Given:Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. To do: We have to find the number of deer in the herd.Solution:Let the total number of deer be $x$This implies, Number of deer grazing in the field $=$ Half of the herd$=\frac{x}{2}$Number of deer playing $=\frac{3}{4}(x-\frac{x}{2})$$=\frac{3}{4}(\frac{x}{2})$$=\frac{3x}{8}$Number of deer drinking water from the pond $=9$Therefore, $x=\frac{x}{2}+\frac{3x}{8}+9$$x-\frac{x}{2}-\frac{3x}{8}=9$$\frac{8(x)-4(x)-3x}{8}=9$$8x-4x-3x=8(9)$$8x-7x=72$$x=72$The total number of deer in the herd is $72$. Read More

Given:Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs. 50 per metre and trouser material that costs him Rs. 90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at \( 12 \% \) and \( 10 \% \) profit respectively. His total sale is \( Rs. 36, 600 \). To do:We have to find the length of the trouser material bought.Solution:Let the length of the shirt material bought by Hasan be $3x\ m$.This implies, The length of the trouser material ... Read More

Given :Shobo's mother's present age is six times Shobo's present age. Shobo's age five years from now will be one third of his mother's present age.To do :We have to find their present ages.Solution :Let the present age of Shobo be $x$.This implies,The present age of Shobo's mother $=6\times x$$=6x$Age of Shobo after 5 years $= x+5$According to the question,$ x+5= \frac{1}{3}(6x)$$x+5 = 2x$$2x-x = 5$$x = 5$The present age of Shobo $=5$ yearsThe present age of Shobo's mother $=6(5)$ years$=30$ yearsThe present age of Shobo is 5 years and the age of her mother is 30 years.

Given: A Positive number is 5 times another number and if 21 is added to both the numbers, then one of the new numbers becomes twice the other new number.To do: We have to find the numbers.Solution: Let the smaller number be $a$This implies,The other number $= 5a$Adding $21$ to both numbers, we get,Smaller number $= a + 21$Other number $= 5a + 21$According to the question,Bigger number $= 2 \times$ Smaller number$5a + 21 = 2 \times (a + 21)$$5a + 21 = 2a + 42$$5a - 2a = 42 - 21$$3a = 21$$a = \frac{21}{3}$$a = 7$Therefore,The smaller number $a=7$The other(bigger) number $= 5a$$= 5 \times 7$$=35$ The required numbers are $7$ and $35$.

Given :A winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3,000.The total number of participants $= 63$.To find :We have to find the number of winners.Solution :Let the number of winners be $x$.This implies, that the number of people who did not win $= 63 - x$Therefore,$Rs. x(100)+ Rs. (63-x)(25) = Rs. 3000$$100x + 63(25) -25x = 3000$$(100-25)x + 1575 = 3000$$75x = 3000 - 1575$$75x = 1425$$x=\frac{1425}{75}$$x=\frac{57}{3}$$x=19$The number of winners is 19.

Given :The total number of coins $= 160$.Sum total of all coins $= Rs. 300$.Coins are in the denomination  $\ Rs. 1, \ Rs. 2, $ and $\ Rs. 5$. The number of $\ Rs. 2$ coins is 3 times the number of $\ Rs. 5$ coins.To do :We have to find the number of coins in each denomination.Solution :Let the number of Rs. 1 coins be $x$ and the number of Rs. 5 coins be $y$.The number of Rs. 2 coins is 3 times the number of Rs. 5 coins.This implies, The number of Rs. 2 coins $= 3y$.Therefore, $x+3y+y=160$$x+4y=160$.................(i)$[x(1)+3y(2)+y(5)] ... Read More

Given: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. To do :We have to find the ages of each of them.Solution :Let the age of Baichung be $x$ and the age of the grandfather be $y$.This implies, Age of Baichung's father $= x+29 = y-26$$y = x+29+26 = x+55$The sum of the ages of all the three $= 135$$x+(x+29)+y = 135$$x+x+29+x+55 = 135$$3x+84 = 135$$3x = 135-84$$3x = 51$$x = \frac{51}{3}$$x = 17$$x+29 = 17+29=46$$y = x+55 = 17+55=72$The age of Baichung is 17 years.The ... Read More

Given:The number of boys and girls are in the ratio of $7:5$.The number of boys is 8 more than the number of girls.To do:We have to find the total strength of the class.Solution:Let the number of boys be $7x$ and the number of girls be $5x$.According to the question,$7x = 5x+8$$7x-5x=8$$2x=8$$x=\frac{8}{2}$$x=4$Therefore,The number of boys $= 7x = 7(4) = 28$.The number of girls $= 5x = 5(4) =20$.The total strength of the class $= 28+20 = 48$.Therefore, the total number of students in the class is $48$. 

Given: 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.To do:We have to find the numbers.Solution:Let the three consecutive integers be $x, x+1$ and $x+2$According to the given question,$2\times x+ 3\times(x+1) + 4\times(x+2) = 74$$2x + 3x + 3 + 4x + 8 = 74$$9x+11=74$$9x = 74-11$$9x = 63$$x = \frac{63}{9}$$x = 7$$\Rightarrow x+1=7+1=8$$x+2 = 7+2 =  9$Therefore, the required numbers are $7, 8$ and $9$. 

Given:Three consecutive integers add up to 51.To do:We have to find the integers.Solution:Let the smallest integer be $x$.This implies,The three consecutive integers are $x, x+1, x+2$.According to the question, $x + x + 1 + x + 2 = 51$$3x + 3 = 51$$3x = 51-3$$3x = 48$$x = \frac{48}{3}$$x = 16$So, the numbers are$x = 16$$x + 1 = 16 + 1 = 17$$x + 2 = 16 + 2 = 18$The required integers are $16, 17$ and $18$.