Q8. Mrs. Jain is 27 years older than her daughter Nilu. After 8 years she will be twice as old as Nilu. Find their present ages.
Sol. Let’s assume that the present age of Nilu be x years.
$$ \therefore \text {The present age of Nilu’s mother }= (x+27) years. $$ Nilu’s age after 8 years =(x+8)
And Mrs. Jain’s age after 8 years =(x + 27) + 8= x + 35 years.
According to question, we get
$$ x+35=2(x+8) $$ $$ \Rightarrow \quad x+35=2 x+16 $$ Transposing x to RHS and 16 to LHS, we get
$$ 35-16=2x-x $$ $$ \Rightarrow \quad x=19 $$ Hence, the present age of Nilu is 19 years.
And the present age of Nilu’s mother = x + 27 = 19 + 27 =46 years.
Q9. A man is 4 times as old as his son. After 16 years, he will be only twice as old as his son. Find the their present ages.
Sol. Let’s assume that the present age of the son be x years.
$$ \therefore \text {The present age of his father = 4x years.} $$ Son’s age after 16 years } =(x+16)
And father’s age after 16 years = (4x + 16) years.
According to question, we get
$$ 4x+16=2(x+16) $$ $$ \Rightarrow \quad 4x+16=2 x+32 $$ Transposing 2x to LHS and 16 to RHS, we get
$$ 4x – 2x=32-16 $$ $$ \Rightarrow \quad 2x=16 $$ Dividing both sides by 2, we get
$$ \frac{2 x}{2}=\frac{16}{2} $$ $$ \Rightarrow \quad x=8 $$ Hence, the present age of the son is 8 years.
$$ \text {And the present age of the father } =4x=4 \times 8 =32 \text { years.} $$Q10. The difference in age between a girl and her younger sister is 4 years. The younger sister in turn is 4 years older than her brother. The sum of the ages of the younger sister and her brother is 16 . How old are the three children?
Sol. Let the age of the girl be x years.
$$ \therefore \text {The age of her younger sister}=(x – 4) \text { years.} $$ Hence, the age of the brother =(x – 4 – 4) = (x – 8) years.
According to question, we get
$$ (x-4)+(x-8)=16 $$ $$ \Rightarrow \quad x + x – 4 – 8=16 $$ $$ \Rightarrow \quad 2x – 12 = 16 $$ Adding 12 to both sides, we get
$$ 2x – 12 + 12 = 16 + 12 $$ $$ \Rightarrow \quad 2x=28 $$ Dividing both sides by 2, we get
$$ \frac{2 x}{2}=\frac{28}{2} $$ $$ \Rightarrow \quad x=14 $$ Hence, the age of the girl is 14 years.
And, the age of the younger sister = x – 4 = 14 – 4 = 10 years.
And, the age of the younger brother = x – 8 = 14 – 8 = 6 years.
Q11. One day, during their vacation at a beach resort, Shella found twice as many sea shells as Anita and Anita found 5 shells more than sandy. Together sandy and Shella found 16 sea shells. How many did each of them find?
Sol. Let’s assume that the number of sea shells found by Sandy be x.
So, the number of sea shells found by Anita = (x + 5)
The number of sea shells found by Shelia =2(x + 5)
According to question, we get
$$ x+2(x+5)=16 $$ $$ \Rightarrow \quad x+2x+10=16 $$ $$ 3 x+10=16 $$ Subtracting 10 from both sides, we get
$$ 3 x+10-10=16-10 $$ $$ \Rightarrow \quad 3x=6 $$ Dividing both sides by 3, we get
$$ \frac{3 x}{3}=\frac{6}{3} $$ $$ \Rightarrow \quad x=2 $$ Hence, the number of sea shells found by Sandy is 2.
The number of sea shells found by Anita =x + 5 = 2 + 5 =7
$$ \text {And the number of sea shells found by Shelia }=2(x+5)=2(2+5)=2 \times 7=14 $$Q12. Andy has twice as many marbles as Pandy, and Sandy has half as many has Andy and Pandy put together. If Andy has 75 marbles more than Sandy. How many does each of them have?
