Linear Equations

Exercise 2.2 Part 2

Question 6: Three consecutive integers add up to 51. What are these integers?

Solution: Let first integer = a
Therefore, second consecutive integer `= a + 1`
And, third consecutive integer `= a + 2`
Since, sum of the given three consecutive number = 51

Therefore, `a + (a +1) + (a + 2) = 51`
`⇒ a + a + 1 + a + 2 = 51`
`⇒ a + a + a + 1 + 2 = 51`
`⇒ 3a + 1 + 2 = 51`
`⇒ 3a + 3 = 51`

By subtracting 3 from both sides, we get

`3a + 3 – 3 = 51 – 3`
`⇒ 3a = 51 – 3`
`⇒ 3a = 48`

After dividing both sides by 3, we get

`(3a)/(3)=(48)/(3)`

Or, `a=16`

Now, second consecutive number `= a + 1`
Therefore, by substituting the value of ‘a’, we get
Second consecutive number `= 16 + 1 = 17`

Similarly, after substituting the value of ‘a’ in third consecutive number, we get
Third consecutive number `= a + 2 = 16 + 2 = 18`

Thus, three required consecutive numbers are 16, 17 and 18 Answer


Question 7: The sum of three consecutive multiples of 8 is 888. Find the multiples.

Solution: Let first multiple of `8 = 8a`
Therefore, second consecutive multiple of `8 = 8 ( a + 1)`
And, third consecutive multiple of `8 = 8 ( a + 2 )`

Since, sum of three consecutive multiples of 8 = 888 as given in the question

Therefore, `8a + [8 (a + 1)] + [8 (a + 2)] = 888`
`⇒ 8a + (8a + 8) + (8a + 16) = 888`
`⇒ 8a + 8a + 8 + 8a + 16 = 888`
`⇒ 8a + 8a + 8a + 8 + 16 = 888`
`⇒ 24a + 24 = 888`

By transposing 24 to RHS, we get
`⇒ 24a = 888 – 24`
`⇒ 24a = 864`

After dividing both sides by 24, we get

`(24a)/(24)=(864)/(24)`

Or, `a=(864)/(24)=36`

Now, since first multiple of `8 = 8a`,
Second multiple of `8 = 8(a + 1)`
And third multiple of `8 = 8 (a + 2)`

Thus, by substituting the value of ‘a’ we get

First multiple of `8 = 8a = 8 xx 36 = 288`
Second multiple of `8 = 8 (a + 1) = 8 (36 + 1) = 8 xx 37 = 296`
Third multiple of `8 = 8 (a + 2) = 8 (36 + 2) = 8 xx 38 = 304`

Thus, three required consecutive multiples of 8 = 288, 296 and 304 Answer


Question 8: Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

Solution: Let first integer = a
Therefore, second consecutive integer `= a + 1`
And, third consecutive integer `= a + 2`

Since, according to question, by taking in increasing number, first integer is multiplied by 2, second consecutive integer is multiplied by 3 and third consecutive integer is multiplied by 4. And sum of all the three integers = 74

Therefore,

`(text(First integer) xx 2)` `+ (text(second consecutive integer) xx 3)` `+ (text(third consecutive integer) xx 4) = 74`

Or `=(axx2)+[(a+1)xx3]+[(a+2)xx4]=74`

Or, `=2a+(3a+3)+(4a+8)=74`

Or, `=2a+3a+3+4a+8=74`

Or, `=2a+3a+4a+3+8=74`

Or, `=9a+11=74`

By subtracting 11 from both sides we get:

`9a+11-11=74-11`

Or, `9a=63`

On dividing both sides by 8 we get:

`(9a)/(9)=(63)/(9)`

Or, `a=7`

By substituting the value of ‘a’ the rest two consecutive integers can be obtained.

Second consecutive integer `= a + 1 = 7 + 1 = 8`
Third consecutive integer `= a + 2 = 7 + 2 = 9`

Thus, three required consecutive integers are 7, 8 and 9 Answer

Question 9: The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

Solution: Let the age of Rahul `=5x`

And age of Haroon `=7x`

After 4 years.

Age of Rahul `=5x+4`

Age of Haroon `=7x+4`

As per question, the sum of ages of Rahul and Haroon after 4 years =56 years

So, `(5x+4)+(7x+4)=56`

Or, `5x+7x+4+4=56`

Or, `12x+8=56`

After transposing 8 to RHS we get:

`12x=56-8=48`

On dividing both sides by 12 we get:

`(12x)/(12)=(48)/(12)`

Or, `x=4`

Substituting the value of x we can find the ages of Rahul and Haroon as follows:

Rahul’s age `=5x=5xx4=20` year

Haroon’s age `=7x=7xx4=28` year

Question 10: The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Solution: Let the number of boys `=7x`

And number of girls `=5x`

As per question, number of girls is 8 less than the number of boys.

So, `7x-8=5x`

After transposing -8 to RHS and 5x to LHS we get

`7x-5x=8`

Or, `2x=8`

On dividing both sides by 2 we get:

`(2x)/(2)=8/2`

Or, `x=4`

Substituting the value of x we can find the numbers of boys and girls as follows:

Number of boys `=7x=7xx4=28`

Number of boys `=5x=5xx4=20`

So, strength of class `=28+20=48`

Question 11: Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Solution: Let the age of Baichung’s father = m year
Since, Baichung’s father is 26 year younger than Baichung’s grand father
Therefore, Age of Baichung’s grandfather = Age of Baichung’s father + 26 = m + 26 year
Since, Baichung’s father is 29 year older than Baichung
Therefore, Age of Baichung = Age of Baichung’s father – 29 = m – 29` year

Now, we have
Age of Baichung `= m – 29` year
Age of Baichung’s father = m year
Age of Bahichung’s grandfather `= m + 26` year

According to question, the sum of ages of all the three = 135 year

Therefore,
Age of Baichung + Age of Baichung’s father + Age of Baichung’s grandfather = 135
`⇒ (m – 29) + m + (m + 26) = 135` year
`⇒ m – 29 + m + m + 26 = 135` year
`⇒ m + m + m – 29 + 26 = 135` year
`⇒ 3m – 3 = 135` year

After transposing – 3 to the RHS, we get
`3m = 135 year + 3`
`⇒ 3m = 138` year

After dividing both sides by 3, we get

`(3m)/(3)=(138)/(3)`

Or, `m=46` year

This means age of Baichung’s father = 46 year

Now, by substituting the value of m, we can calculate the age of Baichung and the age of Baichun’s grandfather.

Therefore,
Age of Baichung `= m – 29 = 46 – 29 = 17` year
Age of Baichung’s grandfather `= m + 26 = 46 + 26 = 72` year

Thus,
Age of Baichung = 17 year
Age of Baichung’s father = 46 year
Age of Baichung’s grandfather = 72 year



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