Solution:
Let the smaller number be \(\ 5x \) and the larger number be \(\ 7x \).
According to question,
\(\ 7x - 12 = 5x \)
\(\ ⇒ 7x - 5x = 12 \)
\(\ ⇒ 2x = 12 \)
\(\ ⇒ x = 6 \)
⸫ The required smaller number is \(\ 5x = 5×2 = 10 \)
and the larger number is \(\ 7x = 7×2 = 14 \)
2. Sumn of three consecutive even numbers is 48. Find the numbers.
Solution:
Let the first numbers be \(\ x-1 \)
second number be \(\ x \)
and the third number be \(\ x+1 \)
According to question,
\(\ x - 1 + x + x + 1 = 48 \)
\(\ ⇒ 3x = 48 \)
\(\ ⇒ x = 16 \)
So, the first number is 16 - 1 = 15
second number is 16
and the third number is 16 + 1 = 17
⸫ The required three consecutive numbers are 15, 16 and 17
3. Divide 17500 among three persons in the ratio 1:2:4. Find the amount that each person will get.
Solution:
Let the amount first person will get be \(\ x \)
second person will get \(\ 2x \)
third person will get \(\ 4x \)
According to question,
\(\ x + 2x + 4x = 17500 \)
\(\ 7x = 17500 \)
\(\ ⇒ x = 2500 \)
⸫ The first person will get an amount of 2500
Second persont will get 2 × 2500 = 5000
And the third person will get 4 × 2500 = 10000
4. The perimeter of a rectangular playground is 280 meter and its length is 2 meters more than twice its breadth. Find the length and breadth of the play ground.
Solution
Let the breadth of the rectangular playground be \(\ x \) m
And the length be \(\ (2x + 2) \) m
According to question,
The perimeter of the rectangular playground = 280
\(\ ⇒ 2[x + (2x + 2)] = 280 \)
\(\ ⇒ 2(x + 2x + 2) = 280 \)
\(\ ⇒ 2 (3x + 2) = 280 \)
\(\ ⇒ 6x + 4 = 280 \)
\(\ ⇒ 6x = 280 - 4 \)
\(\ ⇒ 6x = 276 \)
\(\ ⇒ x = \frac{276}{6} \)
\(\ ⇒ x = 46 \)
⸫ The required breadth of the rectangular playground is 69 m
And the length is (2×46 + 2) m = 94 m
5. The unit place digit of a two digit number is 5. The number is 5 times the sum of the digits. Find the numbers.
Solution
Let the tens place digit of the number be \(\ x \)
Given, unit place of the two digit number is 5
⸫ The number \(\ )= 10(x) + 5(1) = 10x + 5 \)
According to question,
\(\ 10x+5 = 5(x+5) \)
\(\ ⇒ 10x+5 = 5x+25 \)
\(\ ⇒ 10x -5x = 25 -5 \)
\(\ ⇒ 5x = 20 \)
\(\ ⇒ x = \frac{20}{5} \)
\(\ ⇒ x = 4 \)
⸫ The required Number is 10(4) + 5 = 40 + 5 = 45
6. the length of first side of scalene angle triangle is 2 cm more than that of the third side and length of second side is 5 cm less than twice that of the third side. If perimeter of the traingle is 29 cm, determine the length of the sides of the triagle.
Solution:
Let the third side of the triangle be \(\ x \) cm
so, the first side be \(\ (x+2) \) cm
And so the second side be \(\ (2x-5) \) cm
According to question,
Perimeter of the triangle = 29
\(\ ⇒ x+(x+2)+(2x-5) = 29 \)
\(\ ⇒ x+x+2+2x-5 = 29 \)
\(\ ⇒ 4x-3 = 29 \)
\(\ ⇒ 4x = 29+3 \)
\(\ ⇒ 4x = 32 \)
\(\ ⇒ x = \frac{32}{4} \)
\(\ ⇒ x = 8 \)
⸫ The third side of the triangle is 8 cm
And the second side of (8+2) cm = 10 cm
And the first side is (2×8-5) cm = (16-5) cm = 11 cm
⸫ The required sides of the triangle are 11 cm, 10, cm and 8 cm.