Sol. Let’s assume that the number of marbles with Pandy be x.
So, the number of marbles with Andy = 2x
Therefore, the number of marbles with Sandy $$ =\frac{1}{2}(x+2x)=\frac{3x}{2} $$ According to question, we get
$$ \frac{3 x}{2}-115=110 $$ Adding 115 to both sides
$$ \frac{3 x}{2}-115+115=110+115 $$ $$ \Rightarrow \quad \frac{3 x}{2}=225 $$ Multiplying both sides by 2
$$ \frac{3 x}{2} \times 2=225 \times 2 $$ $$ \Rightarrow \quad 3 x=450 $$ Dividing both sides by 3, we get
$$ \frac{3 x}{3}=\frac{450}{3} $$ $$ \Rightarrow \quad x=150 $$ Hence, Pandy has 150 marbles.
No. of marbles Andy having $$ = 2x = 2 \times 150=300 \text { marbles.} $$ And Sandy is having $$ = \frac{3x}{2}=\frac{3 \times 150}{2}=225 \text { marbles.} $$Q13. A bag contains 25 paise and 50 paise coins whose total value is Rs 30. If the number of 25 paise coins is four times that of 50 paise coins, find the number of each type of coins.
Sol. Let’s assume that the number of 50 paise coins be x.
So, Contribution of 50 paise coins =0.5 x
Total number of 25 paise coins in the bag = 4x
Therefore, Contribution of 25 paise coins =0.25(4x) = x
According to the question, we have
$$ 0.5 x+x=30 $$ $$ \Rightarrow \quad 1.5 x=30 $$ Dividing both sides by 1.5, we get
$$ \frac{1.5 x}{1.5}=\frac{30}{1.5} $$ $$ \Rightarrow \quad x=20 $$ Hence, the number of 50 paise coins are 20.
And the number of 25 paise coins = 4x = 4(20) = 80
Q14. The length of a rectangular field is twice its breadth. If the perimeter of the field is 228 metres. Find the dimensions of the field.
Sol. Lets assume that the breadth of the rectangle be x metres.
So, Length of the rectangle = 2x metres.
Perimeter of a rectangle =2 (length + breadth)
$$ \Rightarrow \quad 2(2 x+x)=228 $$ $$ \Rightarrow \quad 2(3 x)=228 $$ $$ \Rightarrow \quad 6 x=228 $$ Dividing both sides by 6, we get
$$ \frac{6 x}{6}=\frac{228}{6} $$ $$ \Rightarrow \quad x=38 $$ Hence, the breadth of the rectangle is 38 metres.
And the length of the rectangle = 2x = 2(38} = 76 metres.
Q15. There are only 25 paise coins in a purse. The value of money in the purse is Rs 17.50. Find the number of coins in the purse.
Sol. Let’s assume that number of 25 paise coins in the purse be x.
Total money in the purse }=0.25 x
According to the question, we have
$$ 0.25 x=17.5 $$ Dividing both sides by 0.25, we get
$$ \frac{0.25 x}{0.25}=\frac{17.5}{0.25} $$ $$ \Rightarrow \quad x=70 $$ Hence, the number of 25 paise coins in the purse are 70.
Q16. In a hostel mess, 50 kg rice are consumed everyday. If each student gets 400 gm of rice day,find the number of students who take meals in the hostel mess.
Sol. Let’s assume that total number of students in the hostel be x.
Quantity of rice consumed by each student = 400 gm.
$$ \Rightarrow \text {Daily rice consumption in the hostel mess } =400x \text { gm} $$ Given daily consumptions = 50 kg = 50(1000 gm) = 50000 gm
According to the question, we have
$$ 400x = 50000 $$ Dividing both sides by 400 , we get
$$ \frac{400 x}{400}=\frac{50000}{400} $$ $$ \Rightarrow \quad x=125 $$ Hence, 125 students have their meals in the hostel mess.
