8th Mathematics JSUNIL
TUTORIAL,SAMASTIPUR
Q. Places A and B are 100kms apart on a highway. One car starts from A and another starts from B at the same time. If the car travels in the same direction at different the speeds, they meet in 5hrs.If they travelled towards each other they meet in 1hr.What are the speeds of the two cars?
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Let the speed of 1st car and 2nd car be x km/h and y km/h.
Respective speed of both cars while they are travelling in same direction = (x-y) km/h
Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = ((x + y) km/h
According to the given information,
First case, in the same direction A and B meet in 5hrs
Distance cover to meet = distance x time
5 (x-y) = 100 Þ x – y =100/5 Þ x- y = 20 Þ x = 20+y
Second case, If they travelled towards each other they meet in 1hr,
1(x+y) = 100 Þ 20+y +y =100
2y= 100-20=80
Y=80/2=40
So, x = 100 – 40= 60
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h
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Q. Divide 123 rupees A and B so that B will get as many 25 paise coins as 50 paise coins a gets.
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Let the number of coins with both be x
Then money with A = x × 50 paise = Rs 0.50 x
and money with B = x × 25 paise = Rs 0.25 x
Now A and B have Rs 123 in total ⇒ 0.50 x + 0.25 x = 123 ⇒ 0.75 x = 123 X = 123/0.75=164
Hence money with A = Rs 0.50 × 164 = Rs 82 And money with B = Rs 0.25 × 164 = Rs 41
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Q. A bag contains a number of 10 paise coins, 3 times as many 25 paise coins as 10 paise coins and five more 50 paise coins, than 25 paise coins. If the total value is Rs 120. How many 10 paise coins are there?
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Let the number of 10 paise coins be x.
Then the number of 25 paise coins are 3x.
And the number of 50 paise coins are 5 + 3x.
Amount of :
10 paise coins = Rs 0.10 × x = Rs 0.x
50 paise coins = Rs 0.50 × (5 + 3x) = Rs 2.5 + 1.5x
According to question : Total amount = Rs 120
⇒ 0.x + 0.75x + 2.5 + 1.5x = 120 ⇒ 2.35x = 120 – 2.5 ⇒ 2.35x = 117.5 ⇒ x = 50
Hence, number of 10 paise coins = x = 50.
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Q. In a 2-digit number, the face value of the digit in the tens place is double the face value of the digits in the ones place. If the sum of the two face values is 12, find the 2-digit number.
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Let the face value of the digit in one's place be x.
Then the face value of the digit in ten's place = 2x
and the number is 10 × 2x + x.
The sum of two face values = 12
⇒ 2x + x = 12
⇒ 3x = 12
Þ x = 4
∴ The required number is 10 × 2 × 4 + 4 = 84
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Q. If 11 is subtracted from one-fourth of a certain number the difference is equal to 1 more than one-sixth of that number. Find the number.
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Let the given number be x.
∴The required number is 144.
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Q. The sum of a 2-digit number is 7. When the digits are reversed the number increases by 27. Find the original number.
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Let the number at one's place be x.
then the number at ten's place = (7 – x)
and the number is (7 – x) 10 + x.
Now According to the question
10x + (7 – x) = (7 – x) 10 + x + 27
⇒ 10x + 7 – x = 70 – 10x + x + 27
⇒ 9x + 7 = –9x + 97
⇒ 9x + 9x – 97 – 7
⇒ 18x = 90
∴ The required number is (7 – 5) 10 + 5 = 25
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Q. Two years ago , a man's age was 3 times the square of his son's age. in 3 yrs time, his age will be 4 times his son's age. find their present ages.
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Let the present age of son be x years.
Before two years age of son was (x – 2) years.
It is given that the age of father before two years = 3(x – 2) 2 years.
Thus, the present age of father [3(x – 2)2 + 2] years
After three years; the age of son will be (x + 3) years and the age of father will be [3(x – 2)2 + 2 + 3] years = [3(x – 2)2 + 5] years
It is given that
3(x – 2)2 + 5 = 4 (x + 3)
⇒ 3(x 2 – 4x + 4)+ 5 = 4x + 12
⇒ 3x2 – 16x+ 5 = 0
⇒ 3x2 – 15x– x+ 5 = 0
⇒ 3x (x – 5) – 1 (x– 5) = 0
⇒ (3x – 1) (x – 5) = 0
⇒ 3x – 1 = 0 or x – 5 = 0
⇒ x = 1/3 or 5
If x = 1/3, then x – 2 = –5/3.
This means the age of son was –5/3 years before two years. This is impossible since age of a person cannot be negative.
If x = 5, the present age of son is 5 years and
the present of his father is =3x(5-2) 2 + 2 years = 29 years.
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