹1000
Withdraw in March = – ₹500
Thus, 5000 +2000 – 700 + 1000 – 500 = 7000 – 700 + 1000 – 500
= 6300 + 1000 – 500
= 7300 – 500
= 6800
Hence, Kalaivani’s bank balance on 01.04.2018 is ₹6800
Challenge Problem Solutions :
(13) Say true or false.
(i) The sum of a positive integer and a negative integer is always a positive integer.
(ii) The sum of two integers can never be zero.
(iii) The product of two negative integers is a positive integer.
(iv) The quotient of two integers having opposite sign is a negative integer.
(v) The smallest negative integer is −1.
Solution :
(i) → False
(ii) → False
(iii) → True
(iv) → True
(v) → False
(14) An integer divided by 7 gives a quotient –3. What is that integer?
Solution :
Let, x be integer
Since, an integer divided by 7 gives a result –3.
Thus, x ÷ 7 = -3
x = -3 × 7 = -21
Hence, the integer is – 21.
(15) Replace the question mark with suitable integer in the equation 72 + (-5) – ? = 72
Solution :
Let, 72 + (-5) – x = 72
x = 72 + (-5) – 72
x = (-5)
Hence, in question mark (-5) is suitable integer.
(18) If the letters in the English alphabets A to M represent the number from 1 to 13 respectively and N represents 0 and the letters O to Z correspond from –1 to 12, find the sum of integers for the names given below
(i) YOUR NAME
(ii) SUCCESS
Solution :
(i) YOUR NAME → sum →
= – 11 – 1 – 7 -4 + 0 + 1 + 13 + 5
= -4
(ii) SUCCESS → sum →
= -5 – 7 + 3 + 3 + 5 – 5 – 5
= -11
(19) From a water tank 100 litres of water is used every day. After 10 days there is 2000 litres of water in the tank. How much water was there in the tank before 10 days?
Solution :
Given, from a water tank 100 litres of water is used every day.
Supposed initially water in a tank = x litters
Every day used water = -100 litres
So, 10 days used water = 10 × (-100) = -1000 litres
Since, after 10 days remaining water = +2000 litres
Therefore, x = 2000 – (-1000)
= 2000 + 1000
= 3000 litres
Hence, 3000 litres water was there in the tank before 10 days.
(20) A dog is climbing down in to a well to drink water. In each jump it goes down 4 steps. The water level is in 20th step. How many jumps does the dog take to reach the water level?
Solution :
Given, a dog is climbing down in to a well to drink water
For each jump it goes down = 4 steps
Let x jumps it goes down = 20th steps
Thus, x = 20 ÷ 4 = 5 jumps
Therefore, 5 jumps does the dog take to reach the water level.
(21) Kannan has a fruit shop. He sells 1 dozen banana at a loss of Rs.2 each because it may get rotten next day. What is his loss?
Solution :
Given, Kannan has a fruit shop
We know 1 dozen = 12
For 1 banana = ₹2 loss
So, for 12 banana = ₹x loss
Thus, ₹ x = 2 * 12 = ₹24
Therefore, Kannan has ₹24 loss
(22) A submarine was situated at 650 feet below the sea level. If it descends 200 feet, what is its new position?
Solution :
According to question,
A submarine was situated at 650 feet below the sea level.
Thus, submarine situated below = -650 feet
It descends = -200 feet
Therefore,
= -650 + (-200)
= – 850
Hence, submarine 850 feet below the sea level
(23) In a magic square given below each row, column and diagonal should have the same sum, Find the values of x, y and z.
Solution :
According to question, in a magic square given below each row, column and diagonal should have the same sum
From, 2nd column
-10 + (-3) +4 = -10 + 1 = -9
Therefore, each row, column and diagonal should have the sum is -9
Thus, from 1st row
1 + (-10) + x = -9
-9 + x = -9
x = -9 + 9 = 0
Now, from 2nd row
y + (-3) + (-2) = -9
y + (-5) = -9
y = -9 + 5 = -4
Also, from 3rd row
-6 + 4 +z = -9
-2 + z = -9
z = -9 + 2
z = -7
Hence, the values of x, y and is 0, -4 and -7 respectively.
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