Sol. Let’s assume that the present age of Nilu be x years.
$$ \therefore \text {The present age of Nilu’s mother }= (x+27) years. $$ Nilu’s age after 8 years =(x+8)
And Mrs. Jain’s age after 8 years =(x + 27) + 8= x + 35 years.
According to question, we get
$$ x+35=2(x+8) $$ $$ \Rightarrow \quad x+35=2 x+16 $$ Transposing x to RHS and 16 to LHS, we get
$$ 35-16=2x-x $$ $$ \Rightarrow \quad x=19 $$ Hence, the present age of Nilu is 19 years.
And the present age of Nilu’s mother = x + 27 = 19 + 27 =46 years.
Q9. A man is 4 times as old as his son. After 16 years, he will be only twice as old as his son. Find the their present ages.
Sol. Let’s assume that the present age of the son be x years.
$$ \therefore \text {The present age of his father = 4x years.} $$ Son’s age after 16 years } =(x+16)
And father’s age after 16 years = (4x + 16) years.
According to question, we get
$$ 4x+16=2(x+16) $$ $$ \Rightarrow \quad 4x+16=2 x+32 $$ Transposing 2x to LHS and 16 to RHS, we get
$$ 4x – 2x=32-16 $$ $$ \Rightarrow \quad 2x=16 $$ Dividing both sides by 2, we get
$$ \frac{2 x}{2}=\frac{16}{2} $$ $$ \Rightarrow \quad x=8 $$ Hence, the present age of the son is 8 years.
$$ \text {And the present age of the father } =4x=4 \times 8 =32 \text { years.} $$Q10. The difference in age between a girl and her younger sister is 4 years. The younger sister in turn is 4 years older than her brother. The sum of the ages of the younger sister and her brother is 16 . How old are the three children?
Sol. Let the age of the girl be x years.
$$ \therefore \text {The age of her younger sister}=(x – 4) \text { years.} $$ Hence, the age of the brother =(x – 4 – 4) = (x – 8) years.
According to question, we get
$$ (x-4)+(x-8)=16 $$ $$ \Rightarrow \quad x + x – 4 – 8=16 $$ $$ \Rightarrow \quad 2x – 12 = 16 $$ Adding 12 to both sides, we get
$$ 2x – 12 + 12 = 16 + 12 $$ $$ \Rightarrow \quad 2x=28 $$ Dividing both sides by 2, we get
$$ \frac{2 x}{2}=\frac{28}{2} $$ $$ \Rightarrow \quad x=14 $$ Hence, the age of the girl is 14 years.
And, the age of the younger sister = x – 4 = 14 – 4 = 10 years.
And, the age of the younger brother = x – 8 = 14 – 8 = 6 years.
Q11. One day, during their vacation at a beach resort, Shella found twice as many sea shells as Anita and Anita found 5 shells more than sandy. Together sandy and Shella found 16 sea shells. How many did each of them find?
Sol. Let’s assume that the number of sea shells found by Sandy be x.
So, the number of sea shells found by Anita = (x + 5)
The number of sea shells found by Shelia =2(x + 5)
According to question, we get
$$ x+2(x+5)=16 $$ $$ \Rightarrow \quad x+2x+10=16 $$ $$ 3 x+10=16 $$ Subtracting 10 from both sides, we get
$$ 3 x+10-10=16-10 $$ $$ \Rightarrow \quad 3x=6 $$ Dividing both sides by 3, we get
$$ \frac{3 x}{3}=\frac{6}{3} $$ $$ \Rightarrow \quad x=2 $$ Hence, the number of sea shells found by Sandy is 2.
The number of sea shells found by Anita =x + 5 = 2 + 5 =7
$$ \text {And the number of sea shells found by Shelia }=2(x+5)=2(2+5)=2 \times 7=14 $$Q12. Andy has twice as many marbles as Pandy, and Sandy has half as many has Andy and Pandy put together. If Andy has 75 marbles more than Sandy. How many does each of them have?