7. Six times of a number is same as three times of a number obtained by adding 12 to the number. Find the number.
Solution:
Let the number be \(\ x \)
According to question,
\(\ 6x = 3(x+12) \)
\(\ ⇒ 6x = 3x+36 \)
\(\ ⇒ 6x-3x = 36 \)
\(\ ⇒ 3x = 36 \)
\(\ ⇒ x = \frac{36}{3} \)
\(\ ⇒ x = 12 \)
⸫ The required number is 12
8. Sum of three consecutive natural numbers is 45. Find the numbers.
Solution:
Let the numbers be \(\ x-1 \), \(\ x \) and \(\ x+1 \)
According to question,
\(\ (x-1)+x+(x+1) = 45 \)
\(\ ⇒ x-1+x+x+1 = 45 \)
\(\ ⇒ 3x = 45 \)
\(\ ⇒ x = \frac{45}{3} \)
\(\ ⇒ x = 15 \)
⸫ The required numbers are 14, 15 and 16
9. Sum of three integers in ascending order and multiplied by 2, 3 and 4 respectively is 119 find the numbers.
Solution:
Let the three consecutive integer be \(\ x \), \(\ x+1 \) and \(\ x+2 \)
Now, when \(\ x \) , \(\ x+1 \) and \(\ x+2 \) get multiplied by 2, 3 and 4 respectively become \(\ 2x \) , \(\ 3(x+1) \) and \(\ 4(x+2) \)
According to question,
\(\ 2x+3(x+1)+4(x+2) = 119 \)
\(\ ⇒ 2x+3x+3+4x+8 = 119 \)
\(\ ⇒ 9x+11 = 119 \)
\(\ ⇒ 9x = 119-11 \)
\(\ ⇒ 9x = 108 \)
\(\ ⇒ x = \frac{108}{9} \)
\(\ ⇒ x = 12 \)
⸫ The required three consecutive integers are 12, 13 and 14.
10. After 20 years, the age of Smita will be 4 years less than 5 times of her present age. What is Smita's present age?
Solution: Let, the present age of Smita be \(\ x \) years
According to question,
\(\ x+20 = 5x-4 \)
\(\ ⇒ x-5x = -4-20 \)
\(\ ⇒ -4x = -24 \)
\(\ ⇒ 4x = 24 \)
\(\ ⇒ x = \frac{24}{4} \)
\(\ ⇒ x = 6 \)
⸫ The required present age of Smita is 6 years.
11. Present age of Raj is twice that of Rashmi 10 years ago, his age was three times the age of Rashmi. Find their present ages.
Solution:
Let the present age of Rashmi be \(\ x \) years
and the present age of Raj be \(\ 2x \) yaers
According to question,
\(\ 2x-10 = 3(x-10) \)
\(\ ⇒ 2x-10 = 3x-30 \)
\(\ ⇒ 2x - 3x = 10-30 \)
\(\ ⇒ -x = -20 \)
\(\ ⇒ x = 20 \)
⸫ The required present age of Rashmi is 20 years
and the presnt age of Raj is (2×20) years = 40 years.
12. Ranu took a 500 rupee note to a shop for change. The shopkeeper gives her a total of 19 notes of Rs. 50 and Rs. 20. How many notes of each was ranu paid?
Solution:
Let the number of Rs. 50 note be \(\ x \)
and the number of Rs. 20 note be \(\ 19-x \)
According to question,
\(\ 50(x)+20(19-x)=500 \)
\(\ ⇒ 50x+380-20x=500 \)
\(\ ⇒ 30x = 500-380 \)
\(\ ⇒ x = \frac{120}{30} \)
\(\ ⇒ x = 4 \)
⸫ The required number of Rs. 50 note is \(\ 4 \)
and the number of Rs. 20 note is \(\ 19-4 = 15 \)
[ This question can also be solved by taking two variables as follows- ]
Let the number of Rs. 50 note be \(\ x \)
and the number of Rs. 20 note be \(\ y \)
According to question,
\(\ x+y = 19 ⇒ x = 19-y~ ...(1) \)
\(\ Also,~ 50x+20y = 500~ ...(2) \)
Now, putting the value of x in eqn (2), we get \(\ 50(19-y)+20y = 500 \)
\(\ ⇒ 950-50y+20y = 500 \)
\(\ ⇒ -50y+20y = 500-950 \)
\(\ ⇒ -30y = -450 \)
\(\ ⇒ 30y = 450 \)
\(\ ⇒ y = \frac{450}{30} \)
\(\ ⇒ y = 15 \)
Again, putting the value of \(\ y \) in eqn(1), we get \(\ x = 19-15 ⇒ x = 4 \)
⸫ The required number of Rs. 50 note is 4
and the number of Rs. 20 note is 15
13. The price of each ticket of a drama show is Rs. 100 for children and Rs. 250 for adults. Rs. 8600 is collected from 50 persons by selling tickets. How many of them are chilren?
Solution: Let the number of children be \(\ x \)
According to question,
\(\ 100x+250(50-x) = 8600 \)
\(\ ⇒ 100x+12500-250x = 8600 \)
\(\ ⇒ 100x-250x = 8600-12500 \)
\(\ ⇒ -150x = -3900 \)
\(\ ⇒ 150x = 3900 \)
\(\ ⇒ x = \frac{3900}{150} \)
\(\ ⇒ x = 26 \)
⸫ The required number of children is 26
14. ⅘ of a number is 6 more than ⅔ rd of that number. What is the number ?
Solution:
Let trhe number be \(\ x \)
According to question,
\(\ \frac{4}{5}x = \frac{2}{3}x+6 \)
\(\ ⇒ \frac{4}{5}x-\frac{2}{3}x = 6 \)
\(\ ⇒ \frac{12x-10x}{15} = 6 \)
\(\ ⇒ 2x = 6×15 \)
\(\ ⇒ 2x = 90 \)
\(\ ⇒ x = \frac{90}{2} \)
\(\ ⇒ x = 45 \)
⸫ The required number is 45
15. Find a rational number which when multiplied by 4/3 and then subtracted ⅖ from the product gives -8/15
Solution:
Let the rational number be \(\ x \)
According to question,
\(\ \frac{4}{3}x-\frac{2}{5} = -\frac{8}{15} \)
\(\ ⇒ \frac{20x-6}{15} = -\frac{8}{15} \)
\(\ ⇒ 20x-6 = -8 \)
\(\ ⇒ 20x = 6-8 \)
\(\ ⇒ 20x = -2 \)
\(\ ⇒ x = \frac{-2}{20} \)
\(\ ⇒ x = -\frac{1}{10} \)
⸫ The required rational number is -1/10
16. Two bases which are at a distance 575 km travels from two places towards each other. Speed of one bus is 60 km per hour and that of another bus is 55 km per hour. At what time will they meet?
Solution:
Given, the distance, \(\ d = 575 \) km
The speed of the 1st bus, \(\ s_1 = 60 \) km/h
and the speed of the 2nd bus, \(\ s_2 = 55 \) km/h
⸫ The total speed, \(\ s = s_1 + s_2 \) = 60 km + 55 km = 115 km/h
we know, \(\ time = \frac{distance}{speed} = \frac{d}{s} = \frac{575}{115} \)
\(\ ⇒ time = 5 hours \)
⸫ At 5 hours two buses will meet.
17. A man spends in buying \(\frac{1}{4}\) th of his totall amount in vegetables, ⅗ in fruits and ⅛ in sweets. He spends the remaining Rs. 8 as bus fare. How much amount had he taken for shopping.
Solution:
Let the total amount be \(\ x \)
According to question,
\(\ x - \left[\frac{1}{4}x+\frac{3}{5}x+\frac{1}{8}\right]=8 \)
\(\ ⇒ x - \left[\frac{10x+24x+5x}{40}\right]= 8 \)
\(\ ⇒ x- \frac{39}{40}x=8 \)
\(\ ⇒ \frac{40x-39x}{40} = 8 \)
\(\ ⇒ \frac{x}{40} = 8 \)
\(\ ⇒ x= 8×40 \)
\(\ ⇒ x = 320 \)
⸫ The man had taken Rs. 320 for shopping.
18. A fraction has denominator 4 more than its numerator. If 6 is added to the numerator and 6 is subtracted from the numerator the fraction is 11/3. Find the fraction.