Sol. Let’s assume that the number of marbles with Pandy be x.
So, the number of marbles with Andy = 2x
Therefore, the number of marbles with Sandy $$ =\frac{1}{2}(x+2x)=\frac{3x}{2} $$ According to question, we get
$$ \frac{3 x}{2}-115=110 $$ Adding 115 to both sides
$$ \frac{3 x}{2}-115+115=110+115 $$ $$ \Rightarrow \quad \frac{3 x}{2}=225 $$ Multiplying both sides by 2
$$ \frac{3 x}{2} \times 2=225 \times 2 $$ $$ \Rightarrow \quad 3 x=450 $$ Dividing both sides by 3, we get
$$ \frac{3 x}{3}=\frac{450}{3} $$ $$ \Rightarrow \quad x=150 $$ Hence, Pandy has 150 marbles.
No. of marbles Andy having $$ = 2x = 2 \times 150=300 \text { marbles.} $$ And Sandy is having $$ = \frac{3x}{2}=\frac{3 \times 150}{2}=225 \text { marbles.} $$Q13. A bag contains 25 paise and 50 paise coins whose total value is Rs 30. If the number of 25 paise coins is four times that of 50 paise coins, find the number of each type of coins.
Sol. Let’s assume that the number of 50 paise coins be x.
So, Contribution of 50 paise coins =0.5 x
Total number of 25 paise coins in the bag = 4x
Therefore, Contribution of 25 paise coins =0.25(4x) = x
According to the question, we have
$$ 0.5 x+x=30 $$ $$ \Rightarrow \quad 1.5 x=30 $$ Dividing both sides by 1.5, we get
$$ \frac{1.5 x}{1.5}=\frac{30}{1.5} $$ $$ \Rightarrow \quad x=20 $$ Hence, the number of 50 paise coins are 20.
And the number of 25 paise coins = 4x = 4(20) = 80
Q14. The length of a rectangular field is twice its breadth. If the perimeter of the field is 228 metres. Find the dimensions of the field.
Sol. Lets assume that the breadth of the rectangle be x metres.
So, Length of the rectangle = 2x metres.
Perimeter of a rectangle =2 (length + breadth)
$$ \Rightarrow \quad 2(2 x+x)=228 $$ $$ \Rightarrow \quad 2(3 x)=228 $$ $$ \Rightarrow \quad 6 x=228 $$ Dividing both sides by 6, we get
$$ \frac{6 x}{6}=\frac{228}{6} $$ $$ \Rightarrow \quad x=38 $$ Hence, the breadth of the rectangle is 38 metres.
And the length of the rectangle = 2x = 2(38} = 76 metres.
Q15. There are only 25 paise coins in a purse. The value of money in the purse is Rs 17.50. Find the number of coins in the purse.
Sol. Let’s assume that number of 25 paise coins in the purse be x.
Total money in the purse }=0.25 x
According to the question, we have
$$ 0.25 x=17.5 $$ Dividing both sides by 0.25, we get
$$ \frac{0.25 x}{0.25}=\frac{17.5}{0.25} $$ $$ \Rightarrow \quad x=70 $$ Hence, the number of 25 paise coins in the purse are 70.
Q16. In a hostel mess, 50 kg rice are consumed everyday. If each student gets 400 gm of rice day,find the number of students who take meals in the hostel mess.
Sol. Let’s assume that total number of students in the hostel be x.
Quantity of rice consumed by each student = 400 gm.
$$ \Rightarrow \text {Daily rice consumption in the hostel mess } =400x \text { gm} $$ Given daily consumptions = 50 kg = 50(1000 gm) = 50000 gm
According to the question, we have
$$ 400x = 50000 $$ Dividing both sides by 400 , we get
$$ \frac{400 x}{400}=\frac{50000}{400} $$ $$ \Rightarrow \quad x=125 $$ Hence, 125 students have their meals in the hostel mess.