Solution:
Let the numerator be \(x \)
And the denominator be \( x+4 \)
⸫ The fraction becomes \( \frac{x}{x+4} \)
According to question,
\(\ ~ \frac{x+6}{x+4-6} = \frac{11}{3} \)
\(\ ⇒ \frac{x+6}{x-2} = \frac{11}{3} \)
\(\ ⇒ 3x+18 = 11x-22 \)
\(\ ⇒ 3x-11x = -22 - 18 \)
\(\ ⇒ -8x = -40 \)
\(\ ⇒ x = \frac{40}{8} \)
\(\ ⇒ x = 5 \)
⸫ The required fraction is \( \frac{5}{5+4} = \frac{5}{9} \)
19. The denominator of a rational number is 5 more than its numerator. If 1 is subtracted from the numerator and 3 separated from the denominator then the new ration number becomes \(\frac{1}{4} \). Find the rational number.
Solution:
Let the numerator of the rational numvber be \(x \)
And the denominator be \(x+5 \)
⸫ The rational number becomes \( \frac{x}{x+5} \)
According to question,
\(\ ~ \frac{x-1}{x+5-3} = \frac{1}{4} \)
\(\ ⇒ \frac{x-1}{x+2} = \frac{1}{4} \)
\(\ ⇒ 4x-4 = x+2 \)
\(\ ⇒ 4x-x = 4+2 \)
\(\ ⇒ 3x = 6\)
\(\ ⇒ x = \frac{6}{3} \)
\(\ ⇒ x = 2 \)
⸫ The required rational number is \( \frac{2}{2+5} = \frac{2}{7} \)
20. Mother is 25 years older than Rohan. After 8 years, the ratio of Rohan's age and mother's age will be 4:9. Find their present ages.
Solution:
Let the age of Ruhon be \(x \) years
And the age of his mother be \(x+25 \) years
According to question,
\(\ ~ \frac{x+8}{x+25+8} = \frac{4}{9} \)
\(\ ⇒ \frac{x+8}{x+33} = \frac{4}{9} \)
\(\ ⇒ 9x+72 = 4x+132 \)
\(\ ⇒ 9x - 4x = 132 - 72 \)
\(\ ⇒ 5x = 60 \)
\(\ ⇒ x = \frac{60}{5} \)
\(\ ⇒ x = 12 \)
⸫ The required age of Ruhon is 12 years
And the age of his mother is \(12+25 = 37 \) years
21. Mandeep sells his car to raktim at a profit of 8%. Raktim spends Rs. 5000 in repairing and then he sells it to retain at Rs. 113 400 without making any profit or loss. At what price Mandeep brought the car?
Solution:
Let, the price at which Mandeep bought the car be \(x \)
According to question,
The price of the car = 113400-5400 = 108000
\(\ ⇒ x +(8 ~percent~ of~ x ) = 108000 \)
\(\ ⇒ x + \frac{8}{100}x = 108000 \)
\(\ ⇒ \frac{ 100x + 8x }{100} = 108000 \)
\(\ ⇒ 108 x = 10800000 \)
\(\ ⇒ x = \frac{10800000}{108} \)
\(\ ⇒ x = 100000 \)
⸫ The price at which Mandeep bought the car is Rs. 100000
22. In a school week, one fifth of the students take part in 100m race and one third take part in 200m race. The driver of the difference of the students taking part in 200m race and 100m race take part in 4× 100 m race. Best 15 students enjoyed only. How many students are there in the playground?
Solution:
Let the number of students in the playground be \(x \)
According to question,
\(\ ~ \frac{1}{5}x+\frac{1}{3}x+2\left(\frac{1}{3}-\frac{1}{5}\right)x+15 = x \)
\(\ ⇒ \frac{x}{5}+\frac{x}{3}+2\left(\frac{5-3}{15}\right)x+15 = x \)
\(\ ⇒ \frac{x}{5}+\frac{x}{3}+2\left(\frac{2}{15}\right)x+15 = x \)
\(\ ⇒ \frac{x}{5}+\frac{x}{3}+\frac{4x}{15}+15 = x \)
\(\ ⇒ \frac{3x+5x+4x+225}{15} = x \)
\(\ ⇒ 12x+225 = 15x \)
\(\ ⇒ 12x-15x = -225 \)
\(\ ⇒ -3x = -225 \)
\(\ ⇒ x = \frac{225}{3} \)
\(\ ⇒ x = 75 \)
⸫ The number of students in the playground is 75
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